Question 15 Marks
If the lines $\frac{\text{x}-1}{-3}=\frac{\text{y}-2}{-2\text{k}}=\frac{\text{z}-3}{2}$ and $\frac{\text{x}-1}{\text{k}}=\frac{\text{y}-2}{1}=\frac{\text{z}-3}{5}$ are perpendicular, find the value of $k$ and, hence find the equation of the plane containing these lines.
AnswerWe know that the lines $\frac{\text{x}-\text{x}_1}{\text{l}_1}=\frac{\text{y}-\text{y}_1}{\text{m}_1}=\frac{\text{z}-\text{z}_1}{\text{n}_1}$ and $\frac{\text{x}-\text{x}_2}{\text{l}_2}=\frac{\text{y}-\text{y}_2}{\text{m}_2}=\frac{\text{z}-\text{z}_2}{\text{n}_2}$ are perpendicular if
$l_1l_2 + m_1m_2 + n_1n_2= 0$
Here,
$l_1 = -3, m_1 = -2k, n_1 = 2, l_2 = k, m_2 = 1, n_2 = 5$
It is given that given are perpendicular.
$\Rightarrow l_1l_2 + m_1m_2 + n_1n_2= 0$
$\Rightarrow (-3)(k) + (-2k)(1) + (2)(5) = 0$
$\Rightarrow -3k - 2k + 10 = 0$
$\Rightarrow -5k = -10$
$\Rightarrow k = 2$
Substituting this value in the given equation of the lines, we get
$\frac{\text{x}-1}{-3}=\frac{\text{y}-2}{-4}=\frac{\text{z}-3}{2}\ ...(\text{i})$
$\frac{\text{x}-1}{2}=\frac{\text{y}-2}{1}=\frac{\text{z}-3}{5}\ ...(\text{ii})$
Finding the equation of the plane
Let the direction ratios of the required plane be proporional to $a, b, c.$
We know from $(i)$ and $(ii)$ that lines $(i)$ and $(ii)$ pass through the point $(1, 2, 3)$ and the direction ratios of $(i)$ and $(ii)$ are proportional to $-3, -4, 2$ and $2, 1, 5$ respectively.
Since the plane contains the lines $(i)$ and $(ii),$
the plane must pass through the point $(1, 2, 3)$ and it must be parallel to the line.
So, the equation of the plane is
$a(x - 1) + b(y - 2) + c(z - 3) = 0 ....(iii)$
$-3a - 4b + 2c = 0 ....(iv)$
$2a + b + 5c = 0 ....(v)$
Solving $(i), (ii)$ and $(iii)$ we get
$\begin{vmatrix}\text{x}-1&\text{y}-2&\text{z}-3\\-3&-4&2\\2&1&5 \end{vmatrix}=0$
$\Rightarrow -22(x - 1) + 19(y - 2) + 5(z - 3) = 0$
$\Rightarrow -22x + 19y + 5z = 31$
View full question & answer→Question 25 Marks
Find the coordinates of the foot of the perependicular drawn from the origin to the plane 2x - 3y + 4z - 6 = 0.
AnswerLet M be the foot of the perpendicular of the origin P(0, 0, 0) in the plane 2x - 3y + 4z - 6 =0.
Then, PM is normal to the plane. So, the direction rations of PM are proportional to 2, -3, 4.
Since PM passes through P(0, 0, 0) and has direction ratios proportional to 2, -3, 4 the equation ot PQ is
$\frac{\text{x}-0}{2}=\frac{\text{y}-0}{-3}=\frac{\text{z}-0}{4}=\text{r (say)}$
Let the coordinted of M be (2r, -3r, 4r)
Since M lies in the plane 2x - 3y + 4z - 6 = 0,
2(2r) - 3(-3r) + 4(4r) - 6 = 0
⇒ 4r + 9r + 16r - 6 = 0
⇒ 29r - 6 = 0
$\Rightarrow\ \text{r}=\frac{6}{29}$
Substituting the value of r in the coordinated of Ml we get
$\text{M}=(2\text{r},-3\text{r},4\text{r})=\bigg(2\Big(\frac{6}{29}\Big),-3\Big(\frac{6}{29}\Big),4\Big(\frac{6}{29}\Big)\bigg)$
$=\Big(\frac{12}{29},\frac{-18}{29},\frac{24}{29}\Big)$
View full question & answer→Question 35 Marks
Find the equation of the plane mid-parallel to the planes $2x - 2y + z + 3 = 0 $ and $2x - 2y + z + 9 = 0$
AnswerWe know that the distance between two planes $ax + by + cz = d_1$ and $ax + by + cz = d_2$ is $\frac{\big|\text{d}_2-\text{d}_1\big|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
The equation of plane thatb is mid-parallel to the planes
$2x - 2y + z + 3 = 0 ...(i)$
$2x - 2y + z + 9 = 0 ....(ii)$
is of the form $2x - 2y + z + k = 0 ...(iii)$
It meance that the distance between $(i)$ and $(iii) =$ distance between $(i)$ and $(ii)$
$\Rightarrow\frac{|\text{k}-3|}{\sqrt{4+4+1}}=\frac{|\text{k}-9|}{\sqrt{4+4+1}}$
$\Rightarrow |k - 3| = |k - 9|$
$\Rightarrow k - 3 = k - 9$ or $ k - 3 = -(k - 9)$
$\Rightarrow 3 = 9 \ ($false$);\ k - 3 = -k + 9$
$\Rightarrow 2k = 12$
$\Rightarrow k = 6$
Substituting this in $(iii)$ we get $2x - 2y + z + 6 = 0,$ which is the required equation of the plane.
View full question & answer→Question 45 Marks
Find the distance between the parallel planes $2x - y + 3z − 4 = 0$ and $6x - 3y + 9z + 13 = 0.$
AnswerMultiplying the first equation of the plane by $3,$ we get
$6x - 3y + 9z - 12 = 0$
$6x - 3y + 9z = 12 ....(i)$
The second equation of the plane is
$6x - 3y + 9z = -13 ...(ii)$
We know that the distance between two planes $ax + by + cz = d_{1 }$ and $ax + by + cz = d_2$ is $\frac{\big|\text{d}_2-\text{d}_1\big|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
So, the required distance
$=\frac{|-13-12|}{\sqrt{6^2+(-3)^2+9^2}}$
$=\frac{|-25|}{\sqrt{39-9+81}}$
$=\frac{25}{\sqrt{126}}$
$=\frac{5}{3\sqrt{14}}$ units
View full question & answer→Question 55 Marks
Find the length and the foot ofo perpendicular from the point $\Big(1,\frac{3}{2},2\Big)$ to the plane 2x - 2y + 4z + 5 = 0
AnswerLet M be the foot of the perpendicular from $\text{P}\Big(1,\frac{3}{2},2\Big)$ on the plane 2x - 2y + 4z + 5 = 0
Then, PM is the normal to the plane. So, its directions rations are proportional to 2, -2, 4.
Since PM passes through $\text{P}\Big(1,\frac{3}{2},2\Big)$, therefore, its equation is
$\frac{\text{x}-1}{2}=\frac{\text{y}-\frac{3}{2}}{-2}=\frac{\text{z}-2}{4}=\lambda\text{ (say)}$
Let the coordinates of M be $\Big(2\lambda+1,-2\lambda+\frac{3}{2},4\lambda+2\Big).$
Now, M lies on the plane 2x - 2y + 4z + 5 = 0.
$\therefore\ 2(2\lambda+1)-2\Big(-2\lambda+\frac{3}{2}\Big)+4(4\lambda+2)+5=0$
$\Rightarrow 24\lambda+12=0$
$\Rightarrow \lambda=-\frac{1}{2}$
So, the coordinates of M are $\Big(2\times\Big(-\frac{1}{2}\big)+1,-2\times\Big(-\frac{1}{2}\Big)+\frac{3}{2},4\times\Big(-\frac{1}{2}\Big)+2\Big)$ or $\Big(0,\frac{5}{2},0\Big)$
Thus, the coordinates of the foot of the perpendicular are $\Big(0,\frac{5}{2},0\Big).$
Now,
$\text{PM}=\sqrt{(1-0)^2+\Big(\frac{3}{2}-\frac{5}{2}\Big)^2+(2-0)^2}$
$=\sqrt{1+1+4}$
$=\sqrt{6}$
Thus, the length of the perpendicular from the given point to the plane is $\sqrt{6}$ units.
View full question & answer→Question 65 Marks
Find the shortest distance between the lines $\frac{\text{x}-1}{2}=\frac{\text{y}-3}{4}=\frac{\text{z}+2}{1}$ and 3x - y - 2z + 4 = 0 = 2x + y + z + 1.
AnswerThe equationg of the plane containing the line 3x - y - 2z + 4 = 0 = 2x + y + z + 1 is
$(3\text{x}-\text{y}-2\text{z}+4)+\lambda(2\text{x}+\text{y}+\text{z}+1)=0$
Or $(3+2\lambda)\text{x}+(\lambda-1)\text{y}+(\lambda-2)\text{z}+(\lambda+4)=0\ .....(\text{i})$
If it is parallel to the line $\frac{\text{x}-1}{2}=\frac{\text{y}-3}{4}=\frac{\text{x}+2}{1},$ then
$2(3+2\lambda)+4(\lambda-1)+(\lambda-2)=0$
$\Rightarrow 9\lambda = 0$
$\Rightarrow\ \lambda=0$
Putting $\lambda=0$ in (i), we get
3x - y - 2z + 4 = 0 ......(ii)
This is the equation of the plane containing the second line and parallel to the first line.
Now, the line $\frac{\text{x}-1}{2}=\frac{\text{y}-3}{4}=\frac{\text{x}+2}{4}$ passes through (1, 3, -2)
$\therefore$ Shortest distance between the given lines
= Lnegth of the perpendicular from (1, 3, -2) to the plane 3x - y - 2z + 4 = 0
$=\bigg|\frac{3\times1-3-2\times(-2)+4}{\sqrt{3^2+(-1)^2+(-2)^2}}\bigg|$
$=\bigg|\frac{3-3+4+4}{\sqrt{9+1+4}}\bigg|$
$=\frac{8}{\sqrt{14}}\text{ units}$
View full question & answer→Question 75 Marks
Find the vector and cartesian equations of the line passing through $(1, 2, 3)$ and parallel to the planes $\vec{\text{r}}\cdot(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=5$ and $\vec{\text{r}}\cdot(3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})=6$
AnswerWe know that equation of a line passing through $(x_1, y_1, z_1)$ is given by
$\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{{\text{y}}-{\text{y}_1}}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}\ ...(\text{i})$
Here, required line is passing through $(1, 2, 3)$, is given by, $[$Using $(i)]$
$\frac{\text{x}-1}{\text{a}_1}=\frac{{\text{y}}-2}{\text{b}_1}=\frac{\text{z}-3}{\text{c}_1}\ ...(\text{ii})$
We know that, line $\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{{\text{y}}-{\text{y}_1}}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}$ is parallel to plane $a_2x + b_2y + c_2z + d_2 = 0$ if
$a_1a_2 + b_1b_2 + c_1c_{2 }= 0 ....(iii)$
Given, line $(ii) $ is parallel to plane $x - y + 2z = 5$
So $, a_1a_2 + b_1b_2 + c_1c_{2 }= 0$
$(a_1)(1) + (b_1)(-1) + (c_1)(2) = 0$
$a_1 + b_1 + 2c_1 = 0 ....(iv)$
Also, given line $(ii)$ is parallel to plane $3x + y + z = 6$
So $, a_1a_2 + b_1b_2 + c_1c_{2 }= 0$
$(a_1)(3) + (b_1)(1) + (c_1)(1) = 0$
$3a_1 + b_1 + c_1 = 0 ....(v)$
Solving $(iv)$ and $(v)$ by cross $-$ multiplication,
$\frac{\text{a}_1}{(-1)(1)-(1)(2)}=\frac{\text{b}_1}{(3)(2)-(1)(1)}=\frac{\text{c}_1}{(1)(1)-(3)(-1)}$
$\frac{\text{a}_1}{-1-2}=\frac{\text{b}_1}{6-1}=\frac{\text{c}_1}{1+3}$
$\frac{\text{a}_1}{-3}=\frac{\text{b}_1}{5}=\frac{\text{c}_1}{4}=\lambda$ say
$\text{a}_1=-3\lambda,\text{b}_1=5\lambda,\text{c}_1=4\lambda$
Put $a_1, b_1, c_1$ in equation $(ii),$
$\frac{\text{x}-1}{-3\lambda}=\frac{\text{y}-2}{5\lambda}=\frac{\text{z}-3}{4\lambda}$
Multiplying by $\lambda,$
$\frac{\text{x}-1}{-3}=\frac{\text{y}-2}{5}=\frac{\text{z}-3}{4}$
Equation of required line is,
$\frac{\text{x}-1}{-3}=\frac{\text{y}-2}{5}=\frac{\text{z}-3}{4}$
The vector equation of the line is
$\vec{\text{r}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}+\lambda(-3\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}})$
View full question & answer→Question 85 Marks
Show that the lines $\frac{\text{x}+4}{3}=\frac{\text{y}+6}{5}=\frac{\text{z}-1}{-2}$ and 3x - 2y + z + 5 = 0 = 2x + 3y + 4z - 4 intersect. Find the equation of the plane in which they lie and also their of intersection.
AnswerThe equation of the given line is
$\frac{\text{x}+4}{3}=\frac{\text{y}+6}{5}=\frac{\text{z}-1}{-2}$
The coordinates of any point on this line are of the form
$\frac{\text{x}+4}{3}=\frac{\text{y}+6}{5}=\frac{\text{z}-1}{-2}=\lambda$
$\Rightarrow\text{x}=3\lambda-4,\text{ y}=5\lambda-6,\text{ z}=-2\lambda+1$
So, the coordinates of the point on the given line are $(3\lambda-4,5\lambda-6,-2\lambda+1).$ Since this point lies on the plane
$3\text{x}-2\text{y}+\text{z}+5=0$
$3(3\lambda-4)-2(5\lambda-6)+(-2\lambda+1)+5=0$
$\Rightarrow9\lambda-12-10\lambda+12-2\lambda+1+5=0$
$\Rightarrow-3\lambda+6=0$
$\Rightarrow\lambda=2$
So, the coordinates of the point are
$(3\lambda-4,5\lambda-6,-2\lambda+1)$
$=\big(3(2)-4,5(2)-6,-2(2)+1\big)$
$=(2,4,-3)$
Substituting this point in another plane equation 2x + 3y + 4z - 4 = 0, we get
2(2) + 3(4) + 4(-3) - 4 = 0
⇒ 4 + 12 - 12 - 4 = 0
⇒ 0 = 0
So, the point (2, 4, -3) lies on another plane too. So, this point pf intersection of the lines.
Finding the plane equation
Let the direction ratios be proportional to a, b, c.
Since the plane contains the line $\frac{\text{x}+4}{3}=\frac{\text{y}+6}{5}=\frac{\text{z}-1}{-2},$ it must pass through the point (-4, -6, 1) and is parallel to this line.
So, the equation of plane is
a(x + 4) + b(y + 6) + c(z - 1) = 0 ....(i)
and 3a + 5b - 2c = 0 ....(ii)
Since the given plane contains the planes 3x - 2y + z + 5 = 0 = 2x + 3y + 4z - 4,
3a - 2b + c = 0 ...(iii)
2a + 3b + 4z = 0 ....(iv)
Solving (iii) and (iv) using cross-multiplication, we get
$\frac{\text{a}}{-11}=\frac{\text{b}}{-10}=\frac{\text{c}}{13}\ ....(\text{v})$
Using (i), (ii) and (v), the equation of plane is
$\begin{vmatrix}\text{x}+4&\text{y}+6&\text{z}-1\\3&-5&-2\\11&10&-13 \end{vmatrix}=0$
⇒ -45(x + 4) + 17(y + 6) - 25(z - 1) = 0
⇒ 45(x + 4) - 17(y + 6) + 25(z - 1) = 0
⇒ 45x - 17y + 25z + 53 = 0
View full question & answer→Question 95 Marks
Find the vector equation of the plane passing through the point (3, 4, 2) and (7, 0, 6) and perpendicular to the plane 2x - 5y - 15 = 0. Also, show that the plane thus obtaines contains the line
AnswerLet the equation of the plane be $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ...(\text{i})$
Plane is passing through (3, 4, 2) and (7, 0, 6)
$\frac{3}{\text{a}}+\frac{4}{\text{b}}+\frac{2}{\text{c}}=1$
$\frac{7}{\text{a}}+\frac{0}{\text{b}}+\frac{6}{\text{c}}=1$
Required plane is perpendicular to 2x - 5y - 15 = 0
$\frac{2}{\text{a}}+\frac{-5}{\text{b}}+\frac{0}{\text{c}}=0$
$\Rightarrow2\text{b}=5\text{a}$
$\therefore\text{ b}=2.5\text{a}$
$\frac{3}{\text{a}}+\frac{4}{\text{2.5a}}+\frac{2}{\text{c}}=1$
$\frac{7}{\text{a}}+\frac{6}{\text{b}}=1$
Solving the above 2 equations,
$\text{a}=3.4=\frac{17}{5},\text{ b}=8.5=\frac{17}{2}$ and $\text{c}=\frac{-34}{6}=-\frac{17}{3}$
Substituting the values in (i)
$\frac{\text{x}}{\frac{17}{5}}+\frac{\text{y}}{\frac{17}{2}}+\frac{\text{z}}{-\frac{17}{3}}=1$
$\Rightarrow\frac{5\text{x}}{17}+\frac{2\text{y}}{17}-\frac{3\text{z}}{17}=1$
$\Rightarrow2\text{x}+2\text{y}-3\text{z}=17$
$\Rightarrow(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(5\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=17$
$\Rightarrow\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=17$
Vector equation of the plane is $\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=17$
The line passes through B(1, 3, -2)
5(1) + 2(3) - 3(-2) = 17
The point B lies on the plane.
$\therefore$ The line $\vec{\text{r}}=\hat{\text{i}}+3\hat{\text{j}}-2\hat{\text{k}}+\lambda(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})$ lies on the plane $\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=17$
View full question & answer→Question 105 Marks
Find the equation of the plane through the line of intersection of the planes $\vec{\text{r}}\cdot(\hat{\text{i}}+3\hat{\text{j}})+6=0$ and $\vec{\text{r}}\cdot(3\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}})=0,$ which is at a unit distance from the origin.
AnswerThe equation of the plane passing through the line intersection of the given planes is,
$\vec{\text{r}}\cdot(\hat{\text{i}}+3\hat{\text{j}})+6+\lambda\big(\vec{\text{r}}\cdot(3\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}})\big)$
$\vec{\text{r}}\cdot\Big[(1+3\lambda)\hat{\text{i}}+(3-\lambda)\hat{\text{j}}-4\lambda\hat{\text{k}}\Big]+6=0\ ...(\text{i})$
$\vec{\text{r}}\cdot\Big[(1+3\lambda)\hat{\text{i}}+(3-\lambda)\hat{\text{j}}-4\lambda\hat{\text{k}}\Big]=-6$
$\vec{\text{r}}\cdot\Big[(-1-3\lambda)\hat{\text{i}}+(\lambda-3)\hat{\text{j}}-4\lambda\hat{\text{k}}\Big]=6$
Dividing both sides by $\sqrt{(-1-3\lambda^2)+(\lambda-3)^2+16\lambda^2},$ we get
$\vec{\text{r}}\cdot\frac{\Big[-1-3\lambda)\hat{\text{i}}+(\lambda-3)\hat{\text{j}}-4\lambda\hat{\text{k}}\Big]}{\sqrt{(-1-3\lambda^2)+(\lambda-3)^2+16\lambda^2}}=\frac{6}{\sqrt{(-1-3\lambda^2)+(\lambda-3)^2+16\lambda^2}}$
Which is the normal form of plane (i), where
The perpendicular distance of plane (i) from the origin
$=\frac{6}{\sqrt{(-1-3\lambda^2)+(\lambda-3)^2+16\lambda^2}}$
$\Rightarrow1=\frac{6}{\sqrt{(-1-3\lambda^2)+(\lambda-3)^2+16\lambda^2}}\text{ (Given})$
$\Rightarrow\sqrt{(-1-3\lambda^2)+(\lambda-3)^2+16\lambda^2}=6$
$\Rightarrow1+9\lambda^2+6\lambda+\lambda^2+9-6\lambda+16\lambda^2=36$
$\Rightarrow26\lambda^2-26=0$
$\Rightarrow\lambda^2=1$
$\Rightarrow\lambda=1,-1$
Case 1: Substituting $\lambda=1$ in (i) we get
$\vec{\text{r}}\cdot\Big[4\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}\Big]+6=0$
Case 2: Substituting $\lambda=-1$ in (i) we get
$\vec{\text{r}}\cdot\Big[-2\hat{\text{i}}+4\hat{\text{j}}+4\hat{\text{k}}\Big]+6=0$
View full question & answer→Question 115 Marks
Find the equatoion of the plane passing through the points $(2, 2, 1)$ and $(9, 3, 6)$ and perpendicular to the plane $2x + 6y + 6z = 1.$
AnswerWe know that, equation of plane passing through the point $(x_1, y_1, z_1)$ is given by,
$a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$
Here, the plane is pasing through $(2, 2, 1)$
$a(x - 2) + b(y - 2) + c(z - 1) = 0 ....(i)$
It is also passing through $(9, 3, 6),$ so it must satisfy the equation $(i),$
$a(9 - 2) + b(3 - 2) + c(6 - 1) = 0$
$7a + b + 5c = 0 ....(ii)$
We know that, plane $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ are perpendicular if
$a_1a_2 + b_1b_2 + c_1c_{2 }= 0 ....(iii)$
Given that, plane (i) is perpendicular to plane
$2x + 6y + 6z = 1 ....(iv)$
Using plane $(i), (iv)$ in equation $(iii),$
$a_1a_2 + b_1b_2 + c_1c_{2 }= 0$
$(a)(2) + (b)(6) + (c)(6) = 0$
$2a + 6b + 6c = 0 ....(v)$
Solving $(ii)$ and $(v)$ by cross $-$ multiplication,
$\frac{\text{a}}{(1)(6)-(5)(6)}=\frac{\text{b}}{(2)(5)-(7)(6)}=\frac{\text{c}}{(7)(6)-(2)(1)}$
$\frac{\text{a}}{6-30}=\frac{\text{b}}{10-42}=\frac{\text{c}}{42-2}$
$\frac{\text{a}}{-24}=\frac{\text{b}}{-32}=\frac{\text{c}}{40}=\lambda(\text{say})$
$\text{a}=-24\lambda,\text{b}=-32\lambda,\text{c}=40\lambda$
Put $a, b, c$ in equation $(i),$
$\text{a}(\text{x}-2)+\text{b}(\text{y}-2)+\text{c}(\text{z}-1)=0$
$(-24\lambda)(\text{x}-2)+(-32\lambda)(\text{y}-2)+(40\lambda)(\text{z}-1)=0$
$-24\lambda\text{x}+48\lambda-32\lambda\text{y}+64\lambda+40\lambda\text{z}-40\lambda=0$
$-24\lambda\text{x}-32\lambda\text{y}+40\lambda\text{z}+72\lambda=0$
Dividing by $(-8\lambda),$
$3x + 4y - 5z - 9 = 0$
Equation of required plane is,
$3x + 4y - 5z = 9$
View full question & answer→Question 125 Marks
Find the equation of the plane passing through the line of intersection of the planes $2x - y = 0$ and $3z - y= 0$ and perpendicular to the plane $4x + 5y - 3z = 8.$
AnswerWe know that, equation of a plane passing through the line of intersection of $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ is given by,
$(\text{a}_1\text{x}+\text{b}_1\text{y}+\text{c}_1\text{z}+\text{d}_1)+\lambda(\text{a}_2\text{x}+\text{b}_2\text{y}+\text{c}_2\text{z}+\text{d}_2)=0$
So, equation of plane passing through the line of intersection of plane $2x - y = 0$ and $3z - y = 0$ is,
$(2\text{x}-\text{y})+\lambda(3\text{z}-\text{y})=0$
$2\text{x}-\text{y}+3\lambda\text{z}-\lambda\text{y}=0$
$\text{x}(2)+\text{y}(-1-\lambda)+\text{z}(3\lambda)=0\ ...(\text{i})$
We know that, two planes are perpendicular if,
$\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0\ ...(\text{ii})$
Given, plane $(i)$ is parpendicular to plane
$4\text{x}+5\text{y}-3\text{z}=8\ ...(\text{iii})$
Using $(i)$ and $(iii)$ in equation $(ii),$
$(2)(4)+(-1-\lambda)(5)+(3\lambda)(-3)=0$
$8-5-5\lambda-9\lambda=0$
$3-14\lambda=0$
$\lambda=\frac{3}{14}$
Put the value of $\lambda$ in equation $(i),$
$2\text{x}+\text{y}(-1-\lambda)+\text{z}(3\lambda)=0$
$2\text{x}+\text{y}\Big(-1-\frac{3}{14}\Big)+3\text{z}\Big(\frac{3}{14}\Big)=0$
$2\text{x}+\text{y}\Big(\frac{-14-3}{14}\Big)+\frac{9\text{z}}{14}=0$
$2\text{x}+\text{y}\Big(-\frac{17}{14}\Big)+\frac{9\text{z}}{14}=0$
Multiplying with $14,$ we get
$28x - 17y + 9z = 0$
Equation of required plane is,
$28x - 17y + 9z = 0$
View full question & answer→Question 135 Marks
Find the equation of the plane that is perpendicular to the plane $5x + 3y + 6z + 8 = 0$ and which contains the line of intersection of the planes $x + 2y + 3z - 4 = 0, 2x + y - z + 5 = 0.$
AnswerWe know that, equation of plane passing through the line of intersection of planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ is given by,
$(\text{a}_1\text{x}+\text{b}_1\text{y}+\text{c}_1\text{z}+\text{d}_1)+\lambda(\text{a}_2\text{x}+\text{b}_2\text{y}+\text{c}_2\text{z}+\text{d}_2)=0$
So, equation of plane passing through the line of intersection of given two planes is $x + 2y + 3z - 4 = 0$ and $2x + y - z + 5 = 0$ is given by,
$(\text{x}+2\text{y}+3\text{z}-4)+\lambda(2\text{x}+\text{y}-\text{z}+5)=0$
$\text{x}(1+2\lambda)+\text{y}(2+\lambda)+\text{z}(3-\lambda)+4+5\lambda=0\ ...(\text{i})$
We know that two planes are perpendicular if
$\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0\ ...(\text{ii})$
Given that plane (i) is perpendicular to plane,
$5\text{x}+3\text{y}+6\text{z}+8=0\ ...(\text{iii})$
Using $(i)$ and $(iii)$ in equation $(ii),$
$(5)(1+2\lambda)+(3)(2+\lambda)+(6)(3-\lambda)=0$
$5+10\lambda+6+3\lambda+18-6\lambda=0$
$29+7\lambda=0$
$7\lambda=-29$
$\lambda=-\frac{29}{7}$
Put the value of $\lambda$ in equation $(i),$
$\text{x}(1+2\lambda)+\text{y}(2+\lambda)+\text{z}(3-\lambda)-4+5\lambda=0$
$\text{x}\Big(1-\frac{58}{7}\Big)+\text{y}\Big(2-\frac{29}{7}\Big)+\text{z}\Big(3+\frac{29}{7}\Big)-4-\frac{145}{7}=0$
$\text{x}\Big(\frac{7-58}{7}\Big)+\text{y}\Big(\frac{14-29}{7}\Big)+\text{z}\Big(\frac{21+29}{7}\Big)\frac{-28-145}{7}=0$
$\text{x}\Big(-\frac{51}{7}\Big)+\text{y}\Big(-\frac{15}{7}\Big)+\text{z}\Big(\frac{50}{7}\Big)-\frac{173}{7}=0$
View full question & answer→Question 145 Marks
Find the image of the point with position vector $3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$ in the plane $\vec{\text{r}}. (2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=4.$ Also, find the position vectors of the foot of the prependicular and the equation of the perpendicular line through $3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}.$
AnswerLet Q be the image the point $\text{P}(3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})$ in the plane $\vec{\text{r}}.(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=4$
Since PQ passes through P and is normal to the given plane, it is parallel to the normal vector $2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$. So, the equation of PQ is
$\vec{\text{r}}=\big(3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$
As Q lies on PQ, let the position vector of Q be $(3+2\lambda)\hat{\text{i}}+(1-\lambda)\hat{\text{j}}+(2+\lambda)\hat{\text{k}}.$
$=\frac{\Big[(3+2\lambda)\hat{\text{i}}+(1-\lambda)\hat{\text{j}}+(2+\lambda)\hat{\text{k}}\Big]+\Big[3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\Big]}{2}$
$=\frac{(6+2\lambda)\hat{\text{i}}+(2-\lambda)\hat{\text{j}}+(4+\lambda)\hat{\text{k}}}{2}$
$=(3+\lambda)\hat{\text{i}}+\Big(1-\frac{\lambda}{2}\Big)\hat{\text{j}}+\Big(2+\frac{\lambda}{2}\Big)\hat{\text{k}}$
Since R lies in the plane $\vec{\text{r}}.\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)=4$
$=\Big[(3+\lambda)\hat{\text{i}}+\Big(1-\frac{\lambda}{2}\Big)\hat{\text{j}}+\Big(2+\frac{\lambda}{2}\Big)\hat{\text{k}}\Big].\Big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\Big)=4$
$\Rightarrow 6+2\lambda-1+\frac{\lambda}{2}+2+\frac{\lambda}{2}=4$
$\Rightarrow 7+2\lambda+\frac{\lambda}{2}+\frac{\lambda}{2}=4$
$\Rightarrow 14+6\lambda=8$
$\Rightarrow 6\lambda=8-14$
$\Rightarrow \lambda=-1$
Putting $\lambda=-1$ in Q, we get
$\text{Q}=(3+2(-1))\hat{\text{i}}+(1-(-1))\hat{\text{j}}+(2+(-1))\hat{\text{k}}$
$=\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{h}}$ or (1, 2, 1)
Therefore, by putting $\lambda=-1$ in R, we get
$\text{R}=(3+(-1))\hat{\text{i}}+\Big(1-\frac{(-1)}{2}\Big)\hat{\text{j}}+\Big(2+\frac{(-1)}{2}\Big)\hat{\text{k}}$
$=2\hat{\text{i}}+\frac{3}{2}\hat{\text{j}}+\frac{3}{2}\hat{\text{k}}$
View full question & answer→Question 155 Marks
Find the equation of the plane passing through the line of intersection of the planes $2x - 7y + 4z = 0, 3x - 5y + 4z + 11 = 0$ and the point $(-2, 1, 3)$.
AnswerWe know that, equation of a plane passing through the line of intersection of two planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ is given by
$(\text{a}_1\text{x}+\text{b}_1\text{y}+\text{c}_1\text{z}+\text{d}_1)+\lambda(\text{a}_2\text{x}+\text{b}_2\text{y}+\text{c}_2\text{z}+\text{d}_2)=0$
Given, equation of plane is,
$2x - 7y + 4z - 3 = 0 $ and
$3x - 5y + 4z + 11 = 0$
So, equation of plane passing through the line of intersection of given two planes is,
$(-2)(2+3\lambda)+(1)(-7-5\lambda)+(3)(4+4\lambda)-3+11\lambda=0$
$-4-6\lambda-7-5\lambda+12+12\lambda-3+11\lambda=0$
$-2+12\lambda=0$
$12\lambda=2$
$\lambda=\frac{2}{12}$
$\lambda=\frac{1}{6}$
Put $\lambda$ in equation $(i),$
$\text{x}(2+3\lambda)+\text{y}(-7+5\lambda)+\text{z}(4+4\lambda)-3+11\lambda=0$
$\text{x}\Big(2+\frac{3}{6}\Big)+\text{y}\Big(-7-\frac{5}{6}\Big)+\text{z}\Big(4+\frac{4}{6}\Big)-3+\frac{11}{6}=0$
$\text{x}\Big(\frac{12+3}{6}\Big)+\text{y}\Big(\frac{-42-5}{6}\Big)+\text{z}\Big(\frac{24+4}{6}\Big)-\frac{18+11}{6}=0$
$\frac{15}{6}\text{x}-\frac{47}{6}\text{y}+\frac{28}{6}\text{z}-\frac{7}{6}=0$
Multiplying by $6,$ we get
$15x - 47y + 28z - 7 = 0$
Therefore, equation of required plane is,
$15x - 47y + 28z = 7$
View full question & answer→Question 165 Marks
Find the equation of the containing the line $\frac{\text{x}+1}{-3}=\frac{\text{y}-3}{2}=\frac{\text{z}+2}{1}$ and the point $(0, 7, -7)$ and show that the line $\frac{\text{x}}{1}=\frac{\text{y}-7}{-3}=\frac{\text{z}+7}{2}$ also lies in the same plane.
AnswerWe know that plane through $(x_1, y_1, z_1)$ is given by
$a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 ...(i)$
Required plane is passing through $(0, 7, -7),$ so
$a(x - 0) + b(y - 7) + c(z + 7) = 0$
$ax + b(y - 7) + c(z + 7) = 0 ....(ii)$
Plane $(ii) $ also contains line $\frac{\text{x}+1}{-3}=\frac{\text{y}-3}{2}=\frac{\text{z}+2}{1}$
so, it passes through point $(-1, 3, -2),$
$a(-1) + b(3 - 7) + c(-2 + 7) = 0$
$-a - 4b + 5c = 0 ....(iii)$
Also, plane $(iii)$ will be parallel to line
So $, a_1a_2 + b_1b_2 + c_1c_2 = 0$
$(a)(-3) + (b)(2) + (c)(1) = 0$
$-3a + 2b + c = 0 .....(iv)$
Solving $(iii)$ and $(iv)$ by cross $-$ multiplication,
$\frac{\text{a}}{(-4)(1)-(5)(2)}=\frac{\text{b}}{(-3)(5)-(-1)(1)}=\frac{\text{c}}{(-1)(2)-(-4)(-3)}$
$\frac{\text{a}}{-4-10}=\frac{\text{b}}{-15+1}=\frac{\text{c}}{-2-12}$
$\frac{\text{a}}{-14}=\frac{\text{b}}{-14}=\frac{\text{c}}{14}=\lambda(\text{say})$
$\text{a}=-14\lambda,\text{b}=-14\lambda,\text{c}=-14\lambda$
Put $a, b, c$ in equation $(ii),$
$\text{ax}+\text{b}(\text{y}-7)+\text{c}(\text{z}+7)=0$
$(-14\lambda)\text{x}+(-14\lambda)(\text{y}-7)+(-14\lambda)(\text{z}+7)=0$
Dividing by $(-14\lambda),$ we get
$\text{x}+\text{y}-7+\text{z}+7=0$
$\text{x}+\text{y}+\text{z}=0$
So, equation of plane containing the given point and line is $x + y + z = 0$
The other line is $\frac{\text{x}}{1}=\frac{\text{y}-7}{-3}=\frac{\text{z}+7}{2}$
So $, a_1a_2 + b_1b_2 + c_1c_2 = 0$
$(1)(1) + (1)(-3) + (1)(2) = 0$
$1 - 3 + 2 = 0$
$0 = 0$
$\text{LHS = RHS}$
So, $\frac{\text{x}}{1}=\frac{\text{y}-7}{-3}=\frac{\text{z}+7}{2}$ lie on plane $x + y + z = 0$
View full question & answer→Question 175 Marks
Find the shortest distance between the lines $\frac{\text{x}-2}{-1}=\frac{\text{y}-5}{2}=\frac{\text{z}-0}{3}$ and $\frac{\text{x}-0}{2}=\frac{\text{y}+5}{-1}=\frac{\text{z}-1}{2}.$
AnswerConsider,
$\text{l}_1:\frac{\text{x}-2}{-1}=\frac{\text{y}-5}{2}=\frac{\text{z}-0}{3}$
$\text{l}_2:\frac{\text{x}-0}{2}=\frac{\text{y}+1}{-1}=\frac{\text{z}-1}{2}$
Clearly line $l_1 $ passes through the point $P(2, 5, 0)$
The equation of a plane containing line $l_2$ is
$a(x - 0) + b(y + 1) + c(z - 1) = 0 .....(i)$
Where $2a - b + 2c = 0$
If it is paralle to line $l_1$ then
$-a + 2b + 3c = 0$
Therefore
$\frac{\text{a}}{-7}=\frac{\text{b}}{-8}=\frac{\text{c}}{3}$
Substituting values of $a, b, c$ in the equation $(i)$ we obtain
$a(x - 0) + b(y + 1) + c(z - 1) = 0$
$-7(x - 0) - 8(y + 1) + 3(z - 1) = 0$
$-7x - 8y - 8 + 3z - 3 = 0$
$7x + 8y - 3z + 11 = 0 .....(ii)$
This is the equation of the plane contaning line $22$ and paralle to line $l1$
Shortest distance between $l_1$ and $l_2 =$ Distance between point $P(2, 5, 0)$ and plane $(ii)$
$=\bigg|\frac{14+40+11}{\sqrt{7^2+8^2+(-3)^2}}\bigg|=\frac{65}{\sqrt{122}}$
View full question & answer→Question 185 Marks
Find the equation of the plane passing through the point (-1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.
AnswerThe equation of any plane passing through (-1, 3, 2) is
Given (x + 1) + b(y - 3) + c(z - 2) = 0 ...(i)
Given that, plane (i) is perpendicular to the planes
x + 2y + 3z = 5
and
3x + 3y + z = 0
Therefore, we have,
a + 2b + 3c = 0 ...(ii)
Solving equations (ii) and (iii) by cross multiplication, we have
$\frac{\text{a}}{2\times1-3\times3}=\frac{\text{b}}{3\times3-1\times1}=\frac{\text{c}}{1\times3-3\times2}=\lambda(\text{say})$
$\Rightarrow\frac{\text{a}}{2-9}=\frac{\text{b}}{9-1}=\frac{\text{c}}{3-6}=\lambda$
$\Rightarrow\frac{\text{a}}{-7}=\frac{\text{b}}{8}=\frac{\text{c}}{-3}=\lambda$
Thus, we have
$\text{a}=-7\lambda,\text{b}=8\lambda$ and $\text{c}=-3\lambda$
Substituting the above values in equation (i), we get
$7\lambda(\text{x}+1)+8\lambda(\text{y}-3)-3\lambda(\text{z}-2)=0$
Since $\lambda\neq0,$ we get
-7(x + 1) + 8(y - 3) - 3(z - 2) = 0
⇒ -7x - 7 + 8y - 24 - 3z + 6 = 0
⇒ -7x + 8y - 3z - 25 = 0
⇒ 7x - 8y + 3z + 25 = 0
Thus the required of the plane is 7x - 8y + 3z + 25 = 0
View full question & answer→Question 195 Marks
Find the equation of the plane passing through the point (2, 1, -1) and (-1, 3, 4) and perpendicular to the plane x - 2y + 4z = 10.
AnswerThe equation of any plane passing through (2, 1, -1) is,
a(x - 2) + b(y - 1) + c(z + 1) = 0 ....(i)
It is given that (i) is passing through (-1, 3, 4). So,
a(-1 - 2) + b(3 - 1) + c(4 + 1) = 0
⇒ -3a + 2b + 5c ....(ii)
It is given that (i) is perpendicular to the plane x - 2y + 4z = 10. So,
a + 2b + 4c = 0 ....(iii)
Solving (i), (ii) and (iii) we get,
$\begin{vmatrix}\text{x}-2&\text{y}-1&\text{z}+1\\-3&2&5\\1&-2&4\end{vmatrix}=0$
⇒ 18(x - 2) + 17(y - 1) + 4(z + 1) = 0
⇒ 18x + 17y + 4z - 49 = 0
View full question & answer→Question 205 Marks
Find the equation of the plane passing through the intersection of the planes $2x + 3y - z + 1 = 0$ and $x + y - 2z + 3 = 0$ and perpendicular to the plane $3x - y - 2z - 4 = 0$.
AnswerWe know that equation of a plane passing through the line of intersection of two planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ is given by,
$(\text{a}_1\text{x}+\text{b}_1\text{y}+\text{c}_1\text{z}+\text{d}_1)+\lambda(\text{a}_2\text{x}+\text{b}_2\text{y}+\text{c}_2\text{z}+\text{d}_2)=0$
So, equation of plane passing through the planes $2x + 3y - z + 1 = 0$ and $x + y - 2z + 3 = 0$ is,
$(2\text{x}+3\text{y}-\text{z}+1)+\lambda(\text{x}+\text{y}-2\text{z}+3)=0$
$\text{x}(2+\lambda)\text{y}(3+\lambda)+\text{z}(-1-2\lambda)+1+3\lambda=0\ ...(\text{i})$
We know that two planes are perpendicular if
$\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0\ ...(\text{ii})$
Given, plane $(i)$ is perpendicular to the plane,
$3\text{x}-\text{y}-2\text{z}-4=0\ ...(\text{iii})$
Using $(i)$ and $(iii)$ in equation $(ii),$
$(3)(2+\lambda)+(-1)(3+\lambda)+(-2)(-1-2\lambda)=0$
$6+3\lambda-3-\lambda+2+4\lambda=0$
$6\lambda+5=0$
$6\lambda=-5$
$\lambda=-\frac{5}{6}$
Put the value of $\lambda$ in equation $(i),$
$\text{x}(2+\lambda)+\text{y}(3+\lambda)+\text{z}(-1-2\lambda)+1+3\lambda=0$
$\text{x}\Big(2-\frac{5}{6}\Big)+\text{y}\Big(3-\frac{5}{6}\Big)+\text{z}\Big(1+\frac{10}{6}\Big)+1-\frac{15}{6}=0$
$\text{x}\Big(\frac{12-5}{6}\Big)+\text{y}\Big(\frac{18-5}{6}\Big)+\text{z}\Big(\frac{-6+10}{6}\Big)+\frac{6-15}{6}=0$
$\frac{7\text{x}}{6}+\frac{13\text{y}}{6}+\frac{4\text{z}}{6}-\frac{9}{6}=0$
View full question & answer→Question 215 Marks
Find the equation of the plane passing through the intersection of the planes $\vec{\text{r}}\cdot(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}})=7,\vec{\text{r}}\cdot(2\hat{\text{i}}+5\hat{\text{j}}+3\hat{\text{k}})=9$ and the point (2, 1, 3).
AnswerThe equation of the plane passing through the line of intersection of the given planes is,
$\vec{\text{r}}\cdot(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}})-7+\lambda\big(\vec{\text{r}}\cdot(2\hat{\text{i}}+5\hat{\text{j}}+3\hat{\text{k}})-9\big)=0$
$\vec{\text{r}}\cdot\Big[(2-2\lambda)\hat{\text{i}}+(1+5\lambda)\hat{\text{j}}+(3+3\lambda)\hat{\text{k}}\Big]-7-9\lambda=0\ ...(\text{i})$
This passes through $(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}})$
$\big(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}\big)\Big[(2-2\lambda)\hat{\text{i}}+(1+5\lambda)\hat{\text{j}}+(3+3\lambda)\hat{\text{k}}\Big]-7-9\lambda=0$
$\Rightarrow4+4\lambda+1+5\lambda+9+9\lambda-7-9\lambda=0$
$\Rightarrow9+\lambda+7=0$
$\Rightarrow\lambda=\frac{-7}{9}$
Substituting this in (i) we get
$\vec{\text{r}}\cdot\Big[\Big(2+2\Big(\frac{-7}{9}\Big)\Big)\hat{\text{i}}+\Big(1+5\Big(\frac{-7}{9}\Big)\Big)\hat{\text{j}}+\Big(3+3\Big(\frac{-7}{9}\Big)\hat{\text{k}}\Big]-7-9\Big(\frac{-7}{9}\Big)=0$
$\Rightarrow\vec{\text{r}}\cdot(4\hat{\text{i}}-26\hat{\text{j}}+6\hat{\text{k}})=0$
$\Rightarrow\vec{\text{r}}\cdot(2\hat{\text{i}}-13\hat{\text{j}}+3\hat{\text{k}})=0$
View full question & answer→Question 225 Marks
Prove that the line of section of the planes $5x + 2y - 4z + 2 = 0 $ and $2x + 8y + 2z - 1 = 0$ is parallel to the plane $4x - 2y - 5z - 2 = 0.$
AnswerFirstly we have to find that line of section of planes $5x + 2y - 4z + 2 = 0$ and $2x + 8y + 2z - 1 = 0$
Let $a_1, b_1, c_1$ be the direction ratios of the line $5x + 2y - 4z + 2 = 0$ and $2x + 8y + 2z - 1 = 0$
Since, line lies in both the planes, so it is perpendicular to both planes, so
$5a_1 + 2b_1- 4c_1 = 0 ...(i)$
$2a_1 + 8b_1 + 2c_1 = 0 ....(ii)$
Solving equation $(i)$ and $(ii),$ by cross $-$ multiplication
$\frac{\text{a}_1}{(2)(2)-(-4)(8)}=\frac{\text{b}_1}{(2)(-4)-(5)(2)}=\frac{\text{c}_1}{(5)(8)-(2)(2)}$
$\frac{\text{a}_1}{4+32}=\frac{\text{b}_1}{-8-10}=\frac{\text{c}_1}{40-4}$
$\frac{\text{a}_1}{36}=\frac{\text{b}_1}{-18}=\frac{\text{c}_1}{36}$
$\frac{\text{a}_1}{2}=\frac{\text{b}_1}{-1}=\frac{\text{c}_1}{2}=\lambda(\text{say})$
$\Rightarrow\text{a}_1=2\lambda,\text{b}_1=-\lambda,\text{c}_1=2\lambda$
We know that, line $\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{{\text{y}}-{\text{y}_1}}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}$ is parallel to plane $a_2x + b_2y + c_2z + d_2= 0$ if
$a_1a_2 + b_1b_2 + c_1c_{2 }= 0 ....(iii)$
Here line with direction ratio $a_1, b_1, c_1$ is parallel to plane $4x - 2y - 5z - 2 = 0,$
$a_1a_2 + b_1b_2 + c_1c_2$
$= (2)(4) + (-1)(-2) + (2)(-5)$
$= 8 + 2 - 10$
$= 0$
Therefore, line of section is parallel to the plane.
View full question & answer→Question 235 Marks
Find the vector equation of the plane passing through three point with position vectors $\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}},2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}.$ Also, find coordinates of the point of intersection of this plane and the line $\vec{\text{r}}=3\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}+\lambda(2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}).$
AnswerLet A(1, 1, -2), B(2, -1, 1) and C(1, 2, 1) be the point represented by the given position vectors.
The required planes through the point A(1, 1, -1) whose position vector is
$\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$ and is normal to the vector $\vec{\text{n}}$ given by
$\vec{\text{n}}=\overrightarrow{\text{AB}}\times\overrightarrow{\text{AC}}$
Clearly, $\overrightarrow{\text{AB}}=\overrightarrow{\text{OB}}-\overrightarrow{\text{OA}}$
$=(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})-(\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}})$
$=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
$\overrightarrow{\text{AC}}=\overrightarrow{\text{OC}}-\overrightarrow{\text{OA}}$
$=(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}})-(\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}})$
$=0\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{n}}=\overrightarrow{\text{AB}}\times\overrightarrow{\text{AC}}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&-2&3\\0&1&3\end{vmatrix}$
$=-9\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}}$
The vector equation of the required plane is
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
$\Rightarrow\vec{\text{r}}\cdot(-9\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}})=(\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}})\cdot(-9\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot(9\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})=14$
To find the point of intersection of the plane
The given equation of the line is
$\vec{\text{r}}=(3\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}})+\lambda(2\hat{\text{i}}-2\text{j}+\hat{\text{k}})$
$\vec{\text{r}}=(3+2\lambda)\hat{\text{i}}+(-1-2\lambda)\hat{\text{j}}+(-1+\lambda)\hat{\text{k}}$
The coordinates of any point on this line are in the form of
$(3+2\lambda)\hat{\text{i}}+(-1-2\lambda)\hat{\text{j}}+(-1+\lambda)\hat{\text{k}}$
Since this point lies on the plane $\vec{\text{r}}\cdot(9\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})=14,$
$\Big[(3+2\lambda)\hat{\text{i}}+(-1-2\lambda)\hat{\text{j}}+(-1+\lambda)\hat{\text{k}}\Big]\cdot\big(9\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}\big)=14$
$\Rightarrow27+18\lambda-3-6\lambda+1-\lambda=14$
$\Rightarrow11\lambda=-11$
$\Rightarrow\lambda=-1$
So, the coordinates of the point are
$\Rightarrow\vec{\text{r}}\cdot(-9\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}})=-9-3-2$
$\Rightarrow\vec{\text{r}}\cdot\Big[-\big(9\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}\big)\Big]=-14$
Or $(3+2\lambda,-1-2\lambda,-1+\lambda)$
$(3+2\lambda,-1-2\lambda,-1+\lambda)$
$=(3-2,-1+2,-1-1)$
$=(1,1,-2)$
View full question & answer→Question 245 Marks
Find the equation of the plane passing through the points whose coordinates are $(-1, 1, 1)$ and $(1, -1, 1)$ and perpendicular to the plane $x + 2y + 2z = 5.$
AnswerWe know that, equation of plane passing through the point $(x_1, y_1, z_1)$ is given by,
$a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$
Here, the required plane is pasing through $(-1, 1, 1)$
$a(x + 1) + b(y - 1) + c(z - 1) = 0 ....(i)$
It is also passing through $(-1, 1, 1),$
so it must satisfy the equation $(i),$
$a(1 + 1) + b(1 - 1) + c(1 - 1) = 0$
$2a - 2b = 0 ....(ii)$
We know that, plane $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ are perpendicular if $a_1a_2 + b_1b_2 + c_1c_{2 }= 0 ....(iii)$
Given that, plane $(i)$ is perpendicular to plane
$x + 2y + 2z = 5 ....(iv)$
Using plane $(i), (iv)$ in equation $(iii),$
$a_1a_2 + b_1b_2 + c_1c_{2 }= 0$
$(a)(1) + (b)(2) + (c)(2) = 0$
$a + 2b + 2c = 0 ....(v)$
Solving $(ii)$ and $(v)$ by cross$-$multiplication,
$\frac{\text{a}}{(-2)(2)-(2)(0)}=\frac{\text{b}}{(1)(0)-(2)(2)}=\frac{\text{c}}{(2)(2)-(1)(-2)}$
$\frac{\text{a}}{-4-0}=\frac{\text{b}}{0-4}=\frac{\text{c}}{4+2}$
$\frac{\text{a}}{-4}=\frac{\text{b}}{-4}=\frac{\text{c}}{6}=\lambda(\text{say})$
$\text{a}=-4\lambda,\text{b}=-4\lambda,\text{c}=6\lambda$
Put $a, b, c$ in equation $(i),$
$\text{a}(\text{x}+1)+\text{b}(\text{y}-1)+\text{c}(\text{z}-1)=0$
$(-4\lambda)(\text{x}+1)+(-4\lambda)(\text{y}-1)+(6\lambda)(\text{z}-1)=0$
$-4\lambda\text{x}+4\lambda-4\lambda\text{y}+4\lambda+6\lambda\text{z}-6\lambda=0$
$-4\lambda\text{x}-4\lambda\text{y}+6\lambda\text{z}-6\lambda=0$
Dividing by $(-2\lambda),$ we get
$2x + 2y - 3z + 3 = 0$
The equation of required plane is,
$2x + 2y - 3z + 3 = 0$
View full question & answer→Question 255 Marks
Find the coordinates of the foot of the perpendicular from the point (2, 3, 7) to the plane 3x - y - z = 7. Also, find the length of the perpendicular.
AnswerLet Q be the foot of the perpendicular.
Here, Direction ratios of normal to plane is 3, -1, -1
⇒ Line PQ is parallel to normal to plane
⇒ Direction ratios of PQ are proportional to 3, -1, -1 and PQ is passing through P(2, 3, 7).
So,
$\frac{\text{x}-\text{x}_1}{\text{a}}=\frac{\text{y}-\text{y}_1}{\text{b}}=\frac{\text{z}-\text{z}_1}{\text{c}}$
$\frac{\text{x}-2}{3}=\frac{\text{y}-3}{-1}=\frac{\text{z}-7}{-1}=\lambda\text{ (say)}$
General ponit on line PQ
$=(3\lambda+2,-\lambda+3,-\lambda+7)$
Coordinates of Q be $(3\lambda+2,-\lambda+3,-\lambda+7)$
Point Q lies on the plane 3x - y - z = 7.
So,
$3(3\lambda+2)-(-\lambda+3)-(-\lambda+7)=7$
$9\lambda+6+\lambda-3+\lambda-7=7$
$11\lambda=7+4$
$11\lambda=11$
$\lambda=\frac{11}{11}$
$\lambda=1$
$\therefore$ Coordinate of Q $=(3\lambda+2,-\lambda+3,-\lambda+7)$
$=\big(3(1)+2,-(1)+3,-(1)+7\big)$
$=(5,2,6)$
Length of the perpendicular PQ
$=\sqrt{(\text{x}_1-\text{x}_2)^2+(\text{y}_1-\text{y}_2)^2+(\text{z}_1-\text{z}_2)^2}$
$=\sqrt{(2-5)^2+(3-2)^2+(7-6)^2}$
$=\sqrt{9+1+1}$
$=\sqrt{11}$
View full question & answer→Question 265 Marks
Find the equation of the plane through the points $(3, 4, 1)$ and $(0, 1, 0)$ and parallel to the line $\frac{\text{x}+3}{2}=\frac{\text{y}-3}{7}=\frac{\text{z}-2}{5}.$
AnswerWe know that equation of a line passing through $(x_1, y_1, z_1)$ is given by
$a(x - x_1) + b(y - y_1) + c(z - z_1) ....(i)$
Given that, requaired equation of plane is passing through $(3, 4, 1),$
so $a(x - 3) + b(y - 4) + c(z - 1) = 0 ....(ii)$
Plane $(ii)$ is also passing through $(0, 1, 0),$
so $a(0 - 3) + b(1 - 4) + c(0 - 1) = 0$
$-3a - 3b - c = 0$
$3a + 3b + c = 0 ....(iii)$
We know that, plane $a_1x + b_1y + c_1z + d_1 = 0$ and line $\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{\text{y}-\text{y}_1}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}$ are parallel if $a_1a_2 + b_1b_2 + c_1c_{2 }= 0$
Here, line $\frac{\text{x}+3}{2}=\frac{\text{y}-3}{7}=\frac{\text{z}-2}{5}$ is parallel to plane $(ii),$
so $(2)(\text{a})+(7)(\text{b})+(\text{c})(5)=0$
$2\text{a}+7\text{b}+5\text{c}=0\ ...(\text{iv})$
Solving $(iii)$ and $(iv)$ by croaa$-$multiplication,
$\frac{\text{a}}{(3)(5)-(7)(1)}=\frac{\text{b}}{(2)(1)-(3)(5)}=\frac{\text{c}}{(3)(7)-(2)(3)}$
$=\frac{\text{a}}{15-7}=\frac{\text{b}}{2-15}=\frac{\text{c}}{21-6}$
$\frac{\text{a}}{8}=\frac{\text{b}}{-13}=\frac{\text{c}}{15}=\lambda(\text{ say})$
$\text{a}=8\lambda,\text{d}=-13\lambda,\text{c}=15\lambda$
Put $a, b, c$ in equation $(ii),$
$\text{a}(\text{x}-3)+\text{b}(\text{y}-4)+\text{c}(\text{z}-1)=0$
$8\lambda(\text{x}-3)+(-13\lambda)(\text{y}-4)+(15\lambda)(\text{z}-1)=0$
$8\lambda\text{x}-24\lambda-13\lambda\text{y}+52\lambda+15\lambda\text{z}-15\lambda=0$
$8\lambda\text{x}-13\lambda\text{y}+15\lambda\text{z}+13\lambda=0$
Dividing by $\lambda,$ equation of required plane is,
$8\text{x}-13\text{y}+15\text{z}+13=0$
View full question & answer→Question 275 Marks
Find the equation of the plane passing through the points (-1, 2, 0), (2, 2, -1) and parallel to the line $\frac{\text{x}-1}{1}=\frac{2\text{y}+1}{2}=\frac{\text{z}+1}{-1}.$
AnswerThe general equation of the plane passing through the point (-1, 2, 0) is given by
a(x + 1) + b(y - 2) + c(z - 0) = 0 ....(i)
If this plane passes through the point (2, 2, -1) we have
a(2 + 1) + b(2 - 2) + c(-1 - 0) = 0
⇒ 3a - c = 0 ....(ii)
Direction ratio's of the normal to the plane (i) are a, b, c
The equation of the given line is $\frac{\text{x}-1}{1}=\frac{2\text{y}+1}{2}=\frac{\text{z}+1}{-1}$
This can be re-written as $\frac{\text{x}-1}{1}=\frac{\text{y}+\frac{1}{2}}{1}=\frac{\text{z}+1}{-1}$
Direction ratio's of line are 1, 1, -1
The required plane is parallel to the given line when the normal to his plane is perpendicular to the line.
$\therefore$ a × 1 + b × 1 + c × (-1) = 0
⇒ a + b - c = 0 ...(iii)
Solving (ii) and (iii) we get
$\frac{\text{a}}{0+1}=\frac{\text{b}}{-1+3}=\frac{\text{c}}{3-0}$
$\Rightarrow\frac{\text{a}}{1}=\frac{\text{b}}{2}=\frac{\text{c}}{3}=\lambda(\text{say})$
$\Rightarrow\text{a}=\lambda ,\text{ b}=2\lambda,\text{ c}=3\lambda$
Putting these values of a, b, c in (i) we get
$\lambda(\text{x}+1)+2\lambda(\text{y}-2)+3\lambda(\text{z}-0)=0$
⇒ x + 1 + 2y - 4 + 3z = 0
⇒ x + 2y + 3z = 3
Thus the equation of the required plane is x + 2y + 3z = 3
View full question & answer→Question 285 Marks
Show that the line whose vector equation is $\vec{\text{r}}=2\hat{\text{i}}+5\hat{\text{j}}+7\hat{\text{k}}+\lambda(\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})$ is parallel to the plane whose vector equation is $\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=7$ Also, find the distance between thetm.
AnswerThe given plane passes through the point with position vector $\vec{\text{a}}=2\hat{\text{i}}+5\hat{\text{j}}+7\hat{\text{k}}$ and is parallel to the vector $\vec{\text{b}}=\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}$
The given plane is $\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=7$ or
So, the normal vector, $\vec{\text{n}}=\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$ and d = 7
Now, $\vec{\text{b}}\cdot\vec{\text{n}}=(\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})\cdot(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})$
$=1+3-4$
$=4-4$
$=0$
So, $\vec{\text{b}}$ is perpendicular to $\vec{\text{n}}$
So, the given line is parallel to the given plane.
The distance between the line and the parallel plane. Then,
d = length of the perpendicular from the point $\vec{\text{a}}=2\hat{\text{i}}+5\hat{\text{j}}+7\hat{\text{k}}$ to the plane $\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}$
$\text{d}=\frac{\big|\vec{\text{a}}\cdot\vec{\text{n}}-\text{d}\big|}{|\vec{\text{n}}|}$
$=\frac{\big|(2\hat{\text{i}}+5\hat{\text{j}}+7\hat{\text{k}})\cdot(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})-7\big|}{|\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}|}$
$=\frac{|2+5-7-7|}{\sqrt{1+1+1}}$
$=\frac{7}{\sqrt{3}}\text{ units}$
View full question & answer→Question 295 Marks
Find the vector and cartesian forms of the plane passing through the point (1, 2, -4) and parallel to the lines $\vec{\text{r}}=(\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}})+\lambda(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}})$ and $\vec{\text{r}}=(\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}})+\mu(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}).$ Also, find the distance of the point (9, -8, -10) from the plane thus obtained.
AnswerThe plane passes through the point $\vec{\text{a}}(1,2,-4)$ a vector in a direction perpendicular to
$\vec{\text{r}}=(\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}})+\lambda(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}})$ and $\vec{\text{r}}=(\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}})+\mu(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})$
is $\vec{\text{n}}=(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}})\times(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})$
$\Rightarrow\vec{\text{n}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{z}}\\2&3&6\\1&1&-1\end{vmatrix}$
$=-9\hat{\text{i}}+8\hat{\text{j}}-\hat{\text{k}}$
Equation of the plane is $(\vec{\text{r}}-\vec{\text{a}})\cdot\vec{\text{n}}=0$
$\big(\vec{\text{r}}-(\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}})\big)\cdot(-9\hat{\text{i}}+8\hat{\text{j}}-\hat{\text{k}})=0$
$\Rightarrow\vec{\text{r}}\cdot(-9\hat{\text{i}}+8\hat{\text{j}}-\hat{\text{k}})=11$
Substituting $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}-\text{z}\hat{\text{k}},$ we get the cartesian form as
$-9\text{x}+8\text{y}-\text{z}=11$
The distance of the point (9, -8, -10) from the plane
$=\bigg|\frac{-9(9)+8(-8)-(-10)-11}{\sqrt{9^2+8^2+1^2}}\bigg|$
$=\frac{146}{\sqrt{146}}$
$=\sqrt{146}$
View full question & answer→Question 305 Marks
Find the vector equation of the plane passing through points A(a, 0, 0), B(0, b, 0) and C(0, 0, c). Reduce in to normal form. If plane ABC is at a distance p from the origin, prov that $\frac{1}{\text{p}^2}=\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}$
AnswerLet A(a, 0, 0), B(0, b, 0) and C(0, 0, c) be three
Points on a plane having their position vector $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ respectively. Then vectors $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{AC}}$ are in the same plane.
Therefore, $\overrightarrow{\text{AB}}\times\overrightarrow{\text{AC}}$ is a vector perpendicular to the plane.
Let $\vec{\text{n}}=\overrightarrow{\text{AB}}\times\overrightarrow{\text{AC}}$
$\overrightarrow{\text{AB}}=(0-\text{a})\hat{\text{i}}+(\text{b}-0)\hat{\text{j}}+(0-0)\hat{\text{k}}$
$\overrightarrow{\text{AB}}=-\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+0\hat{\text{k}}$
Similarly,
$\overrightarrow{\text{AC}}=(0-\text{a})\hat{\text{i}}+(0-0)\hat{\text{j}}+(\text{c}-0)\hat{\text{k}}$
$\overrightarrow{\text{AC}}=-\text{a}\hat{\text{i}}+0\hat{\text{j}}+\text{c}\hat{\text{k}}$
Thus,
$\vec{\text{n}}=\overrightarrow{\text{AB}}\times\overrightarrow{\text{AC}}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\-\text{a}&\text{b}&0\\-\text{a}&0&\text{c}\end{vmatrix}$
$\vec{\text{n}}=\text{bc}\hat{\text{i}}+\text{ac}\hat{\text{j}}+\text{ab}\hat{\text{k}}$
$\hat{\text{n}}=\frac{\text{bc}\hat{\text{i}}+\text{ac}\hat{\text{j}}+\text{ab}\hat{\text{k}}}{\sqrt{\text{b}^2\text{c}^2+\text{a}^2\text{c}^2+\text{a}^2\text{b}^2}}$
The plane passes through the point P with position vector $\vec{\text{a}}=\text{a}\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$
Thus, the vector equation in the normal form is,
$\big\{\vec{\text{r}}-(\text{a}\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}})\big\}\cdot\bigg(\frac{\text{bc}\hat{\text{i}}+\text{ac}\hat{\text{j}}+\text{ab}\hat{\text{k}}}{\sqrt{\text{b}^2\text{c}^2+\text{a}^2\text{c}^2+\text{a}^2\text{b}^2}}\bigg)=0$
$\Rightarrow\vec{\text{r}}\cdot\frac{\big(\text{bc}\hat{\text{i}}+\text{ac}\hat{\text{j}}+\text{ab}\hat{\text{k}}\big)}{\sqrt{\text{b}^2\text{c}^2+\text{a}^2\text{c}^2+\text{a}^2\text{b}^2}}=\frac{\text{abc}}{\sqrt{\text{b}^2\text{c}^2+\text{a}^2\text{c}^2+\text{a}^2\text{b}^2}}$
$\Rightarrow\vec{\text{r}}\cdot\frac{\big(\text{bc}\hat{\text{i}}+\text{ac}\hat{\text{j}}+\text{ab}\hat{\text{k}}\big)}{\sqrt{\text{b}^2\text{c}^2+\text{a}^2\text{c}^2+\text{a}^2\text{b}^2}}=\frac{1}{\sqrt{\frac{\text{b}^2\text{c}^2+\text{a}^2\text{c}^2+\text{a}^2\text{b}^2}{\text{a}^2\text{b}^2\text{c}^2}}}$
$\Rightarrow\vec{\text{r}}\cdot\frac{\big(\text{bc}\hat{\text{i}}+\text{ac}\hat{\text{j}}+\text{ab}\hat{\text{k}}\big)}{\sqrt{\text{b}^2\text{c}^2+\text{a}^2\text{c}^2+\text{a}^2\text{b}^2}}=\frac{1}{\sqrt{\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}}}\ ...(\text{i})$
The vector equation of a plane normal to the unit vector $\hat{\text{n}}$ and at a distance 'd' from the origin is $\vec{\text{r}}\cdot\hat{\text{n}}=\text{d}\ ...(\text{ii})$
Given that the plane is at a distance 'p' from the origin.
Comparing equation (i) and (ii) we get
$\text{d}=\text{p}=\frac{1}{\sqrt{\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}}}$
$\frac{1}{\text{p}^2}=\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}$
View full question & answer→Question 315 Marks
Find the equation of the plane passing through the intersection of the planes x - 2y + z = 1 and 2x + y + z= 8 and parallel to the line with direction ratios proportional to 1, 2, 1. Also, find the perpendicular distance of (1, 1, 1) from this plane.
AnswerThe equation of the plane passing through the intersection of the given planes is
$(\text{x}-2\text{y}+\text{z}-1)+\lambda(2\text{x}+\text{y}+\text{z}-8)=0$
$(1+2\lambda)\text{x}+(-2+\lambda)\text{y}+(1+\lambda)\text{z}-1-8\lambda=0\ ...(\text{i})$
This plane is parallel to the line whose direction ratios are proportional to 1, 2, 1.
So, the normal to the planes is perpendicular to the line whose direction ratios are proportional to 1, 2, 1.
$\Rightarrow(1+2\lambda)1+(-2+\lambda)+(1+\lambda)1=0$
$\Rightarrow1+2\lambda-4+2\lambda+1+\lambda=0$
$\Rightarrow5\lambda-2=0$
$\Rightarrow\lambda=\Big(\frac{2}{5}\Big)$
Substituting this in (i) we get
$\Big(1+2\Big(\frac{2}{5}\Big)\Big)\text{x}+\Big(-2+\Big(\frac{2}{5}\Big)\Big)\text{y}+\Big(1+\Big(\frac{2}{5}\Big)\Big)\text{z}-1-8\Big(\frac{2}{5}\Big)=0$
$\Rightarrow9\text{x}-8\text{y}+7\text{z}-21=0\ ...(\text{ii}),$ which is the required equation of the plane.
Perpendicular distance of plane (ii) from (1, 1, 1)
$=\frac{\big|9(1)-8(1)+7(1)-21\big|}{\sqrt{9^2+(8)^2+7^2}}$
$=\frac{|-13|}{\sqrt{194}}$
$=\frac{13}{\sqrt{194}}\text{ units}$
View full question & answer→Question 325 Marks
Find the coordinates of the foot of the perpendicular and the perpendicular distance of the point P(3, 2, 1) from the plane 2x - y + z + 1 = 0. Also, find the image of the point in the plane.
Answer2x - y + z + 1 = 0
(3, 2, 1)
$=\Big|\frac{6-2+1+1}{\sqrt{4+1+1}}\Big|=\frac{6}{\sqrt{6}}=\sqrt{6}$
Let the foot of perpendicular be (x, y, z). So, DR's are in proportional
$\frac{\text{x}-3}{2}=\frac{\text{y}-2}{-1}=\frac{\text{z}-1}{1}=\text{k}$
$\text{x}=2\text{k}+3$
$\text{y}=-\text{k}+2$
$\text{z}=\text{k}-1$
Subsititute (x, y, z) = (2k + 3, -k + 2, k - 1) in plane equation
2x - y + z + 1 = 0
4k + 6 + k - 2 + k - 1 + 1 = 0
6k = -4
$\text{k}=\frac{-4}{6}=\frac{-2}{3}$
$(\text{x},\text{y},\text{z})=\Big(\frac{5}{3},\frac{8}{3},\frac{-5}{3}\Big)$
View full question & answer→Question 335 Marks
Find the vector equation of the line passing through $(1, 2, 3)$ and parallel to the planes $\vec{\text{r}}\cdot(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=5$ and $\vec{\text{r}}\cdot(3\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=6.$
AnswerWe know that, equation of line passing through $(x_1, y_1, z_1)$ is given by
$\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{\text{y}-\text{y}_1}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}\ ...(\text{i})$
Given that, required line is passing through $(1, 2, 3),$ so
$\frac{\text{x}-1}{\text{a}_1}=\frac{\text{y}-2}{\text{b}_1}=\frac{\text{z}-3}{\text{c}_1}\ ....(\text{ii})$
We know that, line $\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{\text{y}-\text{y}_1}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}$ and plane $a_2x + b_2y + c_2z + d_2= 0$ are parallel if $a_1a_2 + b_1b_2 + c_1c_{2 }= 0 ....(iii)$
Given line $(ii)$ is parallel to
$\vec{\text{r}}\cdot(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=5$
$\text{x}-\text{y}+2\text{z}-5=0,$ so
$\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0$
$(\text{a}_1)(1)+(\text{b}_1)(-1)+(\text{c}_1)(2)=0$
$\text{a}_1-\text{b}_1+2\text{c}_1=0\ ...(\text{iii})$
Line $(ii)$ is also parallel to plane
$\vec{\text{r}}\cdot(3\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=6$
$3\text{x}+\text{y}+\text{z}-6=0,$ so
$\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0$
$(\text{a}_1)(3)+(\text{b}_1)(1)+(\text{c}_1)(1)=0$
$3\text{a}_1+\text{b}_1+\text{c}_1=0\ ...(\text{iv})$
Solving equation $(ii)$ and $(iv)$ by cross $-$ multiplication,
$\frac{\text{a}_1}{(-1)(1)-(2)(1)}=\frac{\text{b}_1}{(3)(2)-(1)(1)}=\frac{\text{c}_1}{(1)(1)-(3)(-1)}$
$\frac{\text{a}_1}{-1-2}=\frac{\text{b}_1}{6-1}=\frac{\text{c}_1}{1+3}$
$\frac{\text{a}_1}{-3}=\frac{\text{b}_1}{5}=\frac{\text{c}_1}{4}=\lambda(\text{say})$
$\text{a}_1=-3\lambda,\text{b}_1=5\lambda,\text{c}_1=4\lambda$
Put $a_1, b_1, c_1$ in equation $(ii),$ so equation line is given by
$\frac{\text{x}-1}{-3\lambda}=\frac{\text{y}-2}{5\lambda}=\frac{\text{z}-3}{4\lambda}$
$\frac{\text{x}-1}{-3}=\frac{\text{y}-2}{5}=\frac{\text{z}-3}{4}$
So, vector equation of required line is,
$\vec{\text{r}}=(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})+\lambda(-3\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}})$
View full question & answer→Question 345 Marks
Find the image of the point (0, 0, 0) in the plane 3x + 4y - 6z + 1 = 0.
Answer3x + 4y - 6z + 1 = 0. Line passing through orgin and perpendicular to plane is given by $\frac{\text{x}}{6}=\frac{\text{y}}{4}=\frac{\text{z}}{-6}=\gamma\text{ say}$ So let the image of (0, 0, 0) is (3r, 4r, -6r) Midpoint of (0, 0, 0) and (3r, 4r, -6r) lies on plane. $3\Big(\frac{3\text{r}}{2}\Big)+2(4\gamma)-3(-6\gamma)+1=0$ $30.5\gamma=-1$ $\gamma=\frac{-2}{61}$So image is $\Big(\frac{-6}{61},\frac{-8}{61},\frac{12}{61}\Big).$
View full question & answer→Question 355 Marks
Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 and twice of its y-intercept is equal to three times its z-intercept.
Answerx + y + z = 1
2x + 3y + 4z = 5
Required equation of plane is $\text{x}+\text{y}-1+\text{z}+\lambda(2\text{x}+3\text{y}+4\text{z}-5)=0$ for same $\lambda.$
i.e., $(1+2\lambda)\text{x}+(1+3\lambda)\text{y}+(1+4\lambda)\text{z}=1+5\lambda$
According to question,
$2\Big(\frac{1+5\lambda}{1+3\lambda}\Big)=3\Big(\frac{1+5\lambda}{1+4\lambda}\Big)$
Solving we get $\lambda=-1$
Thus the equation of required plane is,
-x - 2y - 3z = -4
Or x + 2y + 3z = 4
View full question & answer→Question 365 Marks
Find the distance of the point (1, -2, 4) from plane passing throuhg the point (1, 2, 2) and perpendicular of the planes x - y + 2z = 3 and 2x - 2y + z + 12 = 0
AnswerLet the equation of plane passing through the point (1, 2, 2) be
a(x - 1) + b(y - 2) + c(z - 2) = 0 ....(i)
Here, a, b, c are the direction ratio of the normal to the plane.
The equation of the given planes are x - y + 2z = 3 and 2x - 2y + z + 12 = 0
Plane (i) is perpendicular to the given planes.
a - b + 2c = 0 ...(ii)
2a - 2b + c = 0 ....(iii)
Eliminating a, b and c from (i), (ii) and (iii) we get
$\begin{vmatrix}\text{x}-1&\text{y}-2&\text{z}-2\\1&-1&2\\2&-2&1\end{vmatrix}=0$
$\Rightarrow(\text{x}-1)(-1+4)-(\text{y}-2)(1-4)+(\text{z}-2)(-2+2)=0$
$\Rightarrow3\text{x}+3\text{y}-9=0$
$\Rightarrow\text{x}+\text{y}-3=0$
$\therefore$ Distance of the point (1, -2, 4) from the plane x + y - 3 = 0
$=\bigg|\frac{1-2-3}{\sqrt{1^2+1^2+0^2}}\bigg|$
$=\Big|\frac{-4}{2}\Big|$
$=2\sqrt{2}\text{ units}$
View full question & answer→Question 375 Marks
Find an equation for the set all points that are equidistant from the planes $3x - 4y + 12z = 6$ and $4x + 3z = 7$
AnswerConsider,
$3x - 4y + 12z - 6 = 0 ....(i)$
$4x + 3z - 7 = 0 ....(ii)$
The distance of a point $(x_1, y_1, z_1)$ from plane $3x - 4y + 12z - 6 = 0 $ is,
$\text{D}_1=\Bigg|\frac{\text{ax}_1+\text{by}_1+\text{cz}_1+\text{d}}{\sqrt{\text{a}^2+\text{d}^2+\text{c}^2}}\Bigg|$
$=\Bigg|\frac{3\text{x}_1-4\text{y}_1+12\text{z}_1-6}{\sqrt{3^2+(-4)^2+12^2}}\Bigg|$
$=\Bigg|\frac{3\text{x}_1-4\text{y}_1+12\text{z}_1-6}{\sqrt{169}}\Bigg|$
$=\Bigg|\frac{3\text{x}_1-4\text{y}_1+12\text{z}_1-6}{13}\Bigg|$
The distance of the point $(x_1, y_1, z_1)$ from the plane $4x + 3z - 7 = 0$ is
$\text{D}_2=\Bigg|\frac{\text{ax}_1+\text{by}_1+\text{cz}_1+\text{d}}{\sqrt{\text{a}^2+\text{d}^2+\text{c}^2}}\Bigg|$
$=\bigg|\frac{4\text{x}_1+3\text{z}_1-7}{\sqrt{4^2+3^2}}\bigg|$
$=\bigg|\frac{4\text{x}_1+3\text{z}_1-7}{\sqrt{25}}\bigg|$
$=\bigg|\frac{4\text{x}_1+3\text{z}_1-7}{5}\bigg|$
Since the point $(x_1, y_1, z_1)$ are equidistant from the planes $3x - 4y + 12z - 6 = 0$ and $4x + 3z - 7 = 0$
So, $D_1= D_2$
$\Bigg|\frac{3\text{x}_1-4\text{y}_1+12\text{z}_1-6}{13}\Bigg|=\bigg|\frac{4\text{x}_1+3\text{z}_1-7}{5}\bigg|$
$\frac{3\text{x}_1-4\text{y}_1+12\text{z}_1-6}{13}=\pm\frac{4\text{x}_1+3\text{z}_1-7}{5}$
Taking positive sign
$\frac{3\text{x}_1-4\text{y}_1+12\text{z}_1-6}{13}=\frac{4\text{x}_1+3\text{z}_1-7}{5}$
$15\text{x}_1-20\text{y}_1+60\text{z}_1-30=52\text{x}_1+39\text{z}_1-91$
$37\text{x}_1+20\text{y}_1-21\text{z}_1-61=0$
Taking negative sign,
$\frac{3\text{x}_1-4\text{y}_1+12\text{z}_1-6}{13}=-\frac{4\text{x}_1+3\text{z}_1-7}{5}$
$15\text{x}_1-20\text{y}_1+60\text{z}_1-30=-52\text{x}_1-39\text{z}_1+91$
$67\text{x}_1-20\text{y}_1+99\text{z}_1-121=0$
View full question & answer→Question 385 Marks
Find the corrdinates of the points P where the line throught A(3, -4,-5) and B(2, -3, 1) crosses the plane passing throught three points L(2, 2, 1), M(3, 0, 1) and N(4, -1, 0). Also, find the ratio in which P diveides the line segment AB.
AnswerEquation of the plane passing throught the points L(2, 2, 1), M(3, 0, 1) and N(4, -1, 0) is,
$\Big[\vec{\text{r}}-\big(2\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)\Big]\cdot\Big[\big(\hat{\text{i}}-2\hat{\text{j}})\times\big(\hat{\text{i}}-2\hat{\text{j}}\big)\times(\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\big)\Big]=0$
$\Rightarrow\Big[\vec{\text{r}}-\big(2\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)\Big]\cdot\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=0$
$\Rightarrow\vec{\text{r}}\cdot\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})\cdot\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=4+2+1$
$\Rightarrow\vec{\text{r}}\cdot\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=7\ ...(\text{i})$
The equation of the line segment through A(3, - 4, 5) and B(2, -3, 1) is,
$\frac{\text{x}-3}{2-3}=\frac{\text{y}+4}{-3+4}=\frac{\text{z}+5}{1+5}$
i.e., $\frac{\text{x}-3}{-1}=\frac{\text{y}+4}{1}=\frac{\text{z}+5}{6}$
Any point on this line is of the form $(-\lambda+3,\lambda-4,6\lambda-5)$
This point lies on the plane (i)
$\therefore\Big[(-\lambda+3)\hat{\text{i}}+(\lambda-4)\hat{\text{j}}+(6\lambda-5)\hat{\text{k}}\Big]\cdot\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=7$
$\Rightarrow2(-\lambda+3)+(\lambda-4)+(6\lambda-5)=7$
$\Rightarrow5\lambda=10$
$\Rightarrow\lambda=2$
Thus, the corrdinates of the points P are (-2 + 3, 2 - 4, 6 × 2 - 5) i.e., (1, -2, 7)
Suppose P divides the line segment AB in the ratio $\mu:1$
$\therefore (1,-2,7)=\Big(\frac{2\mu+3}{\mu+1},\frac{-3\mu-4}{\mu+1},\frac{\mu-5}{\mu+1}\Big)$
$\Rightarrow\frac{2\mu+3}{\mu+1}=1,\frac{-3\mu-4}{\mu+1}=-2,\frac{\mu-5}{\mu+1}=7$
$\Rightarrow2\mu=\mu+1,-3\mu-4=-2\mu-2,\mu-5=7\mu+7$
$\Rightarrow\mu=-2$
Thus, the point P divides the line segmenty AB externally in the ratio 2 : 1
View full question & answer→Question 395 Marks
Find the shortest distance between the lines $\frac{\text{x}+1}{7}=\frac{\text{y}+1}{-6}=\frac{\text{z}+1}{0}$ and $\frac{\text{x}-3}{1}=\frac{\text{y}-5}{-2}=\frac{\text{z}-7}{1}.$
AnswerThe given equations of the lines are, $\frac{\text{x}+1}{7}=\frac{\text{y}+1}{-6}=\frac{\text{z}+1}{1}\ .....(\text{i})$ $\frac{\text{x}-3}{1}=\frac{\text{y}-5}{-2}=\frac{\text{z}-7}{1}\ .....(\text{ii})$Clearly (ii) passes through the point P (3, 5, 7).
Let the direction ratios of the plane be proportional to a, b, c.
Since the plane contains line (i), it should pass through (-1, -1, -1) and is parallel to the line (i).
Equation of the plane through (i) is
a(x + 1) + b(y + 1) + c(z + 1) = 0 .....(iii),
Where 7a - 6b + c = 0 .....(iv)
Since the plane is parallel to the line (ii),
a - 2b + c = 0...(v)
Solving (iv) and (v)
Using cross-multiplication, we get
$\frac{\text{a}}{-4}=\frac{\text{b}}{-6}=\frac{\text{c}}{\text{-8}}$
$\Rightarrow\frac{\text{a}}{2}=\frac{\text{b}}{3}=\frac{\text{c}}{\text{4}}$
Substituting a, b and c in (3), we get
2(x + 1) + 3(v + 1) + 4(z + 1) = 0
2x + 3y + 4z + 9 = 0... (vi)
Which is the equation of the plane containing line (i) and parallel to line (ii).
Shortest distance between (i) and (ii)
= Distance between the point P(3, 5, 7) and plane (vi)
$=\bigg|\frac{2(3)+3(5)+4(7)+9}{\sqrt{4+9+16}}\bigg|$
$=\frac{58}{\sqrt{29}}$
$=2\sqrt{29} \text{ units}$
View full question & answer→Question 405 Marks
Show that the plane vector equation is $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})=1$ and the line whose vector equation is $\vec{\text{r}}=(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})+\lambda(2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}})$ are parallel. Also, find the distance between them.
AnswerThe given plane passes through the point with position vector $\vec{\text{a}}=-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ and is parallel to the vector $\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}$
The given plane is $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})=1$ or $\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}.$
So, normal vector $\vec{\text{n}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$ and d = 1
Now, $\vec{\text{b}}\cdot\vec{\text{n}}=(2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}})\cdot(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})$
$=2+2-4=0$
So, $\vec{\text{b}}$ is perpendicular to $\vec{\text{n}}.$
So, the given line is parallel to the given plane.
$\text{d}=\frac{\big|\vec{\text{a}}\cdot\vec{\text{n}}-\text{d}\big|}{|\vec{\text{n}}|}$
$=\frac{(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})\cdot(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})-1}{|\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}|}$
$=\frac{|-1+2-1-1|}{\sqrt{1+4+1}}$
$=\frac{1}{\sqrt{6}}\text{ units}$
View full question & answer→Question 415 Marks
Find the distance of the point (1, -2, 3) from the plane x - y + z = 5 measured along a line parallel to $\frac{\text{x}}{2}=\frac{\text{y}}{3}=\frac{\text{z}}{-6}.$
AnswerHere, we have to find distance of the point P(1, -2, 3) from the plane
x - y + z = 5 measured parallel to line AB, $\frac{\text{x}}{2}=\frac{\text{y}}{3}=\frac{\text{z}}{-6}$
Let Q be the mid point of the line joining P to plane.
Here PQ is parallel to line AB
⇒ Direction ratios of line PQ are proportional to direction ratios of line AB.
⇒ Direction ratios of line PQ are 2, 3, -6 and PQ is passing through P(1, -2, 3)
So equation of PQ is given by
$\frac{\text{x}-\text{x}_1}{\text{a}}=\frac{\text{y}-\text{y}_1}{\text{b}}=\frac{\text{z}-\text{z}_1}{\text{c}}$
$\frac{\text{x}-1}{2}=\frac{\text{y}+2}{3}=\frac{\text{z}-3}{-6}=\lambda\ (\text{say})$
General point on line PQ is $(2\lambda+1,3\lambda-2,-6\lambda+3)$
Suppose coordinates of Q be $(2\lambda+1,3\lambda-2,-6\lambda+3)$
General point on line PQ is $(2\lambda+1,3\lambda-2,-6\lambda+3)$
Suppose coordinates off Q be $(2\lambda+1,3\lambda-2,-6\lambda+3)$
Since Q lies on the plane x - y + z = 5
$(2\lambda+1)-(3\lambda-2)+(-6\lambda+3)=5$
$2\lambda+1-3\lambda+2-6\lambda+3=5$
$-7\lambda=5-6$
$-7\lambda=-1$
$\lambda=\frac{1}{7}$
Coordinate of $\text{Q}=(2\lambda+1,3\lambda-2,-6\lambda+3)=\Big(\frac{9}{7},\frac{-11}{7},\frac{15}{7}\Big)$
Distance Between (1, -2, 3) and plane = PQ
$=\sqrt{(\text{x}_1-\text{x}_2)^2+(\text{y}_1-\text{y}_2)^2+(\text{z}_1-\text{z}_2)^2}$
$=\sqrt{\Big(1-\frac{9}{7}\Big)^2+\Big(-2+\frac{11}{7}\Big)^2+\Big(3-\frac{15}{7}\Big)^2}$
$=\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}$
$=\sqrt{\frac{49}{49}}$
$=1$
Required distance = 1 unit.
View full question & answer→Question 425 Marks
Find the coordinates of the point where the line $\frac{\text{x}-2}{3}=\frac{\text{y}+1}{4}=\frac{\text{z}-2}{2}$ intersectscts the plane x - y + z - 5 = 0. Also, find the angle between the line and the plane.
AnswerLet $\frac{\text{x}-2}{3}=\frac{\text{y}+1}{4}=\frac{\text{z}-2}{2}=\lambda(\text{say})$
$\text{x}=3\lambda+2,\text{ y}=4\lambda-1,\text{ z}=2\lambda+2\ ...(\text{i})$
Since (x, y, z) intersects the plane x - y + z - 5 = 0,
$3\lambda+2-(4\lambda-1)+2\lambda+2-5=0$
$3\lambda+2-4\lambda+1+2\lambda+2-5=0$
$\lambda=0$
Substituting this in (i) we get
$\text{x}=2,\text{ y}=-1,\text{ z}=2$
So, $(\text{x},\text{y},\text{z})=(2,-1,2)$
Finding the angle
The given line is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}}$ and the given plane is normal to the vector $\vec{\text{n}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
We know that the angle $\theta$ between a line and a plane is given by
$\sin\theta=\frac{\vec{\text{b}}\cdot\vec{\text{n}}}{|\vec{\text{b}}||\vec{\text{n}}|}$
$=\frac{(3\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}})\cdot(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})}{|3\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}}||\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}|}$
$=\frac{3-4+2}{\sqrt{9+16+4}\sqrt{1+1+1}}=\frac{1}{\sqrt{87}}$
$\theta=\sin^{-1}\Big(\frac{1}{\sqrt{87}}\Big)$
View full question & answer→Question 435 Marks
Find the equation of the plane through the intersection of the planes 3x - 4y + 5z = 10 and 2x + 2y - 3z = 4 and parallel to the line x = 2y = 3z.
AnswerThe equation of the plane passing through the intersection of the given planes is
$(3\text{x}-4\text{y}+5\text{z}-10)+\lambda(2\text{x}+2\text{y}-3\text{z}-4)=0$
$\Rightarrow(3+2\lambda)\text{x}+(-4+2\lambda)\text{y}+(5-3\lambda)\text{z}-10-4\lambda=0\ ...(\text{i})$
The given line is
$\text{x}=2\text{y}=3\text{z}$
Dividing this equation by 6, we get
$\frac{\text{x}}{6}=\frac{\text{y}}{3}=\frac{\text{z}}{2}$
The direction ratios of this line are proportional to 6, 3, 2
So, the normal to the plane is perpendicular to the line whose direction ratios are proportional to 6, 3, 2.
$\Rightarrow(3+2\lambda)6+(-4+2\lambda)3+(5-3\lambda)2=0$
$\Rightarrow18+12\lambda-12+6\lambda+10-6\lambda=0$
$\Rightarrow12\lambda+16=0$
$\Rightarrow\lambda=\Big(\frac{-4}{3}\Big)$
Substituting this in (i) we get
$\Big(3+2\Big(\frac{-4}{3}\Big)\Big)\text{x}+\Big(-4+2\Big(\frac{-4}{3}\Big)\Big)\text{y}+\Big(5-3\Big(\frac{-4}{3}\Big)\Big)\text{z}-10-4\Big(\frac{-4}{3}\Big)=0$
$\Rightarrow\text{x}-20\text{y}+27\text{z}=14$
View full question & answer→Question 445 Marks
Find the equation of the passing throught the point (1, 2, 1) and perpendicular to the joining the point (1, 4, 2) and (2, 3, 5). Find also the perpendicular distance of the origin from this plane.
AnswerHere, we have to find equation a plane passing throught A(1, 2, 1) and perpendicular to line joining B(1, 4, 2) and C(2, 3, 5)
We know that, the vector equation of a plane passing through a point $\vec{\text{a}}$ and perpendicular to vector $\vec{\text{n}}$ is given by,
$(\vec{\text{r}}-\vec{\text{a}})\vec{\text{n}}=0\ ...(\text{i})$
Here, $\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{n}}=\overrightarrow{\text{BC}}$
= Position vector of C - Position vector of B
$=2\hat{\text{i}}+3\hat{\text{j}}+5\hat{\text{k}}-\hat{\text{i}}-4\hat{\text{j}}-2\hat{\text{k}}$
$\vec{\text{n}}=\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$
Put, $\vec{\text{a}}$ and $\vec{\text{n}}$ in equation (i),
vector equation of plane is
$\Big[\vec{\text{r}}-(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}})\Big]\cdot(\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})=0$
$\vec{\text{r}}\cdot(\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})-(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}})(\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})=0$
$\vec{\text{r}}\cdot(\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})-\big[(1)(1)+(2)(-1)+(1)(3)\big]=0$
$\vec{\text{r}}\cdot(\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})-\big[1-2+3\big]=0$
$\vec{\text{r}}\cdot(\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})-(4-2)=0$
$\vec{\text{r}}\cdot(\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})-2=0$
$\vec{\text{r}}\cdot(\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})=2\ ....(\text{ii})$
$|\vec{\text{n}}|=\sqrt{(1)^2+(-1)^2+(3)^2}$
$=\sqrt{1+1+9}$
$=\sqrt{11}$
Dividing equation (i) by $\sqrt{11},$
$\vec{\text{r}}\cdot\Big(\frac{1}{\sqrt{11}}\hat{\text{i}}-\frac{1}{\sqrt{11}}\hat{\text{j}}+\frac{3}{\sqrt{11}}\hat{\text{k}}\Big)=\frac{2}{\sqrt{11}}$
$\hat{\text{r}}\cdot\hat{\text{n}}={\text{d}}$
So, perpendicular distance of plane from origin $=\frac{2}{\sqrt{11}}\text{ units}$
Equation of plane, $\vec{\text{r}}(\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})=2$
Equation of plane, x - y + 3z - 2 = 0
View full question & answer→Question 455 Marks
Find the coordinates of the foot of the perpendicular from the point (1, 1, 2) to the plane 2x - 2y + 4z + 5 = 0. Also, find the length of the perpendicular.
AnswerLet M be the foor of the perpendicular of the point P(1, 1, 2) in the plane 2x - 2y + 4z + 5 = 0
Then, PM is normal to the plane. So, the direction ratios of PM are proportional to 2, -2, 4.
Since PM passes through P(1, 1, 2) and has direction ratios proportional to 2, -2 and 4, equation of PQ is
$\frac{\text{x}-1}{2}=\frac{\text{y}-1}{-2}=\frac{\text{z}-2}{4}=\text{r (say)}$
Let the coordiantes of M be (2r + 1, -2r + 1, 4r + 2).
Since M lies in the plane 2x - 2y + 4z + 5 = 0
2(2r + 1) - 2(-2r + 1) + 4(4r + 2) + 5 = 0
⇒ 4r + 2 + 4r - 2 + 16r + 8 + 5 = 0
⇒ 24r + 13 = 0
$\Rightarrow\ \text{r}=\frac{-13}{24}$
Substituting this in the coordinates of M, we get
$\text{M}=(2\text{r}+1,-2\text{r}+1,4\text{r}+2)\\=\Big(2\Big(\frac{-13}{24}\Big)+1,-2\Big(\frac{-13}{24}\Big)+1,4\Big(\frac{-13}{24}\Big)+2\Big)\\=\Big(\frac{-1}{12},\frac{25}{12},\frac{-1}{6}\Big)$
Now, the length of the perpendicular from P onto the given plane
$=\frac{|2(1)-2(1)+4(2)+5|}{\sqrt{4+4+16}}$
$=\frac{13}{\sqrt{24}}\text{ units}$
View full question & answer→Question 465 Marks
State when the line $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$ is parallel to the plane $\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}.$ Show that the line $\vec{\text{r}}=\hat{\text{i}}+\hat{\text{j}}+\lambda(3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})$ is parallel to the plane $\vec{\text{r}}\cdot(2\hat{\text{i}}+\hat{\text{k}})=3.$ Also, find the distance between the line and the plane.
AnswerWe know that line $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$ is paralle to plane $\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}$ if
$\vec{\text{b}}\cdot\vec{\text{n}}=0$
Given, line is $\vec{\text{r}}=(\hat{\text{i}}+\hat{\text{j}})+\lambda(3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})$ and plane is $\vec{\text{r}}\cdot(2\hat{\text{i}}+\hat{\text{k}})=3,$
$\vec{\text{b}}=3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}},\vec{\text{a}}=(\hat{\text{i}}+\hat{\text{j}})$ and $\vec{\text{n}}=(2\hat{\text{j}}+\hat{\text{k}})$
Now, $\vec{\text{b}}\cdot\vec{\text{n}}=(3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})(2\hat{\text{j}}+\hat{\text{k}})$
$=(3)(0)+(-1)(2)+(2)(1)$
$=0-2+2$
$=0$
Since, $\vec{\text{b}}\cdot\vec{\text{n}}=0$ So line is parallel to plane.
Distance between point $\vec{\text{a}}$ and plane $\vec{\text{r}}\cdot\vec{\text{n}}-\text{d}=0$ is given by
$\text{D}=\Bigg|\frac{\vec{\text{a}}\vec{\text{n}}-\text{d}}{|\vec{\text{n}}|}\Bigg|\ ...(\text{i})$
$\vec{\text{a}}$ is a point on the line. So diatance between line and plane is equal to the distance between $\vec{\text{a}}=(\hat{\text{i}}+\hat{\text{j}})$ and plane $\vec{\text{r}}\cdot(2\hat{\text{i}}+\hat{\text{k}})=3,$ so using (i)
$\text{D}=\Bigg|\frac{(\hat{\text{i}}+\hat{\text{j}})(2\hat{\text{j}}+\hat{\text{k}})-3}{\sqrt{(2)^2+(1)^2}}\Bigg|$
$=\bigg|\frac{(1)(0)+(1)(2)+(0)(1)-3}{\sqrt{4+1}}\bigg|$
$=\Big|\frac{0+2+0-3}{\sqrt{5}}\Big|$
$=\Big|\frac{-1}{\sqrt{5}}\Big|$
$=\frac{1}{\sqrt{5}}\text{ unit}$
So, required distance $=\frac{1}{\sqrt{5}}\text{ unit}$
View full question & answer→Question 475 Marks
Find the equation of the plane which contains two parallel lines $\frac{\text{x}-4}{1}=\frac{\text{y}-3}{-4}=\frac{\text{z}-2}{5}$ and $\frac{\text{x}-3}{1}=\frac{\text{y}+2}{-4}=\frac{\text{z}}{5}.$
AnswerWe know that the equation of the plane containing two parallel lines $\frac{\text{x}-\text{x}_1}{\text{a}}=\frac{\text{y}-\text{y}_1}{\text{b}}=\frac{\text{z}-\text{z}_1}{\text{c}}$ and $\frac{\text{x}-\text{x}_2}{\text{a}}=\frac{\text{y}-\text{y}_2}{\text{b}}=\frac{\text{z}-\text{z}_2}{\text{c}}$ is
$\begin{vmatrix}\text{x}-\text{x}_1&\text{y}-\text{y}_1&\text{z}-\text{z}_1\\\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{a}&\text{b}&\text{c} \end{vmatrix}=0$
Here, $\text{x}_1=4,\text{ y}_1=3,\text{ z}_1=2,\text{ x}_2=3,\text{ y}_2=-2,\text{ z}_2=0$
$\text{l}_1=1,\text{ m}_1=-4,\text{ n}_1=5,\text{ l}_2=1,\text{ m}+2=-4,\text{ n}_2=5$
Now, $\begin{vmatrix}\text{x}-4&\text{y}-3&\text{z}-2\\3-4&-2-3&0-2\\1&-4&5 \end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-4&\text{y}-3&\text{z}-2\\3-4&-2-3&0-2\\1&-4&5 \end{vmatrix}=0$
$\Rightarrow-33(\text{x}-4)+3(\text{y}-3)+9(\text{z}-2)=0$
$\Rightarrow11(\text{x}-4)-(\text{y}-3)-3(\text{z}-2)=0$
$\Rightarrow11\text{x}-\text{y}-3\text{z}=35$
View full question & answer→Question 485 Marks
Find the equation of the plane that contains the line of intersection of the planes $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})-4=0$ and $\vec{\text{r}}\cdot(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})+5=0$ and which is perpendicular to the plane $\vec{\text{r}}\cdot(5\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}})+8=0.$
AnswerThe equation of the plane passing through the line of intersection of the given planes is
$\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})-4+\lambda\Big[\vec{\text{r}}\cdot(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})+5\Big]=0$
$\vec{\text{r}}\cdot\Big[(1+2\lambda)\hat{\text{i}}+(2+\lambda)\hat{\text{j}}+(3-\lambda)\hat{\text{k}}\Big]-4+5\lambda=0\ ...(\text{i})$
The plane is perpendicular to $\vec{\text{r}}(5\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}})+8=0$ So,
$5(1+2\lambda)+3(2+\lambda)-6(3-\lambda)=0 \ (\because a_1a_2 + b_1b_2 + c_1c_{2 }= 0)$
$\Rightarrow5+10\lambda+6+3\lambda-18+6\lambda=0$
$\Rightarrow19\lambda-7=0$
$\Rightarrow\lambda=\frac{7}{19}$
Substituting this in $(i)$ we get
$\vec{\text{r}}\cdot\Big[\Big(1+2\Big(\frac{7}{19}\Big)\Big)\hat{\text{i}}+\Big(2+\frac{7}{19}\Big)\hat{\text{j}}+\Big(3-\frac{7}{19}\Big)\hat{\text{k}}\Big]-4+5\Big(\frac{7}{19}\Big)=0$
$\Rightarrow\vec{\text{r}}(33\hat{\text{i}}+45\hat{\text{j}}+50\hat{\text{k}})-41=0$
$\Rightarrow(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(33\hat{\text{i}}+45\hat{\text{j}}+50\hat{\text{k}})-41=0$
$\Rightarrow33\text{x}+45\text{y}+50\text{z}-41=0$
View full question & answer→Question 495 Marks
Show that the line $\frac{\text{x}+3}{-3}=\frac{\text{y}-1}{1}=\frac{\text{z}-5}{5}$ and $\frac{\text{x}+1}{-1}=\frac{\text{y}-2}{2}=\frac{\text{z}-5}{5}$ are coplanar. Hence, find the equation of the plane containing these lines.
AnswerThe lines $\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{\text{y}-\text{y}_1}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}$ and $\frac{\text{x}-\text{x}_2}{\text{a}_2}=\frac{\text{y}-\text{y}_2}{\text{b}_2}=\frac{\text{z}-\text{z}_2}{\text{c}_2}$ are coplanar if
$\begin{vmatrix}\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2\end{vmatrix}=0$
The given lines are $\frac{\text{x}+3}{-3}=\frac{\text{y}-1}{1}=\frac{\text{z}-5}{5}$ and $\frac{\text{x}+1}{-1}=\frac{\text{y}-2}{2}=\frac{\text{z}-5}{5}$
Now, $\begin{vmatrix}-1-(-3)&2-1&5-5\\-3&1&5\\-1&2&5\end{vmatrix}=\begin{vmatrix}2&1&0\\-3&1&5\\-1&2&5\end{vmatrix}$
$=2(5-10)-1(-15+15)+0=-10+10+0=0$
So, the given lines are coplanar.
The equation of the containing the given lines is
$\begin{vmatrix}\text{x}-(-3)&\text{y}-1&\text{z}-5\\-3&1&5\\-1&2&5\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}+3&\text{y}-1&\text{z}-5\\-3&1&5\\-1&2&5\end{vmatrix}=0$
$\Rightarrow(\text{x}+3)(5-10)-(\text{y}-1)(-15+5)+(\text{z}-5)(-6+1)=0$
$\Rightarrow-5(\text{x}+3)+10(\text{y}-1)-5(\text{z}-5)=0$
$\Rightarrow\text{x}-2\text{y}+\text{z}=0$
Thus, the equation of the containing the given lines is x - 2y + z = 0.
View full question & answer→Question 505 Marks
Show that the lines $\frac{\text{x}+1}{-3}=\frac{\text{y}-3}{2}=\frac{\text{z}+2}{1}$ and $\frac{\text{x}}{1}=\frac{\text{y}-7}{-3}=\frac{\text{z}+7}{2}$ are coplanar. Also, find the equation of the plane containing them.
AnswerWe know that lines $\frac{\text{x}-\text{x}_1}{\text{l}_1}=\frac{\text{y}-\text{y}_1}{\text{m}_1}=\frac{\text{z}-\text{z}_1}{\text{n}_1}$ and $\frac{\text{x}-\text{x}_2}{\text{l}_2}=\frac{\text{y}-\text{y}_2}{\text{m}_2}=\frac{\text{z}-\text{z}_2}{\text{n}_2}$ are coplanar if
$\begin{vmatrix}\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{l}_1&\text{m}_1&\text{n}_1\\\text{l}_2&\text{m}_2&\text{n}_2\end{vmatrix}=0$
And equation of plane containing them is
$\begin{vmatrix}\text{x}-\text{x}_1&\text{y}-\text{y}_1&\text{z}-\text{z}_1\\\text{l}_1&\text{m}_1&\text{n}_1\\\text{l}_2&\text{m}_2&\text{n}_2\end{vmatrix}=0$
Here, equation of lines are
$\frac{\text{x}+1}{-3}=\frac{\text{y}-3}{2}=\frac{\text{z}+2}{1}$ and $\frac{\text{x}}{1}=\frac{\text{y}-7}{-3}=\frac{\text{z}+7}{2}$
So, $\text{x}_1=-1,\text{ y}_1=3,\text{ z}_1=-2,\text{ l}_1=-3,\text{ m}_1=2,\text{ n}_1=1$
$\text{x}_2=0,\text{ y}_2=7,\text{ z}_2=-7,\text{ l}_2=1,\text{ m}_1=-3,\text{ n}_1=2$
So, $\begin{vmatrix}\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{l}_1&\text{m}_1&\text{n}_1\\\text{l}_2&\text{m}_2&\text{n}_2 \end{vmatrix}=0$
$=\begin{vmatrix}0+1&7-3&-7+2\\-3&2&1\\1&-3&2\end{vmatrix}$
$=\begin{vmatrix}1&4&-5\\-3&2&1\\1&-3&2\end{vmatrix}$
$=1(4+3)-4(6-1)-5(9-2)$
$=7+28-35$
$=0$
So, lines are coplanar.
Equation of plane containing line is
$\begin{vmatrix}\text{x}+1&\text{y}-3&\text{z}+2\\-3&2&1\\1&-3&2\end{vmatrix}=0$
$(\text{x}+1)(4+3)-(\text{y}-3)(-6-1)+(\text{z}+2)(9-2)=0$
$7\text{x}+7+7\text{y}-21+7\text{z}+14=0$
$7\text{x}+7\text{y}+7\text{z}=0$
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