Question 14 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 = d1 and ax + by + cz = d2 is $\frac{\big|\text{d}_2-\text{d}_1\big|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
So, the required distance
$=\frac{|-13-12|}{\sqrt{6^2+(-3)^2+9^2}}$
$=\frac{|-25|}{\sqrt{39-9+81}}$
$=\frac{25}{\sqrt{126}}$
$=\frac{5}{3\sqrt{14}}\text{ units}$
View full question & answer→Question 24 Marks
If the axes are rectangular and p is the point (2, 3, -1), find the equation of the plane throught p at right angles to OP.
AnswerThe normal is passing through the points O(0, 0, 0) and P(2, 3, -1). So, $\vec{\text{n}}=\overrightarrow{\text{OP}}$
$=(2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})-(0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}})$
$=2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}$
Since the plane passes through the point (2, 3, -1)
$=2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}$
We know that the vector equation of the plane passing through a point
$\vec{\text{a}}$ and normal to $\vec{\text{n}}$ is $\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
Substituting
$\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}$ and $\vec{\text{n}}=2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}},$ we get $\vec{\text{r}}\cdot(2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})=(2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})\cdot(2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot(2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})=4+9+1$
$\Rightarrow\vec{\text{r}}\cdot(2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})=14$
Substituting
$\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ in the vector equation, we get $(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})=14$
$\Rightarrow2\text{x}+3\text{y}-\text{z}=14$
View full question & answer→Question 34 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 44 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 54 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 (x1, y1, z1) 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 a2x + b2y + c2z + d2 = 0 if
a1a2 + b1b2 + c1c2 = 0 ....(iii)
Given, line (ii) is parallel to plane x - y + 2z = 5
So, a1a2 + b1b2 + c1c2 = 0
(a1)(1) + (b1)(-1) + (c1)(2) = 0
a1 + b1 + 2c1 = 0 ....(iv)
Also, given line (ii) is parallel to plane 3x + y + z = 6
So, a1a2 + b1b2 + c1c2 = 0
(a1)(3) + (b1)(1) + (c1)(1) = 0
3a1 + b1 + c1 = 0 ....(v)
Solving (iv) and (v) by cross-multiplication,
$\frac{\text{a}_1}{(-1)(1)-(1)(2)}=\frac{\text{b}_1}{(3)(2)-(1)(1)}=\frac{\text{c}_1}{(1)(1)-(3)(-1)}$
$\frac{\text{a}_1}{-1-2}=\frac{\text{b}_1}{6-1}=\frac{\text{c}_1}{1+3}$
$\frac{\text{a}_1}{-3}=\frac{\text{b}_1}{5}=\frac{\text{c}_1}{4}=\lambda(\text{ say})$
$\text{a}_1=-3\lambda,\text{b}_1=5\lambda,\text{c}_1=4\lambda$
Put a1, b1, c1 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 64 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 74 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
l1l2 + m1m2 + n1n2 = 0
Here,
l1 = -3, m1 = -2k, n1 = 2, l2 = k, m2 = 1, n2 = 5
It is given that given are perpendicular.
⇒ l1l2 + m1m2 + n1n2 = 0
⇒ (-3)(k) + (-2k)(1) + (2)(5) = 0
⇒ -3k - 2k + 10 = 0
⇒ -5k = -10
⇒ 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
$\begin{vmatrix}\text{x}-1&\text{y}-2&\text{z}-3\\-3&-4&2\\2&1&5 \end{vmatrix}=0$
⇒ -22(x - 1) + 19(y - 2) + 5(z - 3) = 0
⇒ -22x + 19y + 5z = 31
View full question & answer→Question 84 Marks
Find the cartesian form of the equations of the following planes.
$\vec{\text{r}}=(\hat{\text{i}}-\hat{\text{j}})+\text{s}(-\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})+\text{t}(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}})$
Answer $\vec{\text{r}}=(\hat{\text{i}}-\hat{\text{j}})+\text{s}(-\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})+\text{t}(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}})$ We know that the equation $\vec{\text{r}}=\vec{\text{a}}+\text{s}\vec{\text{b}}+\text{t}\vec{\text{c}}$ represents a plane passing through a point whose position vector is $\vec{\text{a}}$ and parallel to the vectors $\vec{\text{b}}$ and $\vec{\text{c}}.$
Here, $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+0\hat{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}},\vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$
Normal vector, $\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{c}}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&1&2\\1&2&1\end{vmatrix}$
$=-3\hat{\text{i}}+3\hat{\text{j}}-3\hat{\text{k}}$
The vector equation of the plane in scalar product form is,
$\vec{\text{i}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
$\Rightarrow\vec{\text{r}}\cdot(-3\hat{\text{i}}+3\hat{\text{j}}-3\hat{\text{k}})=(\hat{\text{i}}-\hat{\text{j}}+0\hat{\text{k}})\cdot(-3\hat{\text{i}}+3\hat{\text{j}}-3\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot\Big[-3(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})\Big]=-3-3+0 $
$\Rightarrow\vec{\text{r}}\cdot\Big[-3(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})\Big]=-6$
$\Rightarrow\vec{\text{r}}\cdot(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=2$
For cartesian form, let us substitute $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ here. Then we get
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=2$
$\text{x}-\text{y}+\text{z}=2$
View full question & answer→Question 94 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 104 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 114 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 124 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 134 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 144 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 154 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 (x1, y1, z1) 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 a2x + b2y + c2z + d2 = 0 are parallel if a1a2 + b1b2 + c1c2 = 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 a1, b1, c1 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 164 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 174 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 184 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 194 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 (x1, y1, z1) 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 (x1, y1, z1) 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 (x1, y1, z1) are equidistant from the planes 3x - 4y + 12z - 6 = 0 and 4x + 3z - 7 = 0
So, D1 = D2
$\Bigg|\frac{3\text{x}_1-4\text{y}_1+12\text{z}_1-6}{13}\Bigg|=\bigg|\frac{4\text{x}_1+3\text{z}_1-7}{5}\bigg|$
$\frac{3\text{x}_1-4\text{y}_1+12\text{z}_1-6}{13}=\pm\frac{4\text{x}_1+3\text{z}_1-7}{5}$
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 204 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 214 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 224 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 234 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 244 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 254 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 264 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 274 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 284 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 294 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 304 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 a1a2 + b1b2 + c1c2 = 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 314 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 324 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
$\begin{vmatrix}\text{x}+1&\text{y}-3&\text{z}+2\\-3&2&1\\1&-3&2\end{vmatrix}=0$
$(\text{x}+1)(4+3)-(\text{y}-3)(-6-1)+(\text{z}+2)(9-2)=0$
$7\text{x}+7+7\text{y}-21+7\text{z}+14=0$
$7\text{x}+7\text{y}+7\text{z}=0$
View full question & answer→Question 334 Marks
Show that the normal vector to the plane 2x + 2y + 2z = 3 is equally inclined to the coordinate axes.
AnswerGiven, equation of plane is,
2x + 2y + 2z = 3
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})(2\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})=3$
$\vec{\text{r}}\cdot(2\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})=3$
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{d}}$
Normal to the plane $\vec{\text{n}}=2\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
Direction ratio of $\vec{\text{n}}=2,2,2$
Direction cosine of $\vec{\text{n}}=\frac{2}{|\vec{\text{n}}|},\frac{2}{|\vec{\text{n}}|},\frac{2}{|\vec{\text{n}}|}$
$|\vec{\text{n}}|=\sqrt{(2)^2+(2)^2+(2)^2}$
$=\sqrt{4+4+4}$
$=\sqrt{12}$
$|\vec{\text{n}}|=2\sqrt{3}$
Direction cosine of $|\vec{\text{n}}|=\frac{2}{2\sqrt{3}},\frac{2}{2\sqrt{3}},\frac{ 2}{ 2\sqrt{3}}$
$=\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$
So, $\text{l}=\frac{1}{\sqrt{3}},\text{ m}=\frac{1}{\sqrt{3}},\text{ n}=\frac{1}{\sqrt{3}}$
Let $\alpha,\beta,\gamma$ be the angle that normal $\vec{\text{n}}$ makes with the coordinate axes respectively.
$\text{l}=\cos\alpha=\frac{1}{\sqrt{3}}$
$\alpha=\cos^{-1}\Big(\frac{1}{\sqrt{3}}\Big)\ ...(\text{i})$
$\text{m}=\cos\beta=\frac{1}{\sqrt{3}}$
$\beta=\cos^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$
$\text{n}=\cos\gamma=\frac{1}{\sqrt{3}}$
$\gamma=\cos^{-1}\Big(\frac{1}{\sqrt{3}}\Big)\ ...(\text{ii})$
From equation (i), (ii), (iii),
$\alpha=\beta=\gamma$
So, normal to the plane, $\vec{\text{n}}$ is equally inclined with the coordinate axes.
View full question & answer→Question 344 Marks
Find the distance between the point (7, 2, 4) and the plane determined by the points A(2, 5, -3), B(-2, -3, 5) nad C(5, 3, -3).
AnswerThe given points are A(2, 5, -3), B(-2, -3, -3). The equation of the plane ABC is given by
$\begin{vmatrix}\text{x}-\text{x}_1&\text{y}-\text{y}_1&\text{z}-\text{z}_1\\\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{x}_3-\text{x}_1&\text{y}_3-\text{y}_1&\text{z}_3-\text{z}_1\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-2&\text{y}-5&\text{z}-(-3)\\-2-2&-3-5&5-(-3)\\5-2&3-5&-3-(-3)\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-2&\text{y}-5&\text{z}+3\\-4&-8&8\\3&-2&0\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-2&\text{y}-5&\text{z}+3\\1&2&-2\\3&-2&0\end{vmatrix}=0$
$\Rightarrow-4(\text{x}-2)-6(\text{y}-5)-8(\text{z}+3)=0$
$\Rightarrow2(\text{x}-2)+3(\text{y}-5)+4(\text{z}+3)=0$
$\Rightarrow2\text{x}+3\text{y}+4\text{z}-7=0$
Distance between the point (7, 2, 4) and the plane 2x + 3y + 4z - 7 = 0
Distance between the point (7, 2, 4) to the plane 2x + 3y + 4z - 7 = 0
$=\bigg|\frac{2\times7+3\times2+4\times4-7}{\sqrt{2^2+3^2+4^2}}\bigg|$
$=\bigg|\frac{14+16-16-7}{\sqrt{4+9+16}}\bigg|$
$=\Big|\frac{29}{\sqrt{29}}\Big|$
$=\sqrt{29}\text{ units}$
Thus, the required distance between the given point is $\sqrt{29}\text{ units}.$
View full question & answer→Question 354 Marks
Find the length and the foot of the perpendicular from the point (1, 1, 2) to the plane $\vec{\text{r}}.\big(\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}\big)+5=0.$
AnswerThe Cartesian equation of the given plane is 2x - 2y + 4z + 5 = 0
Let P(x1, y1, z1) be the foot of perpendicular formula (1, 1, 2) to the plane 2x - 2y + 4z + 5 = 0
Direction ratios of the line PQ are proportional to the direction ratios of the given plane
$\frac{\text{x}_1-1}{1}=\frac{\text{y}_1-1}{-2}=\frac{\text{z}_1-2}{4}=\lambda$
$\Rightarrow\ \text{x}_1=2\lambda+1,\text{y}_1=-2\lambda+1,\text{z}_1=4\lambda+2$
$\text{P}(2\lambda+1,-2\lambda+1,4\lambda+2)$ lies on the plane 2x - 2y + 4z + 5 = 0
$\therefore\ 2(2\lambda+1) -2(2\lambda+1)+4(4\lambda+2)+5=0$
$\Rightarrow 4\lambda+2+4\lambda-2+16\lambda+8+5=0$
$\Rightarrow 24\lambda+13=0$
$\Rightarrow\lambda=-\frac{13}{24}$
$\therefore\ \text{x}_1=2\Big(\frac{-13}{24}\Big)+1=-\frac{-13}{12}+\frac{-1}{12}$
$\text{y}_1=-2\Big(\frac{-13}{24}\Big)+1=\frac{13}{12}+1=\frac{25}{12}$
$\text{z}_1=4\lambda+2=4\Big(\frac{-13}{24}\Big)+4=\frac{-7}{6}$
$\therefore$ Coordinates of foot of perpendicular are $\Big(\frac{-1}{12},\frac{25}{12},\frac{-7}{6}\Big)$
Length of perpendicular from (1, 1, 2) to the plane 2x - 2y + 4z + 5 = 0
$=\Bigg|\frac{2\times1-2\times1+4\times2+5}{\sqrt{(2)^2+(-2)^2+(4)^2}}\Bigg|\ \begin{pmatrix} \text{Length of perpendicular from P}(\text{x}_1,\text{y}_1,\text{z}_1)\text{ to the plane}\\ \text{ax}+\text{by}+\text{cz}+\text{d}=0=\Bigg|\frac{\text{ax}_1+\text{by}_1+\text{cz}_1+\text{d}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}\bigg| \end{pmatrix}$
$=\Big|\frac{2-2+8+5}{\sqrt{24}}\Big|$
$=\frac{13}{\sqrt{24}}$
View full question & answer→Question 364 Marks
Find the distance between the planes $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})+7=0$ and $\vec{\text{r}}\cdot(2\hat{\text{i}}+4\hat{\text{j}}+6\hat{\text{k}})+7=0$
AnswerThe given plane are,
$\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})=-7$
$\Rightarrow\text{x}+2\text{y}+3\text{z}=-7$
Multiplying this equation of the plane by 2, we get
$2\text{x}+4\text{y}+6\text{z}=-14\ ...(\text{i})$
and
$\vec{\text{r}}\cdot(2\hat{\text{i}}+4\hat{\text{j}}+6\hat{\text{k}})=-7$
$2\text{x}+4\text{y}+6\text{z}=-7\ ...(\text{ii})$
We know that distance between two planes ax + by + cz = d1 and ax + by + cz = d2 is $\frac{\big|\text{d}_2-\text{d}_1\big|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
So, the required distance
$=\frac{|-7-(-14)|}{\sqrt{2^2+4^2+6^2}}$
$=\frac{|7|}{\sqrt{4+16+36}}$
$=\frac{7}{\sqrt{56}}\text{ units}$
View full question & answer→Question 374 Marks
Find the image of the point (1, 3, 4) in the plane 2x - y + z + 3 = 0.
AnswerLet Q be the image of the point P(1, 3, 4) in the plane 2x - y + z + 3 = 0.
Then PQ is normal to the plane. So, the direction ratios or PQ are proportional to 2, -1, 1.
Since PQ passes through P(1, 3, 4) and has direction ratios proportional to 2, -1 and 1. equation ot PQ is
$\frac{\text{x}-1}{2}=\frac{\text{y}-3}{-1}=\frac{\text{z}-4}{1}=\text{r (say)}$
Let the coordinates of Q be (2r + 1, -r + 3, r + 4)
Let R be the mid-point or PQ. Then,
$\text{R}=\Big(\frac{2\text{r}+1+1}{2},\frac{-\text{r}+3+3}{2},\frac{\text{r}+4+4}{2}\Big)=\Big(\text{r}+1,\frac{-\text{r}+6}{2},\frac{\text{r}+8}{2}\Big)$
Since R lies in the plane 2x - y + z + 3 = 0
$2(\text{r}+1)-\Big(\frac{-\text{r}+6}{2}\Big)+\frac{\text{r}+8}{2}+3=0$
$\Rightarrow 4\text{r}+4+\text{r}-6+\text{r}+8+6=0$
$\Rightarrow 6\text{r}+12=0$
$\Rightarrow \text{r}=-2$
Substituting this in the coordinates of Q, we get
$\text{Q}=(2\text{r}+1,-\text{r}+3,\text{r}+4)$
$=(2(-2))+1,2+3,-2+4)=(-3,5,2)$
View full question & answer→Question 384 Marks
Find the distance of the point (2, 12, 5) from the point of intersection of the line $\vec{\text{r}}=2\hat{\text{i}}-4\hat{\text{j}}+2\hat{\text{k}}+\lambda(3\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}})$ and $\vec{\text{r}}\cdot(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}})=0.$
AnswerThe equation of the given line is $\vec{\text{r}}=2\hat{\text{i}}-4\hat{\text{j}}+2\hat{\text{k}}+\lambda(3\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}})$
The position vector of any point on the line is
$\vec{\text{r}}=(2+3\lambda)\hat{\text{i}}+(-4+4\lambda)\hat{\text{j}}+(2-2\lambda)\hat{\text{k}}$
If this lies on the plane $\vec{\text{r}}\cdot(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}})=0,$ then
$\Big[(2+3\lambda)\hat{\text{i}}+(-4+4\lambda)\hat{\text{j}}+(2-2\lambda)\hat{\text{k}}\Big]\cdot(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}})=0$
$\Rightarrow(2+3\lambda)-2(-4+4\lambda)+(2+2\lambda)=0$
$\Rightarrow2+3\lambda+8-8\lambda+2+2\lambda=0$
$\Rightarrow3\lambda=12$
$\Rightarrow\lambda=4$
View full question & answer→Question 394 Marks
Find the distance of the point with position vector $-\hat{\text{i}}-5\hat{\text{j}}-10\hat{\text{k}}$ from the point of intersection of the line $\vec{\text{r}}=(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})+\lambda(3\hat{\text{i}}+4\hat{\text{j}}+12\hat{\text{k}})$ with the plane $\vec{\text{r}}.(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=5.$
AnswerThe given equation of the line is,
$\vec{\text{r}}=(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})+\lambda(3\hat{\text{i}}+4\hat{\text{j}}+12\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}=(2+3\lambda)\hat{\text{i}}+(-1+4\lambda)\hat{\text{j}}+(2+2\lambda)\hat{\text{k}}$
The coordinated of any point on line are of the form $(2+3\lambda)\hat{\text{i}}+(-1+4\lambda)\hat{\text{j}}+(2+2\lambda)\hat{\text{k}}$ or $(2+3\lambda,-1+4\lambda,2+2\lambda)$
Since this point lies on the plane $\vec{\text{r}}.(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=5$
$\big[(2+3\lambda)\hat{\text{i}}+(-1+4\lambda)\hat{\text{j}}+(2+2\lambda)\hat{\text{k}}\big].(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=5$
$\Rightarrow 2+3\lambda+1-4\lambda+2+2\lambda-5=0$
$\Rightarrow \lambda=0$
So, the coordinates ofthe point are
$(2+3\lambda,-4+4\lambda,2+2\lambda)$
$=(2+1,-1+0,2+0)$
$=(2,-1,2)$
The coordinated of the point corresponding to the position vector $-\hat{\text{i}}-5\hat{\text{j}}-10\hat{\text{k}}$ are (-1, -5, -10).
Distance between (2, -1, 2) and (-1, -5, -10)
$=\sqrt{(-1-2)^2+(-5+1)^2+(-10-2)^2}$
$=\sqrt{9+16+144}$
$=13\text{ units}$
View full question & answer→Question 404 Marks
Show that the plane whose vector equation is $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})=3$ contains the line whose vector equation is $\vec{\text{r}}=\hat{\text{i}}+\hat{\text{j}}+\lambda(2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}).$
AnswerThe line $\vec{\text{r}}=(\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}})+\lambda(2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}})\ ...(\text{i})$
Passes through a point whose posotion vector is $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}$ and is parallel to the vector $\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}$
If the plane $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})=3$ contains the given line, then
It should passes through the point $\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}.$
It should be parallel to the line.
Now, the plane passes through the point $\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}.$
So, the plane vector to the given plane is $\vec{\text{n}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}.$
We observe that
$\vec{\text{b}}\cdot\vec{\text{n}}=(2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}})\cdot(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})$
$=2+2-4=0$
Therefore, the plane is parallel to the line.
Hence, the given plane contains the given line.
View full question & answer→Question 414 Marks
Find the reflection of the point (1, 2, -1) in the plane 3x - 5y + 4z = 5.
AnswerHere, we have to find reflection of the point P(1, 2, -1) in the plane 3x - 5y + 4z = 5
Let Q be the reflection of the point P and R be the mid-point of PQ.
Then, R lien on the plane 3x - 5y + 4z = 5
Direction rations or PQ are propaortional to 3, -5, 4 and PQ is passing through (1, 2, -1).
So, equation of PQ is given by,
$\frac{\text{x}-1}{3}=\frac{\text{y}-2}{-5}=\frac{\text{z}+1}{4}=\lambda \ (\text{Say})$
Let Q be $(3\lambda+1, -5\lambda+2, 4\lambda-1)$
The coordinated of R are $\Big(\frac{3\lambda+1+1}{2},\frac{-5\lambda+2+2}{2},\frac{4\lambda-1-1}{2}\Big)=\Big(\frac{3\lambda+2}{2},\frac{-5\lambda+4}{2},\frac{4\lambda-2}{2}\Big)$
Since, R lies on the given plane 3x - 5y + 4z = 5
$\therefore\ 3\Big(\frac{3\lambda+2}{2}\Big)-5\Big(\frac{-5\lambda+4}{2}\Big)+4\Big(\frac{4\lambda-2}{2}\Big)=5$
View full question & answer→Question 424 Marks
Show that the lines $\frac{5-\text{x}}{-4}=\frac{\text{y}-7}{4}=\frac{\text{z}+3}{-5}$ and $\frac{\text{x}-8}{7}=\frac{2\text{y}-8}{2}=\frac{\text{z}-5}{3}$ are coplanar.
Answer$\frac{5-\text{x}}{-4}=\frac{\text{y}-7}{4}=\frac{\text{z}+3}{-5}$
$\frac{\text{x}-5}{4}=\frac{\text{y}-7}{4}=\frac{\text{z}+3}{-5}\ ...(\text{i})$
$\frac{\text{x}-8}{7}=\frac{2\text{y}-8}{2}=\frac{\text{z}-5}{3}$
$\frac{\text{x}-8}{7}=\frac{\text{y}-4}{1}=\frac{\text{z}-5}{3}\ ....(\text{ii})$
Here, a1 = 4, b1 = 4, c1 = -5
a2 = 7, b2 = 1, c2 = 3
x1 = 5, y1 = 7, z1 = -3
x2 = 8, y2 = 4, z2 = 5
Condition for two lines to be coplanar,
$\begin{vmatrix}\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2 \end{vmatrix}=0$
$\therefore\begin{vmatrix}8-5&4-7&5+3\\4&4&-5\\7&1&3\end{vmatrix}$
$=\begin{vmatrix}3&-3&8\\4&4&-5\\7&1&3\end{vmatrix}$
$=3(12+5)+3(12+35)+8(4-28)$
$=3\times17+3\times47+8\times(-24)$
$=51+141-192$
$=192-192$
$=0$
$\therefore$ The lines are coplanar to each other.
View full question & answer→Question 434 Marks
If O be the origin and the coordinates of P be (1, 2, -3), then find the equation of the plane passing through P and perpendicular to OP.
AnswerThe normal is passing through the points O(0, 0, 0) and P(1, 2, -3) So,
$\vec{\text{n}}=\overrightarrow{\text{OP}}$
$=(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})-(0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}})$
$=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$
Since the plane passes through P(1, 2, -3), $\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$
We know that the vector equation of the plane passing through a point $\vec{\text{a}}$ and normal to $\vec{\text{n}}$ is,
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
Substituting $\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}$ and $\vec{\text{n}}=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$ in the relation, we get
$\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})\cdot(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=1+4+9$
$\Rightarrow\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=14$
$\Rightarrow\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=14$
Substituting $\overrightarrow{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ in the vector equation, we get
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=14$
$\Rightarrow\text{x}+2\text{y}-3\text{z}=14$
View full question & answer→Question 444 Marks
Find the vector equations of the following planes in scalar product form $(\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}):$
$\vec{\text{r}}=(2\hat{\text{i}}-\hat{\text{k}})+\lambda\hat{\text{i}}+\mu(\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}})$
AnswerHere, $\vec{\text{r}}=(2\hat{\text{i}}-\hat{\text{k}})+\lambda\hat{\text{i}}+\mu(\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}})$
We know that, $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}+\mu\vec{\text{c}}$ represent a plane passing throught a point having position vector $\vec{\text{a}}$ and parallel to vectors $\vec{\text{b}}$ and $\vec{\text{c}}$
Here, $\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}},\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}}$
The given plane is perpendicular to a vector
$\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{c}}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&0&0\\1&-2&-1\end{vmatrix}$
$=\hat{\text{i}}(0-0)-\hat{\text{j}}(-1-0)+\hat{\text{k}}(-2-0)$
$=0\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$
$\vec{\text{n}}=\hat{\text{j}}-2\hat{\text{k}}$
We know that vector equation of plane in scalar product form is,
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}\ ...(\text{i})$
Put $\vec{\text{n}}$ and $\vec{\text{a}}$ in equation (i),
$\vec{\text{r}}\cdot(\hat{\text{j}}-2\hat{\text{k}})=(2\hat{\text{i}}-\hat{\text{k}})(\hat{\text{j}}-2\hat{\text{k}})$
$\vec{\text{r}}\cdot(\hat{\text{j}}-2\hat{\text{k}})=(2)(0)+(0)+(1)+(-1)(-2)$
$\vec{\text{r}}\cdot(\hat{\text{j}}-2\hat{\text{k}})=0+0+2$
$\vec{\text{r}}\cdot(\hat{\text{j}}-2\hat{\text{k}})=2$
The equation in required form is,
$\vec{\text{r}}\cdot(\hat{\text{j}}-2\hat{\text{k}})=2$
View full question & answer→Question 454 Marks
Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-5\hat{\text{k}})+9=0.$
AnswerLet a, b, c be the direction ratios of the given line.
Since the line passes through the point (1, 2, 3) is,
$\frac{\text{x}-1}{\text{a}}=\frac{\text{y}-2}{\text{b}}=\frac{\text{z}-3}{\text{c}}\ ...(\text{i})$
Since this line is perpendicular to the planer $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-5\hat{\text{k}})+9=0$ or x + 2y - 5z + 9 = 0, this line is parallel to the normal of the plane.
So, the direction ratios of the line are proportional to the direction ratios of the given plane.
So, $\frac{\text{a}}{1}=\frac{\text{b}}{2}=\frac{\text{c}}{-5}=\lambda$
$\text{a}=\lambda,\text{ b}=2\lambda,\text{ c}=-5\lambda$
Substituting these value in (i) we get
$\frac{\text{x}-1}{1}=\frac{\text{y}+1}{2}=\frac{\text{z}-2}{-5},$ which is the cartesian form of the line.
vector from
The given line passes through a point whose position vector is $\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ and is parallel to the vector $\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-5\hat{\text{k}}.$ So, its equation in vector form is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$\vec{\text{r}}=(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})+\lambda(\hat{\text{i}}+2\hat{\text{j}}-5\hat{\text{k}})$
View full question & answer→Question 464 Marks
Find the equation of the plane passing through (a, b, c) and parallel to the plane $\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=2.$
AnswerLet the equation of a plane parallel to the given plane be
$\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=\text{k}$
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=\text{k}\ ...(\text{i})$
This passes through (a, b, c). So,
$(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=\text{k}$
$\Rightarrow\text{k}=\text{a}+\text{b}+\text{c}$
Substituting this in (i), we get
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=\text{a}+\text{b}+\text{c}$
$\text{x}+\text{y}+\text{z}=\text{a}+\text{b}+\text{c},$ Which is the equation of the required plane.
View full question & answer→Question 474 Marks
A plane passes through the point (1, -2, 5) and is perpendicular to the line joining the origin to the point $2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}.$ Find the vector and cartesian forms of the equation of the plane.
AnswerThe normal is passing throught the point A(0, 0, 0) and B(3, 1, -1). So,
$\vec{\text{n}}=\overrightarrow{\text{OP}}$
$=(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})-(0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}})$
$=3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
Since the plane passes throught (1, -2, 5)
$\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+5\hat{\text{k}}$
We know that the vector equation of the plane passing through a point $\vec{\text{a}}$ and normal to $\vec{\text{n}}$ is,
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
Substituting $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{n}}=4\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}},$ we get
$\vec{\text{r}}\cdot(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=(\hat{\text{i}}-2\hat{\text{j}}+5\hat{\text{k}})\cdot(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=3-2-5$
$\Rightarrow\vec{\text{r}}\cdot(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=-4$
Substituting $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ in the vector equation we get
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=-4$
$\Rightarrow3\text{x}+\text{y}-\text{z}=-4$
View full question & answer→Question 484 Marks
Find the vector equations of the following planes in scalar product form $(\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}):$
$\vec{\text{r}}=(1+\text{s}-\text{t})\hat{\text{t}}+(2-\text{s})\hat{\text{j}}+(3-2\text{s}+2\text{t})\hat{\text{k}}$
AnswerHere, $\vec{\text{r}}=(1+\text{s}-\text{t})\hat{\text{t}}+(2-\text{s})\hat{\text{j}}+(3-2\text{s}+2\text{t})\hat{\text{k}}$
$\vec{\text{r}}=(\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}})+\text{s}(\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}})+\text{t}(-\hat{\text{i}}+2\hat{\text{k}})$
We know that, $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}+\mu\vec{\text{c}}$ represent a plane passing through a point having position vector $\vec{\text{a}}$ and parallel to vectors $\vec{\text{b}}$ and $\vec{\text{c}}.$
Here, $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}},\vec{\text{c}}=-\hat{\text{i}}+2\hat{\text{k}}$
The given plane is perpendicular to a vector
$\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{c}}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&-1&-2\\-1&0&2\end{vmatrix}$
$=\hat{\text{i}}(-2-0)-\hat{\text{j}}(2-2)+\hat{\text{k}}(0-1)$
$\vec{\text{n}}=-2\hat{\text{i}}-\hat{\text{k}}$
We know that vector equation of plane in scalar product form is,
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}\ ...(\text{i})$
Put $\vec{\text{n}}$ and $\vec{\text{a}}$ in equation (i),
$\vec{\text{r}}\cdot(-2\hat{\text{i}}-\hat{\text{k}})=(-2\hat{\text{i}}-\hat{\text{k}})(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})$
$\vec{\text{r}}\cdot(-2\hat{\text{i}}-\hat{\text{k}})=(-2)(1)+(0)(2)+(-1)(3)$
$\vec{\text{r}}\cdot(-2\hat{\text{i}}-\hat{\text{k}})=-2+0-3$
$\vec{\text{r}}\cdot(-2\hat{\text{i}}-\hat{\text{k}})=-5$
Multiplying both the sides by (-1),
$\vec{\text{r}}\cdot(2\hat{\text{i}}+\hat{\text{k}})=5$
The equation in required form,
$\vec{\text{r}}\cdot(2\hat{\text{i}}+\hat{\text{k}})=5$
View full question & answer→Question 494 Marks
Find the vector and Cartesian equations of the plane that passes through the point (5, 2, -4) and is perpendicular to the line with direction ratios 2, 3, -1.
AnswerWe know that the vector equation of the plane passing through a point $\vec{\text{a}}$ and normal to $\vec{\text{n}}$ is,
$\overrightarrow{\text{r}}\cdot\overrightarrow{\text{n}}=\overrightarrow{\text{a}}\cdot\overrightarrow{\text{n}}$
Substituting $\overrightarrow{\text{a}}=5\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}$ and $\overrightarrow{\text{n}}=2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}$ (because the direction ratios of $\vec{\text{n}}$ are 2, 3, -1), we get
$\overrightarrow{\text{r}}\cdot(2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})=(5\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}})\cdot(2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot(2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})=10+6+4$
$\Rightarrow\vec{\text{r}}\cdot(2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})=20$
For cartesian form, we need to substitute $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ in this equation. Then, we get
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})=20$
$\Rightarrow2\text{x}+3\text{y}-\text{z}=20$
View full question & answer→Question 504 Marks
If the straight lines $\frac{\text{x}-1}{2}=\frac{\text{y}+1}{\text{k}}=\frac{\text{z}}{2}$ and $\frac{\text{x}+1}{2}=\frac{\text{y}+1}{2}=\frac{\text{z}}{\text{k}}$ are coplanar, find the equation of the planes containing them.
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 $\frac{\text{x}-1}{2}=\frac{\text{y}+1}{\text{k}}=\frac{\text{z}}{2}$ and $\frac{\text{x}+1}{2}=\frac{\text{y}+1}{2}=\frac{\text{z}}{\text{k}}$ are coplanar.
$\therefore\ \begin{vmatrix} -1-1&-1-(-1)&0-0\\2&\text{k}&2\\2&2&\text{k}\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix} -2&0&0\\2&\text{k}&2\\2&2&\text{k}\end{vmatrix}=0$
$\Rightarrow-2(\text{k}^2-4)-0+0=0$
$\Rightarrow\text{k}^2-4=0$
$\Rightarrow\text{k}=\pm2$
The equation of the plane containing the given lines is $\begin{vmatrix} \text{x}-1&\text{y}+1&\text{z}\\2&\text{k}&2\\2&2&\text{k}\end{vmatrix}=0$
For k = 2, $\begin{vmatrix} \text{x}-1&\text{y}+1&\text{z}\\2&\text{k}&2\\2&2&\text{k}\end{vmatrix}=\begin{vmatrix} \text{x}-1&\text{y}+1&\text{z}\\2&\text{2}&2\\2&2&\text{2}\end{vmatrix}=0$
So, no plane exists for k = 2
For k = -2
$\begin{vmatrix} \text{x}-1&\text{y}+1&\text{z}\\2&\text{k}&2\\2&2&\text{k}\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix} \text{x}-1&\text{y}+1&\text{z}\\2&-\text{2}&2\\2&2&-\text{2}\end{vmatrix}=0$
$\Rightarrow(\text{x})(4-4)-(\text{y}+1)(-4-4)+\text{z}(4+4)=0$
$\Rightarrow8(\text{y}+1)+8\text{z}=0$
$\Rightarrow\text{y}+\text{z}+1=0$
Thus, the equation of the plane containing the given lines is y + z + 1 = 0
View full question & answer→