Question 15 Marks
Prove that the line through A(0, –1, –1) and B(4, 5, 1) intersects the line through C(3, 9, 4) and D(–4, 4, 4).
AnswerEquation of line $\vec{\text{AB}}$
$\vec{\text{r}} = (-\hat{\text{j}} - \hat{\text{k}}) + \lambda (4\hat{\text{i}} + 6\hat{\text{j}} + 2\hat{\text{k}})$
Equation of line $\vec{\text{CD}}$
$\vec{\text{r}} = (3\hat{\text{i}} + 9\hat{\text{j}} + 4\hat{\text{k}}) + \mu (-7\hat{\text{i}} - 5\hat{\text{j}})$
$\vec{\text{a}}_{2} - \vec{\text{a}}_{1} = 3\hat{\text{i}} + 10\hat{\text{j}} + 5 \hat{\text{k}}$
$\vec{\text{b}}_{1} \times \vec{\text{b}}_{2} = \begin{vmatrix} \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\ 4 & 6 & 2 \\ -7 & -5 & 0 \end{vmatrix} = 10\hat{\text{i}} - 14\hat{\text{j}} + 22\hat{\text{k}}$
$(\vec{\text{a}}_{2} - \vec{\text{a}}_{1}). (\vec{\text{b}}_{1} \times \vec{\text{b}}_{2}) = 30 – 140 + 110 = 0$
$\Rightarrow$ Lines intersect.
View full question & answer→Question 25 Marks
Find the perpendicular distence of the point (1, 0, 0) from the line $\frac{\text{x}-1}{2}=\frac{\text{y}+1}{-3}=\frac{\text{z}+10}{8}.$ Also, find the coordinates of the perpendicular and the equation of the perpendicular.
AnswerLet foot of the perpemdicular drawn from the point P(1, 0, 0) to the line $\frac{\text{x}-1}{2}=\frac{\text{y}+1}{-3}=\frac{\text{z}+10}{8}$ is Q. we have to find lengh of PQ.
Q is a genelar point on the line,
$\frac{\text{x}-1}{2}=\frac{\text{y}+1}{-3}=\frac{\text{z}+10}{8}=\lambda$ (say)
coordinate of Q $=\big(2\lambda+1,-3\lambda-1,8\lambda-10\big)$
Direction ratios line PQ are
$=\big(2\lambda+1-1\big),\big(-3\lambda-1-0\big),\big(8\lambda-10-0\big)$
$=(2\lambda),\big(-3\lambda-1\big),\big(8\lambda-10\big)$
Since, line PQ is perpendicular to the given line, so
$\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0$
$(2)(2\lambda)+(-3)(-3\lambda-1)+8(8\lambda-10)=0$
$4\lambda+9\lambda+3+64\lambda-80=0$
$77\lambda-77=0$
$\lambda=1$
Therefore, coordinate of Q is $\big(2\lambda+1,-3\lambda-1,8\lambda-10\big)$
$=\big(2(1)+1,-3(1)-1,8(1)-10\big)$
$=(3,-4,-2)$
$\text{PQ}=\sqrt{(\text{x}_1-\text{x}_2)^2+(\text{y}_1-\text{y}_2)^2+(\text{z}_1-\text{z}_2)^2}$
$=\sqrt{(1-3)^2+(0+4)^2+(0+2)^2}$
$=\sqrt{4+16+4}$
$=\sqrt{24}$
$=2\sqrt{6}$
So, foot of perpendicular $=(3,-1,-2)$
length of perpendicular $=2\sqrt{6}\text{ units}$
View full question & answer→Question 35 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 45 Marks
Find the angle between the pairs of lines with direction ratios proportional to $1, 2, -2$ and $-2, 2, 1$
AnswerWe know that, angle $(\theta)$ between two 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 $\frac{\text{x}-\text{x}_2}{\text{a}_2}=\frac{\text{y}-\text{y}_2}{\text{b}_2}=\frac{\text{z}-\text{z}_2}{\text{c}_2}$
is given by,
$\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{{\text{a}_1}^2+{{\text{b}_1}^2}+{{\text{c}_1}^2}}\sqrt{{\text{a}_2}^2+{{\text{b}_2}^2}+{{\text{c}_2}^2}}}\dots(1)$
Here, $a_1= 1, b_1= 2, c_1= -2$
$a_2= -2, b_2= 2, c_2= 1$
Let $\theta$ be required angle, so using equation (1),
$\cos\theta=\frac{(1)(-2)+(2)(2)+(-2)(1)}{\sqrt{(1)^2+(2)^2+(-2)^2}\sqrt{(-2)^2+(2)^2+(1)^2}}$
$=\frac{-2+4-2}{3.3}$
$=\frac{0}{9}$
$\cos\theta=0$
$\theta=\frac{\pi}{2}$
View full question & answer→Question 55 Marks
Find the equation of the plane through the points $(2, 1, 0), (3, -2, -2)$ and $(3, 1, 7)$.
AnswerWe know that, the equation of a plane passing through three non-collinear points $(x_1, x_1, x_1), (x_2, x_2, x_2)$ and $(x_3, x_3, x_3)$ is
$\begin{vmatrix}\text{x}-\text{x}_1&\text{y}-\text{y}_1&\text{z}-\text{z}_1 \\\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{x}_3-\text{x}_1&\text{y}_3-\text{y}_1&\text{z}_3-\text{z}_1 \end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-2&\text{y}-1&\text{z}-0 \\3-2&-2-1&-2-0\\3-2&1-1&7-0 \end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-2&\text{y}-1&\text{z} \\1&-3&-2\\1&0&7 \end{vmatrix}=0$
$\Rightarrow(\text{x}-2)(-21+0)-(\text{y}-1)(7+2)+\text{z}(3)=0$
$\Rightarrow-21\text{x}+42-9\text{y}+9+3\text{z}=0$
$\Rightarrow-21\text{x}-9\text{y}+3\text{z}=-51$
$\therefore7\text{x}+3\text{y}-\text{z}=17$
So, the required equation of plane is $7\text{x}+3\text{y}-\text{z}=17.$
View full question & answer→Question 65 Marks
Find the angle between the follwing pairs of lines:$\vec{\text{r}}=\lambda\big(\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)$ and $\vec{\text{r}}=2\hat{\text{j}}+\mu\big\{\big(\sqrt{3}-1\big)\hat{\text{i}}-\big(\sqrt{3}+1\big)\hat{\text{j}}+4\hat{\text{k}}\big\}$
Answer$\vec{\text{r}}=\lambda\big(\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)$ and $\vec{\text{r}}=2\hat{\text{j}}+\mu\big\{\big(\sqrt{3}-1\big)\hat{\text{i}}-\big(\sqrt{3}+1\big)\hat{\text{j}}+4\hat{\text{k}}\big\}$
Let $\vec{\text{b}}_1$ and $\vec{\text{b}}_2$ be vectors parallel to the given lines.
Now,
$\vec{\text{b}}_1=\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{b}}_2=\big(\sqrt{3}-1\big)\hat{\text{i}}-\big(\sqrt{3}+1\big)\hat{\text{j}}+4\hat{\text{k}}$
If $\theta$ is the angle between the given lines, then
$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$
$=\frac{\big(\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big).\big(\big(\sqrt{3}-1\big)\hat{\text{i}}-\big(\sqrt{3}+1\big)\hat{\text{j}}+4\hat{\text{k}}\big)}{\sqrt{1^2+1^2+2^2}\sqrt{(\sqrt{3}-1)^2+(\sqrt{3}+1)^2+4^2}}$
$=\frac{\big(\sqrt{3}-1\big)-\big(\sqrt{3}+1\big)+8}{\sqrt{6}\sqrt{24}}$
$=\frac{6}{12}$
$=\frac{1}{2}$
$\Rightarrow\theta=\frac{\pi}{3}$
View full question & answer→Question 75 Marks
Find the coordinates of the point where the line through $(5, 1, 6)$ and $(3, 4, 1)$ crosses the $ZX$-plane.
AnswerGiven: A line through the points $A(5, 1, 6)$ and $B(3, 4, 1)$
$\therefore$ Direction ratios of this line $AB$ are $x_2 - x_1, y_2 - y_1, z_2 - z_1$
$\Rightarrow 3 - 5, 4 - 1, 1 - 6$
$\Rightarrow -2, 3, -5 = a, b, c$
Equation of the line AB is $\frac{\text{x}-\text{x}_1}{\text{a}}=\frac{\text{y}-\text{y}_1}{\text{b}}=\frac{\text{z}-\text{z}_1}{\text{c}}$
$\Rightarrow\ \ \frac{\text{x}-5}{-2}=\frac{\text{y}-1}{3}=\frac{\text{z}-6}{-5}\ \ \ .....(\text{i})$
Now we have to find the coordinates of the point where this line AB crosses the ZX-plane
i.e., $y = 0 .......(ii)$
Putting y = 0 in eq. (i), we get
$\frac{\text{x}-5}{-2}=\frac{-1}{3}=\frac{\text{z}-6}{-5}$
$\Rightarrow\ \ \frac{\text{x}-5}{-2}=\frac{-1}{3}\ \text{and}\ \frac{\text{z}-6}{-5}=\frac{-1}{3}$
$\Rightarrow 3x - 15 = 2$ and $3z - 18 = 5$
$\Rightarrow 3x = 17$ and $3z = 23$
$\Rightarrow\ \ \ \text{x}=\frac{17}{3}\ \text{and}\ \text{z}=\frac{23}{3}$
Thus, required point is $\text{P}\Big(\ \frac{17}{3},\ 0,\frac{23}{3}\Big).$
View full question & answer→Question 85 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 $P(x_1, y_1, z_1)$ 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 95 Marks
Prove that the line $\vec{\text{r}}=\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)+\lambda\big(3\hat{\text{i}}-\hat{\text{j}}\big)$ and $\vec{\text{r}}=\big(4\hat{\text{i}}-\hat{\text{k}}\big)+\mu\big(2\hat{\text{i}}+3\hat{\text{k}}\big)$ intersect and find their point of intersection.
AnswerThe position vectors of two arbitrary points on the given lines are
$\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)+\lambda\big(3\hat{\text{i}}-\hat{\text{j}}\big)=(1+3\lambda)\hat{\text{i}}+(1-\lambda)\hat{\text{j}}-\hat{\text{k}}$
$\big(4\hat{\text{i}}-\hat{\text{k}}\big)+\mu\big(2\hat{\text{i}}+3\hat{\text{k}}\big)=(4+2\mu)\hat{\text{i}}+0\hat{\text{j}}+(3\mu-1)\hat{\text{k}}$
If the lines intersect, then they have a common point. so, for some values of $\lambda$ and $\mu,$ we must have
$(1+3\lambda)\hat{\text{i}}+(1-\lambda)\hat{\text{j}}-\hat{\text{k}}=(4+2\mu)\hat{\text{i}}+0\hat{\text{j}}+(3\mu-1)\hat{\text{k}}$
Equating the coefficients of $\hat{\text{i}},\hat{\text{j}}$ and $\hat{\text{k}},$ we get
$1+3\lambda=4+2\mu\dots(1)$
$1-\lambda=0\dots(2)$
$3\mu-1=-1\dots(3)$
Solving (2) and (3), we get
$\lambda=1$
$\mu=0$
Substituting the valuse $\lambda=1$ and $\mu=0$ in (1), we get
$\text{LHS}=1+3\lambda$
$=1+3(1)$
$=4$
$\text{RHS}=4+2\mu$
$=4+2(0)$
$=4$
$\Rightarrow\text{LHS}=\text{RHS}$
Since $\lambda=1$ and $\mu=0$ satisfy (3), the given lines intersect.
Substituting $\mu=0$ in the second line, we get $\vec{\text{r}}=4\hat{\text{i}}+0\hat{\text{j}}-\hat{\text{k}}$ as the position vector of the point of intersection.
Thus, the coordinates of the point of intersection are (4, 0, -1).
View full question & answer→Question 105 Marks
Find the shortest distance between the lines
$\vec{\text{r}}=\Big(4\hat{\text{i}}-\hat{\text{j}}\Big)+\lambda\Big(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\Big)$ $\text{and}\ \vec{\text{r}}=\Big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\Big)+\mu\Big(2\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}\Big).$
Answer$\vec{\text{r}}=\Big(4\hat{\text{i}}-\hat{\text{j}}\Big)+\lambda\Big(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\Big)$
$\vec{\text{a}}_1=4\hat{\text{i}}-\hat{\text{j}}\ \ \ \vec{\text{b}}_1=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$
$\vec{\text{r}}=\Big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\Big)+\mu\Big(2\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}\Big)$
$\vec{\text{a}}_2=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}},\ \vec{\text{b}}_2=2\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}$
$\text{S.D.}=\begin{vmatrix}\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\cdot\big(\vec{\text{b}}_2\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1-\vec{\text{b}}_2\big|}\end{vmatrix}$
$\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}} &\hat{\text{j}}&\hat{\text{k}}\\1 & 2&-3\\2&4&-5 \end{vmatrix}$
$=\hat{\text{i}}(-10+12)-\hat{\text{j}}(-5+6)+\hat{\text{k}}(4-4)$
$=2\hat{\text{i}}-\hat{\text{j}}$
$\text{S.D.}=\begin{vmatrix}\frac{\big(-3\hat{\text{i}}+0\hat{\text{j}}+2\hat{\text{k}}\big)\cdot\big(2\hat{\text{i}}-\hat{\text{j}}\big)}{\sqrt{4+1}}\end{vmatrix}$
$=\begin{vmatrix}\frac{-6}{\sqrt{5}}\end{vmatrix}=\frac{6}{\sqrt{5}}$
View full question & answer→Question 115 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 $ax + by + cz = d_1$ 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
$=\frac{|-7-(-14)|}{\sqrt{2^2+4^2+6^2}}$
$=\frac{|7|}{\sqrt{4+16+36}}$
$=\frac{7}{\sqrt{56}}\text{ units}$
View full question & answer→Question 125 Marks
Find the equation of the plane which is perpendicular to the plane $5x + 3y + 6z + 8 = 0$ and which contains the line of intersection of the planes $x + 2y + 3z - 4 = 0$ and $2x + y - z + 5 =0$
AnswerEquation of plane, containing the line of intersection of Planes:$ x + 2y + 3z - 4 = { 0 } \text{ and 2x} + y - z + 5 = \text{ 0 is}$
${(x + 2y + 3z - 4)} + \lambda {(2x + y - z + 5}) = 0$
$\Rightarrow {x (1 + 2\lambda ) + y(2 + \lambda) + z (3 - \lambda) + ( - 4 + 5\lambda )} = 0 \dots\dots\dots\dots\text{(i})$
$\text{(i) is} \perp \text{to 5x} + 3y + 6z + 8 = 0$
$\therefore 5 ( 1 + 2\lambda ) + 3 ( 2 + \lambda ) + 6 ( 3 - \lambda) = 0 \Rightarrow \lambda = - \frac{29}{7}$
$\therefore$ Requuired equation of plane is:
$x \bigg( 1 - \frac{58}{7}\bigg) + y \bigg(2 -\frac{29}{7}\bigg)+\text{z}\Big(3+\frac{29}{7}\Big)- 4 - \frac{145}{7} = 0$
$\Rightarrow 51 x + 15 y- 50 z + 173 = 0$
View full question & answer→Question 135 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 145 Marks
A(1, 0, 4) B(0, -11, 1), C(2, -3, 1) are three points and D is the fool of perpendicular from A on BC. Find the coordinates of D.
AnswerPoint D is the foot of the perpendicular drawn from the point A(1, 0, 4) to the line BC.
The coordinates of a general point on the line BC are given by
$\frac{\text{x}-0}{2-0}=\frac{\text{y}+11}{-3+11}=\frac{\text{z}-3}{1-3}=\lambda$
$\Rightarrow\text{x}=2\lambda$
$\text{y}=8\lambda-11$
$\text{z}=-2\lambda+3$
Let the coordinates of D be $(2\lambda,8\lambda-11,-2\lambda+3).$
The direction ratios of AD are proportional to
$2\lambda-1,8\lambda-11,-2\lambda+3-4,$
i.e. $2\lambda-1,8\lambda-11,-2\lambda-1.$
The direction ratios of the line BC are proportional to 2, 8 -2, but AD is perpendicular to the line BC.
$\therefore2(2\lambda-1)+8(8\lambda-11)-2(-2\lambda-1)=0$
$\Rightarrow\lambda=\frac{11}9{}$
Substituting $\lambda=\frac{11}{9}$ in $(2\lambda,8\lambda-11,-2\lambda+3),$ we get the coordinates of D as $\Big(\frac{22}{9},-\frac{11}{9},\frac{5}{9}\Big).$
View full question & answer→Question 155 Marks
Find the distance of the point (1, -5, 9) from the plane x - y + z = 5 measured along the line x = y = z.
AnswerThe equation of line parallel to the line x = y = z and passing through the point (1, -5, 9) is
$\frac{\text{x}-1}{1}=\frac{\text{y}+5}{1}=\frac{\text{z}-9}{1}\ ...(\text{i})$
Any point on this line is of the form (k + 1, k - 5, k + 9)
If (k + 1, k - 5, k + 9) be the point of intersection of line (i) and the given plane, then
(k + 1) - (k - 5) + (k + 9) = 5
⇒ k = -10
So, the point of intersection of line (i) and the given plane is (-10 + 1, -10 - 5, -10 + 9) i.e., (-9, -15, -1).
$\therefore$ Required distance = Distance between (1, -5. 9) and (-9, -15, -1)
$=\sqrt{(1+9)^2+(-5+15)^2+(9+1)^2}$
$=\sqrt{3\times10^2}$
$=10\sqrt{3}\text{ units}$
View full question & answer→Question 165 Marks
Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-5\hat{\text{k}})+9=0.$
View full question & answer→Question 175 Marks
By computing the shortest distance determine whether the following pairs of lines intersect or not:
$\vec{\text{r}}=\big(\hat{\text{i}}-\hat{\text{j}}\big)+\lambda\big(2\hat{\text{i}}+\hat{\text{k}}\big)$ and $\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}\big)+\mu\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
AnswerGiven equations of lines are,
$\vec{\text{r}}=\big(\hat{\text{i}}-\hat{\text{j}}\big)+\lambda\big(2\hat{\text{i}}+\hat{\text{k}}\big)$
$\Rightarrow\vec{\text{a}}_1=\big(\hat{\text{i}}-\hat{\text{j}}\big),\vec{\text{b}}_1=\big(2\hat{\text{i}}+\hat{\text{k}}\big)$
and, $\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}\big)+\mu\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
$\Rightarrow\vec{\text{a}}_2=\big(2\hat{\text{i}}-\hat{\text{j}}\big),\vec{\text{b}}_2=\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
We know that, shortest distance between lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\lambda\vec{\text{b}}_2$ is given by
$\text{S.D.}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|\dots(1)$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)=\big(2\hat{\text{i}}-\hat{\text{j}}\big)-\big(\hat{\text{i}}-\hat{\text{j}}\big)$
$=2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{i}}+\hat{\text{j}}$
$=\hat{\text{i}}$
$\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&0&1\\1&1&-1 \end{vmatrix}$
$=\hat{\text{i}}(0-1)-\hat{\text{j}}(-2-1)+\hat{\text{k}}(2-0)$
$=-\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=(\hat{\text{i}})\big(-\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}\big)$
$=(1)(-1)+(0)(3)+(0)(2)$
$=-1+0+0$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=-1$
$\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\sqrt{(-1)^2+(3)^2+(2)^2}$
$=\sqrt{1+9+4}$
$\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\sqrt{14}$
So, shortest distance between the given lines using equation (1) is,
$\text{S.D.}=\Big|\frac{-1}{\sqrt{14}}\Big|$
$=\frac{1}{\sqrt{14}}\text{ units}$
$\text{S.D.}\neq0$
Since, shortest distance between lines is not zero, so lines are not intersecting.
View full question & answer→Question 185 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 195 Marks
Find the coordinates of the foot of perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, -1, 3) and C(2, -3, -1).
AnswerThe cartesian equation of the line joining points B(0, -1, 3) and C(2, -3, -1) is
$\frac{\text{x}-0}{2-0}=\frac{\text{y}-(-1)}{-3-(-1)}=\frac{\text{z}-3}{-1-3}$
Or $\frac{\text{x}}{2}=\frac{\text{y}+1}{-2}=\frac{\text{z}-3}{-4}$
Let L be the foot of the perpendicular drawn from the point A(1, 8, 4) to the line $\frac{\text{x}}{2}=\frac{\text{y}+1}{-2}=\frac{\text{z}-3}{-4}.$
The coordinates of general point on the line $\frac{\text{x}}{2}=\frac{\text{y}+1}{-2}=\frac{\text{z}-3}{-4}$ are given by
$\frac{\text{x}}{2}=\frac{\text{y}+1}{-2}=\frac{\text{z}-3}{-4}=\lambda$
Or $\text{x}=2\lambda,\text{y}=-2\lambda-1,\text{z}=-4\lambda+3$
Let the coordinates of L be $(2\lambda,-2\lambda-1,-4\lambda+3).$ Therefore, the direction ratios of AL are proportional to
$2\lambda-1,-2\lambda-1-8,-4\lambda+3-4$ or $2\lambda-1,-2\lambda-9,-4\lambda-1$
Direction ratios of the given line are proportional to 2, -2, -4.
But, AL is perpendicular to the given line.
$\therefore2\times(2\lambda-1)+(-2)\times(-2\lambda-9)+(-4)\times(-4\lambda-1)=0$
$\Rightarrow4\lambda-2+4\lambda+18+16\lambda+4=0$
$\Rightarrow24\lambda+20=0$
$\Rightarrow\lambda=-\frac{5}{6}$
Puting $\lambda=-\frac{5}{6} $ in $(2\lambda,-2\lambda-1,-4\lambda+3),$ we get
$\Big(2\times\Big(-\frac{5}{6}\Big),-2\times\Big(-\frac{5}{6}\Big)-1,-4\times\Big(-\frac{5}{6}\Big)+3\Big)=\Big(-\frac{5}{3},\frac{2}{3},\frac{19}{3}\Big)$
Thus, the required coordinates of the foot of the perpendicular are $\Big(-\frac{5}{3},\frac{2}{3},\frac{19}{3}\Big).$
View full question & answer→Question 205 Marks
Show that the lines $\vec{\text{r}}=(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})+\lambda(3\hat{\text{i}}-\hat{\text{j}})\ \text{and}\ \vec{\text{r}}=({4\hat{\text{i}}-\hat{\text{k}})+\mu(2\hat{\text{i}}+3\hat{\text{k}}})$intersect. Also find their point of intersection.
AnswerGeneral points on the lines are
$(1+3\lambda)\hat{\text{i}}+(1-\lambda)\hat{\text{j}}-\hat{\text{k}}\ \ \&\ \ ({4+2\mu)\hat{\text{i}}+(3\mu-1)\hat{\text{k}}}$
lines intersect if
$1+3\lambda=4+2\mu\ \ \ \ \dots(1);$ $1-\lambda=0\ \ \ \ \dots(2);$ $3\mu-1=-1\ \ \ \ \dots(3)$ $\text{ for some }\lambda\ \&\ \mu$
From (2) & (3) λ =1, μ = 0
substituting in equation (1)
Since, 1 + 3(1) = 4 + 2 (0) is true $\therefore$ lines interset
Point of intersection is : $4\hat{\text{i}}-\hat{\text{k}}$ or (4, 0, -1)
View full question & answer→Question 215 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, $a_1 = 4, b_1 = 4, c_1 = -5$
$a_2 = 7, b_2 = 1, c_2= 3$
$x_1= 5, y_1 = 7, z_1 = -3$
$x_2 = 8, y_2= 4, z_2 = 5$
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 225 Marks
Find the angle between the following pairs of lines:$\frac{-\text{x}+2}{-2}=\frac{\text{y}-1}{7}=\frac{\text{z}+3}{-3}$ and $\frac{\text{x}+2}{-1}=\frac{2\text{y}-8}{4}=\frac{\text{z}-5}{4}$
Answer$\frac{-\text{x}+2}{-2}=\frac{\text{y}-1}{7}=\frac{\text{z}+3}{-3}$ and $\frac{\text{x}+2}{-1}=\frac{2\text{y}-8}{4}=\frac{\text{z}-5}{4}$
The equation of the given lines can be re-written as
$\frac{\text{x}-2}{2}=\frac{\text{y}-1}{7}=\frac{\text{z}+3}{-3}$ and $\frac{\text{x}+2}{-1}=\frac{\text{y}-4}{2}=\frac{\text{z}-5}{4}$
Let $\vec{\text{b}}_1$ and $\vec{\text{b}}_2$ be vectors parallel to the given line.
Now,
$\vec{\text{b}}_1=2\hat{\text{i}}+7\hat{\text{j}}-3\hat{\text{k}}$
$\vec{\text{b}}_2=-1\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}$
If $\theta$ is the angle between the given lines, then
$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$
$=\frac{\big(2\hat{\text{i}}+7\hat{\text{j}}-3\hat{\text{k}}\big).\big(-1\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}\big)}{\sqrt{2^2+7^2+(-3)^2}\sqrt{(-1)^2+2^2+4^2}}$
$=\frac{-2+14-12}{\sqrt{62}\sqrt{21}}$
$=0$
$\Rightarrow\theta=\frac{\pi}{2}$
View full question & answer→Question 235 Marks
Find the angle between the following pair of lines:
$\frac{\text{-x + 2}}{-2}=\frac{\text{y - 1}}{7}=\frac{\text{z + 3}}{-3}$ and $\frac{\text{x + 2}}{-1}=\frac{\text{2y - 8}}{4}=\frac{\text{z -5 }}{4}$
and check whether the lines are parallel or perpendicular.
Answer$\frac{\text{x}-2}{2}=\frac{\text{y}-1}{7}=\frac{\text{z}+3}{-3}$
$\frac{\text{x}+2}{-1}=\frac{\text{y}-4}{2}=\frac{\text{z}-5}{4}$
The direction ratios of given lines are
2, 7, -3 and -1, 2, 4
Let $\theta$be the angle between these lines, then
$\cos\theta=\frac{2(-1)+7(2)+(-3)4}{\sqrt{4+49+9}\cdot\sqrt{1+4+16}}=0$
$\Rightarrow\theta=\frac{\pi}{2}$
Hence the lines are perpendicular to each other.
View full question & answer→Question 245 Marks
Show that the lines $\frac{\text{x}}{1}=\frac{\text{y}-2}{2}=\frac{\text{z}+3}{3}$ and $\frac{\text{x}-2}{2}=\frac{\text{y}-6}{3}=\frac{\text{z}-3}{4}$ intersect and find their point of intersection.
AnswerThe coordinates of any point on the first line are given by
$\frac{\text{x}}{1}=\frac{\text{y}-2}{2}=\frac{\text{z}+3}{3}=\lambda$
$\Rightarrow\text{x}=\lambda$
$\text{y}=2\lambda+2$
$\text{z}=3\lambda-3$
The coordinales of a general point on the first line are $\big(\lambda,2\lambda+2,3\lambda-3\big)$
Also, the coordinates of any point on the second line are given by
$\frac{\text{x}-2}{2}=\frac{\text{y}-6}{3}=\frac{\text{z}-3}{4}=\mu$
$\Rightarrow\text{x}=2\mu+2$
$\text{y}=3\mu+6$
$\text{z}=4\mu+3$
The coordinates of a general point on the second line are $\big(2\mu+2,3\mu+6,4\mu+3\big)$
It the lines intersect, then they have a common point. so, for some veluse of $\lambda$ and $\mu,$ we must have
$\lambda=2\mu+2,2\lambda+2=3\mu+6,3\lambda-3=4\mu+3$
$\Rightarrow\lambda-2\mu=2\dots(1)$
$2\lambda-3\mu=4\dots(2)$
$3\lambda-4\mu=6\dots(3)$
Solving (1) and (2), we get
$\lambda=2$ and $\mu=0$
Substituting $\lambda=2$ and $\mu=0$ in (3), we get
$\text{LHS}=3\lambda-4\mu$
$=3(2)-4(0)$
$=6$
$=\text{RHS}$
Since $\lambda=2$ and $\mu=0$ satisty the thied equation, the given lines intersect at (2, 6, 3).
View full question & answer→Question 255 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 265 Marks
Find the points on the curve $y = x^3$ at which the slope of the tangent is equal to the y-coordinate of the point.
AnswerEquation of curve is $y = x^3 $
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{3x}^{2}$......................(i)
= y – coordinate of the point
$\Rightarrow 3x^2 = y = x^3$
$\Rightarrow x^2 (x – 3) = 0$
x = 0, x = 3
When x = 0, y = 0, when x = 3, y = 27
The points are (0,0), (3, 27).
View full question & answer→Question 275 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 285 Marks
If O is the origin and the coordinates of A are (a, b, c) Find the direction cosines of OA and the equation of the plane through A at right angles to OA.
AnswerIt is given that O is the origin and the coordinates of A are (a, b, c)
The direction of OA are proportional to
a - 0, b - 0, c - 0 or a, b, c
$\therefore$ Direction cosines of OA are
$\frac{\text{a}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}},\frac{\text{b}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}},\frac{\text{c}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
The normal vector to the required plane is $(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})$
The vector equation of the plane through A(a, b, c) and perpendicular to OA is
$\big[\vec{\text{r}}-(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})\big]\cdot(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})=0$
$\Rightarrow\vec{\text{r}}-(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})=(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})\cdot(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}-(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})=\text{a}^2+\text{b}^2+\text{c}^2$
The cartesian equation of this plane is
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})=\text{a}^2+\text{b}^2+\text{c}^2$
Or $\text{ax}+\text{by}+\text{cz}=\text{a}^2+\text{b}^2+\text{c}^2$
View full question & answer→Question 295 Marks
Find the vector equations of the coordinate planes.
AnswerWe have to find vector equation of coordinate planes.
For xy-plane.
It passes through origin and is perpendicular to z-axis, so
Put $\vec{\text{a}}=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$ and $\vec{\text{n}}=\hat{\text{k}}$ in the vector equation of plane passing through point $\vec{\text{a}}$ and perpendicular to vector $\vec{\text{n}}$
$(\vec{\text{r}}-\vec{\text{n}})\vec{\text{n}}=0$
$(\vec{\text{r}}-0\hat{\text{i}}-0\hat{\text{j}}-0\hat{\text{k}})\hat{\text{k}}=0$
$\vec{\text{r}}\cdot\vec{\text{k}}=0\ ...(\text{i})$
For xz-plane,
It passes throught origin and perpendicular to y-axis, so
$\vec{\text{a}}=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$ and $\vec{\text{n}}=\hat{\text{j}}$
Equation of xz-plane is given by
$(\vec{\text{r}}-\vec{\text{a}})\vec{\text{n}}=0$
$(\vec{\text{r}}-0\hat{\text{i}}-0\hat{\text{j}}-0\hat{\text{k}})\hat{\text{j}}=0$
$\vec{\text{r}}\cdot\hat{\text{j}}=0$
For yz-plane,
It passes throught origin and is perpendicular to x-axis, so
$\vec{\text{a}}=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}},\vec{ \text{n}}=\hat{\text{i}}$
$(\vec{\text{r}}-\vec{\text{a}})\vec{\text{n}}=0$
$(\vec{\text{r}}-0\hat{\text{i}}-0\hat{\text{j}}-0\hat{\text{k}})\hat{\text{i}}=0$
$\vec{\text{r}}\cdot\hat{\text{i}}=0$
Hence, equation of xy, yz, zx-plane are given by
$\vec{\text{r}}\cdot\vec{\text{k}}=0$
$\vec{\text{r}}\cdot\hat{\text{i}}=0$
$\vec{\text{r}}\cdot\hat{\text{j}}=0$
View full question & answer→Question 305 Marks
Find the vector equation of the plane passing through the points (1, 1, 1), (1, -1, 1) and (-7, -3, -5)
Answer
Let A(1, 1, 1), B(1, -1, 1) and C(-7, -3, -5) be the coordinates.
The required plane passes through the point A(1, 1, 1)
Whose position vector is $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ and is normal to the vector $\vec{\text{n}}$ given by
$\vec{\text{n}}=\overrightarrow{\text{AB}}\times\overrightarrow{\text{AC}}$
Clearly, $\overrightarrow{\text{AB}}=\overrightarrow{\text{OB}}-\overrightarrow{\text{OA}}$
$=(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})-(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})$
$=0\hat{\text{i}}-2\hat{\text{j}}+0\hat{\text{k}}$
$\overrightarrow{\text{AC}}=\overrightarrow{\text{OC}}-\overrightarrow{\text{OA}}$
$=(-7\hat{\text{i}}-3\hat{\text{j}}-5\hat{\text{k}})-(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})$
$=-8\hat{\text{i}}-4\hat{\text{j}}-6\hat{\text{k}}$
$\vec{\text{n}}=\overrightarrow{\text{AB}}\times\overrightarrow{\text{AC}}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\0&-2&0\\-8&-4&-6\end{vmatrix}$
$=12\hat{\text{i}}+0\hat{\text{j}}-16\hat{\text{k}}$
The vector equation of the required plane is,
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
$\Rightarrow\vec{\text{r}}\cdot(12\hat{\text{i}}+0\hat{\text{j}}-16\hat{\text{k}})=(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})(12\hat{\text{i}}+0\hat{\text{j}}-16\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot\Big[4(3\hat{\text{i}}-4\hat{\text{k}})\Big]=12+0-16$
$\Rightarrow\vec{\text{r}}\cdot\Big[4(3\hat{\text{i}}-4\hat{\text{k}})\Big]=-4$
$\Rightarrow\vec{\text{r}}\cdot(3\hat{\text{i}}-4\hat{\text{k}})=-1$
$\Rightarrow\vec{\text{r}}\cdot(3\hat{\text{i}}-4\hat{\text{k}})+1=0$ View full question & answer→Question 315 Marks
Find the angle between the pairs of lines with direction ratios proportional toa, b, c and b - c, c - a, a - b.
Answera, b, c and b - c, c - a, a - b are direction ratios these are the vectors with above direction ratios $\hat{\text{x}}=\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}},\hat{\text{y}}=(\text{b}-\text{c})\hat{\text{i}}+(\text{c}-\text{a})\hat{\text{j}}+(\text{a}-\text{b})\hat{\text{k}}$ are the vectors parallel to two given lines $\therefore$ angle between the lines with above direction ratios are $\hat{\text{x}}$ and $\hat{\text{y}}\rightarrow\cos\theta=\frac{\hat{\text{x}}.\hat{\text{y}}}{|\hat{\text{x}}||\hat{\text{y}}|}$ $\cos\theta=\frac{\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).\big((\text{b}-\text{c})\hat{\text{i}}+(\text{c}-\text{a})\hat{\text{j}}+(\text{a}-\text{b})\hat{\text{k}}\big)}{\big|\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big)\big|\big|\big((\text{b}-\text{c})\hat{\text{i}}+(\text{c}-\text{a})\hat{\text{j}}+(\text{a}-\text{b})\hat{\text{k}}\big)\big|}$ $=\frac{\text{a}(\text{b}-\text{c})+\text{b}(\text{c}-\text{a})+\text{c}(\text{a}-\text{b})}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}\sqrt{(\text{b}-\text{c})^2+(\text{c}-\text{a})^2+(\text{a}-\text{b})^2}}$ $=\frac{\text{ab}-\text{ac}+\text{bc}-\text{ba}+\text{ca}-\text{cb}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}\sqrt{(\text{b}-\text{c})^2+(\text{c}-\text{a})^2+(\text{a}-\text{b})^2}}=0$$\cos\theta=0\rightarrow\theta=90^\circ$
View full question & answer→Question 325 Marks
Find the distance between the point (-1, -5, -10) and the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 5.$
AnswerAny point on the line $\frac{\text{x} - 2}{3} = \frac{\text{y} + 1}{4} = \frac{\text{z} - 2}{12} = \text{is} (3\lambda + 2, 4\lambda - 1, 12\lambda + 2)$
If this is the point of intersection with plane $\text{x - y + z = 5}$
$\text{then 3} \lambda + 2 - 4\lambda + 1 + 12\lambda + 2 - 5 = 0 \Rightarrow \lambda = 0$
$\therefore$ Point of intersection is (2, -1, 2)
Required distance = $\sqrt{(2 + 1)^{2} +(-1 + 5)^{2} + (2 + 10)^{2} } = 13$
View full question & answer→Question 335 Marks
Find the vector equation of the following planes in non-parametric form.
$\vec{\text{r}}=(\lambda-2\mu)\hat{\text{i}}+(3-\mu)\hat{\text{j}}+(2\lambda+\mu)\hat{\text{k}}$
AnswerThe given equation of the plane is,
$\vec{\text{r}}=(\lambda-2\mu)\hat{\text{i}}+(3-\mu)\hat{\text{j}}+(2\lambda+\mu)\hat{\text{k}}$
$\Rightarrow\vec{\text{r}}=(0\hat{\text{i}}+3\hat{\text{j}}+0\hat{\text{k}})+\lambda(\hat{\text{i}}+0\hat{\text{j}}+2\hat{\text{k}})+\mu(-2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})$
We know that equation $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}+\mu\vec{\text{c}} $ represents a plane passing through a point whose position vector is $\vec{\text{a}}$ and parallel to t.
Here, $\vec{\text{a}}=0\hat{\text{i}}+3\hat{\text{j}}+0\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+0\hat{\text{j}}+2\hat{\text{k}},\hat{\text{c}}=-2\hat{\text{i}}-\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&0&2\\-2&-1&1\end{vmatrix}$
$=2\hat{\text{i}}-5\hat{\text{j}}-\hat{\text{k}}$
The vector equation of the plane in scalar product form is,
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
$\Rightarrow\vec{\text{r}}\cdot(2\hat{\text{i}}-5\hat{\text{j}}-\hat{\text{k}})=(0\hat{\text{i}}+3\hat{\text{j}}+0\hat{\text{k}})(2\hat{\text{i}}-5\hat{\text{j}}-\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot(2\hat{\text{i}}-5\hat{\text{j}}-\hat{\text{k}})=0-15+0$
$\Rightarrow\vec{\text{r}}\cdot(2\hat{\text{i}}-5\hat{\text{j}}-\hat{\text{k}})+15=0$
View full question & answer→Question 345 Marks
Find the equation of the plane through the points $(2, 2, -1)$ and $(3, 4, 2)$ and parallel to the line whose direction ratios are $7, 0, 6.$
AnswerThe equation of the plane through $(2, 2, -1)$ is
$a(x - 2) + b(y - 2) + c(z + 1) = 0 ....(i)$
This plane passes through (3, 4, 2). so,
$a(3 - 2) + b(4 - 2) + c(2 + 1) = 0$
$\Rightarrow a + 2b + 3c = 0 ....(ii)$
Again plane (i) is parallel to the line whose direction ratios are 7, 0, 6.
It means that the normal of plane (i) is perpendicular to the line whose direction ratios are $7, 0, 6$
$\Rightarrow 7a + 0b + 6c = 0$ (Because $a_1a_2 + b_1b_2 + c_1c_2= 0)$
Solving (i), (ii) and (iii), we get
$\begin{vmatrix}\text{x}-2&\text{y}-2&\text{z}+1\\1&2&3\\7&0&6\end{vmatrix}=0$
$\Rightarrow 12(x - 2) + 15(y - 2) - 14(z + 1) = 0$
$\Rightarrow 12x + 15y - 14z - 68 = 0$
View full question & answer→Question 355 Marks
If $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\ 2\hat{\text{i}}+5\hat{\text{J}},\ 3\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$ and $\hat{\text{i}}-6\hat{\text{j}}-\hat{\text{k}}$ respectively are the position vectors of points A, B, C and D, then find the angle between the straight lines AB and CD. Find whether $\vec{\text{AB}}$ and $\vec{\text{CD}}$ are collinear or not.
AnswerGiven:
The position vector of A is $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}.$
The position vector of B is $2\hat{\text{i}}+5\hat{\text{j}}.$
Therefore, $\vec{\text{AB}}=(2-1)\hat{\text{i}}+(5-1)\hat{\text{j}}+(0-1)\hat{\text{k}}=\hat{\text{i}}+\hat4{\text{j}}-\hat{\text{k}}$
The position vector of C is $3\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$ and
The postion vector of D is $\hat{\text{i}}-6\hat{\text{j}}-\hat{\text{k}}.$
Therefore, $\vec{\text{CD}}=(1-3)\hat{\text{i}}+(-6-2)\hat{\text{j}}\\+(-1+3)\hat{\text{k}}=-2\hat{\text{i}}-\hat8{\text{j}}+2\hat{\text{k}}$
$\cos\theta=\frac{\vec{\text{AB}}.\vec{\text{CD}}}{|\vec{\text{AB}}||\vec{\text{CD}}|}$
$\Rightarrow\cos\theta=\frac{-2-32-2}{\sqrt{18}\sqrt{72}}=-1$
$\Rightarrow\theta=180^\circ$
Since, angle between Line AB and CD is 180°, therefore $\vec{\text{AB}}$ and $\vec{\text{CD}}$ are collinear.
View full question & answer→Question 365 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 375 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 385 Marks
Find the angle between the following pairs of lines:$\frac{5-\text{x}}{-2}=\frac{\text{y}+3}{1}=\frac{1-\text{z}}{3}$ and $\frac{\text{x}}{3}=\frac{1-\text{y}}{-2}=\frac{\text{z}+5}{-1}$
Answer$\frac{5-\text{x}}{-2}=\frac{\text{y}+3}{1}=\frac{1-\text{z}}{3}$ and $\frac{\text{x}}{3}=\frac{1-\text{y}}{-2}=\frac{\text{z}+5}{-1}$The equations of the given lines can be re-written as
$\frac{\text{x}-5}{2}=\frac{\text{y}+3}{1}=\frac{\text{z}-1}{-3}$ and $\frac{\text{x}}{3}=\frac{\text{y}-1}{2}=\frac{\text{z}+5}{-1}$
Let $\vec{\text{b}}_1$ and $\vec{\text{b}}_2$ be vectors parallel to the given lines.
Now,
$\vec{\text{b}}_1=2\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}$
$\vec{\text{b}}_2=3\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
If $\theta$ is the angle between the given lines, then
$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$
$=\frac{\big(2\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}\big).\big(3\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}\big)}{\sqrt{2^2+1^2+(-3)^2}\sqrt{3^2+2^2+(-1)^2}}$
$=\frac{6+2+3}{\sqrt{14}\sqrt{14}}$
$=\frac{11}{14}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{11}{14}\Big)$
View full question & answer→Question 395 Marks
$\text{If }\overrightarrow{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{7k}};\text{ and }\overrightarrow{b}=\hat{\text{5i}}-\hat{\text{j}}+\lambda\hat{\text{k}},$ then find the value of $\lambda,$ so that $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$ are perpendicular vectors.
AnswerHere $\text{If }\overrightarrow{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{7k}};\text{ }\overrightarrow{b}=\hat{\text{5i}}-\hat{\text{j}}+\lambda\hat{\text{k}},$$\therefore \overrightarrow{a}+\overrightarrow{b}=6\hat{\text{i}}-2\hat{\text{j}}+\text{(7}+\lambda)\hat{\text{k}};\text{ } \overrightarrow{a}-\overrightarrow{b}=-4\hat{\text{i}}+\text{(7}-\lambda)\hat{\text{k}}$
$\therefore\text{ }\overrightarrow{(a}+\overrightarrow{b)}\text{ is perpendicular to }\overrightarrow{(a}-\overrightarrow{b)}$
$\Rightarrow \text{ }\overrightarrow{(a}+\overrightarrow{b)}.\overrightarrow{(a}-\overrightarrow{b)}=0\text{ }\Rightarrow \text{ -24 + (7}+\lambda).\text{(7} - \lambda)=0$
$\Rightarrow$ –24 + 49 – $\lambda^2 $ = 0 $\Rightarrow$ $\lambda^2 $ = 25
$\Rightarrow$ $\lambda $ = ± 5.
View full question & answer→Question 405 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 415 Marks
Find the shortest distance between the lines whose vector equations are:
$\vec{\text{r}}=\Big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\Big)+\lambda\Big(\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}\Big)$
$\text{and}\ \vec{\text{r}}=4\hat{\text{i}}+5\hat{\text{j}}+6\hat{\text{k}}+\mu\Big(2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}\Big)$
AnswerEquation of the first line is $\vec{\text{r}}=\Big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\Big)+\lambda\Big(\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}\Big)$
Comparing this equation with $\vec{\text{r}}=\vec{\text{a}_1}+\lambda\vec{\text{b}_1},$
$\vec{\text{a}_1}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\ \text{and}\ \vec{\text{b}_1}=\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}$
Again equation of second line $\vec{\text{r}}=\Big(4\hat{\text{i}}+5\hat{\text{j}}+6\hat{\text{k}}\Big)+\mu\Big(2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}\Big)$
Comparing this equation with $\vec{\text{r}}=\vec{\text{a}_2}+\mu\vec{\text{b}_2},$
$\vec{\text{a}_2}=4\hat{\text{i}}+5\hat{\text{j}}+6\hat{\text{k}}\ \text{and}\ \vec{\text{b}_2}=2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$
Now shortest distance $(\text{d})=\frac{\Big|\Big(\vec{\text{a}_2}-\vec{\text{a}_1}\Big).\Big(\vec{\text{b}_1}\times\vec{\text{b}_2}\Big)\Big|}{\Big|\vec{\text{b}_1}\times\vec{\text{b}_2}\Big|}\ \ \ \ ...(\text{i})$
Here $\vec{\text{a}_2}-\vec{\text{a}_1}=\Big(4\hat{\text{i}}+5\hat{\text{j}}+6\hat{\text{k}}\Big)-\Big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\Big)$
$=3\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{b}_1}\times\vec{\text{b}_2}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&-3&2\\2&3&1\end{vmatrix}$
$=(-3-6)\hat{\text{i}}-(1-4)\hat{\text{j}}+(3+6)\hat{\text{k}}=-9\hat{\text{i}}+3\hat{\text{j}}+9\hat{\text{k}}$
$\Big|\vec{\text{b}}_1\times\vec{\text{b}}_2\Big|=\sqrt{(-9)^2+(3)^2+(9)^2}=\sqrt{171}=3\sqrt{19}$
$\Big(\vec{\text{a}_2}-\vec{\text{a}_1}\Big).\Big(\vec{\text{b}_1}\times\vec{\text{b}_2}\Big)=3\times(-9)+(3\times3)+(3\times9)$
$=-27+9+27=9$
Putting these values in eq.(i),
Shortest distance $(\text{d})=\frac{|9|}{3\sqrt{19}}=\frac{9}{3\sqrt{19}}=\frac{3}{\sqrt{19}}.$
View full question & answer→Question 425 Marks
Find the vector equation of the plane through the line of intersection of the planes $x + y + z = 1$ and $2x + 3y + 4z = 5$ which is perpendicular to the plane $x - y + z = 0$.
AnswerThe equation of the plane passing through the line of intersection of the given planes is,
$\text{x}+\text{y}+\text{z}-1+\lambda(2\text{x}+3\text{y}+4\text{z}-5)=0$
$(1+2\lambda)\text{x}+(1+3\lambda)\text{y}+(1+4\lambda)\text{z}-1-5\lambda=0\ ...(\text{i})$
This plane is perpendicular to x - y + z = 0. So,
$1+2\lambda-1(1+3\lambda)+1+4\lambda=0$ (Because $a_1a_2 + b_1b_2 + c_1c_2= 0$)
$\Rightarrow1+2\lambda-1-3\lambda+1+4\lambda=0$
$\Rightarrow3\lambda+1=0$
$\Rightarrow\lambda=\frac{-1}{3}$
Substituting this in (i), we get
$\Big(1+2\Big(\frac{-1}{3}\Big)\Big)\text{x}+\Big(1+3\Big(\frac{-1}{3}\Big)\Big)\text{y}+\Big(1+4\Big(\frac{-1}{3}\Big)\text{z}-1-5\Big(\frac{-1}{3}\Big)\Big)=0$
$\Rightarrow\text{x}-\text{z}+2=0$
View full question & answer→Question 435 Marks
$\text{Let } \vec{\text a} = \hat{\text{i}} + \hat{\text{j}} + \hat{\text{k}}, \vec{\text{b}} = \hat{\text{i}} \text{ and } \vec{\text{c}} = \text{c}_{1} \hat{\text{i}} + \text{c}_{2} \hat{\text{j}} + \text{c}_{3} \hat{\text{k}}, \text{then}$
- Let $c_1 = 1$ and $c_2 = 2$, find $c_3$ which makes $\vec{\text{a}}, \vec{\text{b}} \text{ and }\vec{\text{c}} \text{ coplanar.}$
- If $c_2 = –1$ and $c_3 = 1$, show that no value of $c_1$ can make $\vec{\text{a}}, \vec{\text{b}} \text{ and } \vec{\text{c}} \text{ coplanar}.$
Answer$[\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ \text{c}_{1} & \text{c}_{2} & \text{c}_{3} \end{vmatrix} = \text{c}_{2} - \text{c}_{3}$
- $\text{c}_{1} = 1, \text{ c}_{2} = 2$
$[\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = 2 - \text{c}_{3}$
$\therefore \vec{\text{a}}, \text{ }\vec{\text{b}}, \text{ } \vec{\text{c}} \text{ are coplanar} [\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = 0 \Rightarrow \text{c}_{3} = 2$
- $\text{c}_{2} = -1, \text{c}_{3} = 1$
$[\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = \text{c}_{2} - \text{c}_{3} = -2 \neq 0$
$\Rightarrow \text{No value of }\text{c}_{1} \text{can make }\vec{\text{a}}, \text{ }\vec{\text{b}}, \text{ } \vec{\text{c}} \text{ coplanar}.$ View full question & answer→Question 445 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 455 Marks
Find the angle between the following pairs of lines:
$\frac{\text{x}+4}{3}=\frac{\text{y}-1}{5}=\frac{\text{z}+3}{4}$ and $\frac{\text{x}+1}{1}=\frac{\text{y}-4}{1}=\frac{\text{z}-5}{2}$
Answer$\frac{\text{x}+4}{3}=\frac{\text{y}-1}{5}=\frac{\text{z}+3}{4}$ and $\frac{\text{x}+1}{1}=\frac{\text{y}-4}{1}=\frac{\text{z}-5}{2}$ Let $\vec{\text{b}}_1$ and $\vec{\text{b}}_2$ be vectors parallel to the given lines. $\vec{\text{b}}_1=3\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}$ $\vec{\text{b}}_2=\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$ If $\theta$ is the angle between the given lines, then $\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$ $=\frac{\big(3\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}\big).\big(\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)}{\sqrt{3^2+5^2+4^2}\sqrt{1^2+1^2+2^2}}$ $=\frac{3+5+8}{10\sqrt{3}}$ $=\frac{8}{5\sqrt{3}}$$\Rightarrow\theta=\cos^{-1}\Big(\frac{8}{5\sqrt{3}}\Big )$
View full question & answer→Question 465 Marks
Find the vector and cartesian equation of the line through the point (5, 2, -4) and which is parallel to the vector $3\hat{\text{i}}+2\hat{\text{j}}-8\hat{\text{k}}.$
AnswerWe know that the vector equation of a line passing through a point with position vector $\vec{\text{a}}$ and parallel to vector $\vec{\text{b}}$ is $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}.$
Here,
$\vec{\text{a}}=5\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}$
$\vec{\text{b}}=3\hat{\text{i}}+2\hat{\text{j}}-8\hat{\text{k}}$
Vector equation of the required line is given by
$\vec{\text{r}}=\big(5\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}\big)+\lambda\big(3\hat{\text{i}}+2\hat{\text{j}}-8\hat{\text{k}}\big)\dots(1)$
Here, $\lambda$ is a parameter.
reducing (1) to cartesian from, we get
$\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}=\big(5\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}\big)+\lambda\big(3\hat{\text{i}}+2\hat{\text{j}}-8\hat{\text{k}}\big)$ $\big[\text{putting }\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}\text{ in}(1)\big]$
$\Rightarrow\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}=(5+3\lambda)\hat{\text{i}}+(2+2\lambda)\hat{\text{j}}+(-4-8\lambda)\hat{\text{k}}$
Comparing the cofficients of $\hat{\text{i}},\hat{\text{j}}$ and $\hat{\text{k}},$ we get
$\text{x}=5+3\lambda,\text{y}=2+2\lambda,\text{z}=-4-8\lambda$
$\Rightarrow\frac{\text{x}-5}{3}=\lambda,\frac{\text{y}-2}{2}=\lambda,\frac{\text{z+4}}{-8}=\lambda$
$\Rightarrow\frac{\text{x}-5}{3}=\frac{\text{y}-2}{2}=\frac{\text{z}+4}{-8}=\lambda$
Hence, the cartesian form of (1) is
$\frac{\text{x}-5}{3}=\frac{\text{y}-2}{2}=\frac{\text{z}+4}{-8}$
View full question & answer→Question 475 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→Question 485 Marks
In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.
$4x + 8y + z - 8 = 0$ and $y + z - 4 = 0$
AnswerThe direction ratios of normal to the plane, $L_1: a_1x + b_1y + c_1z = 0,$
are $a_1, b_1, c_1$ and $L_2: a_2x + b_2y + c_2z = 0$ are $a_2, b_2, c_2$
$\text{L}_1||\text{L}_2,\ \text{if }\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\text{L}_1\perp\text{L}_2,\ \text{if}\ \text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0$
The angle between L_1 and L_2 is given by,
$\text{Q}=\cos^{-1}\Bigg|\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}.\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}_2^2}}$
The equations of the given planes are $4x + 8y + z - 8 = 0$ and $y + z - 4 = 0$
Here, $a_1 = 4, b_1 = 8, c_1 = 1$ and $a_2 = 0, b_2 = 1, c_2 = 1$
$a_1a_2 +b_1b_2 + c_1c_2 = 4\times 0 + 8 \times 1 + 1$
$=9\neq0$
$\frac{\text{a}_1}{\text{a}_2}=\frac{4}{0},\ \frac{\text{b}_1}{\text{b}_2}=\frac{8}{1}=8\text{ and } \frac{\text{c}_1}{\text{c}_2}=\frac{1}{1}=1$
$\therefore\ \ \frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
Therefore, the given lines are not parallel to each other.
The angle between the planes is given by,
$\text{Q}=\cos^{-1}\Bigg|\frac{4\times0+8\times1+1\times1}{\sqrt{4^2+8^2+1^2}\times\sqrt{0^2+1^2+1^2}}\Bigg|=\cos^{-1}\Bigg|\frac{9}{9\sqrt{2}}\Bigg|$
$=\cos^{-1}\Big(\frac{1}{\sqrt{2}}\Big)=45^{\circ}.$
View full question & answer→Question 495 Marks
Show that the lines $\vec{\text{r}}=(2\hat{\text{i}}-3\hat{\text{k}})+\lambda(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})$ and $\vec{\text{r}}=(2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}})+\mu(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})$ are coplanar. Also, find the equation of the plane containing them.
AnswerWe know that the lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2$ are coplanar if
$\vec{\text{a}}_1\cdot(\vec{\text{b}}_1\times\vec{\text{b}}_2)=\vec{\text{a}}_2\cdot(\vec{\text{b}}_1\times\vec{\text{b}}_2)$ and the equation of the plane containing them is
$\vec{\text{r}}\cdot(\vec{\text{b}}_1\times\vec{\text{b}}_2)=\vec{\text{a}}_1\cdot(\vec{\text{b}}_1\times\vec{\text{b}}_2)$
Here,
$\vec{\text{a}}_1=0\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}},\vec{\text{b}}_1=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{a}}_2=2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}},\vec{\text{b}}_2=2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&2&3\\2&3&4\end{vmatrix}$
$=-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
$\vec{\text{a}}_1\cdot(\vec{\text{b}}_1\times\vec{\text{b}}_2)=(0\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})\cdot(-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})$
$=0+4+3=7$
$\vec{\text{a}}_2\cdot(\vec{\text{b}}_1\times\vec{\text{b}}_2)=(2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}})\cdot(-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})$
$=-2+12-3=7$
Clearly, $\vec{\text{a}}_1\cdot(\vec{\text{b}}_1\times\vec{\text{b}}_2)=\vec{\text{a}}_2\cdot(\vec{\text{b}}_1\times\vec{\text{b}}_2)$
Hence, the given lines are coplanar.
The equation of the plane containing the given lines is
$\vec{\text{r}}\cdot(\vec{\text{b}}_1\times\vec{\text{b}}_2)=\vec{\text{a}}_1\cdot(\vec{\text{b}}_1\times\vec{\text{b}}_2)$
$\Rightarrow\vec{\text{r}}\cdot(-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})=(0\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})\cdot(-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot(-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})=7$
$\Rightarrow\vec{\text{r}}\cdot(-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})+7=0$
View full question & answer→Question 505 Marks
Show that the vectors $\overrightarrow{\text{a}},\overrightarrow{\text{b}} \text{and}{\overrightarrow{\text{c}}}$ are coplanar if $\overrightarrow{\text{a}} +\overrightarrow{\text{b}}, \overrightarrow{\text{b}}+\overrightarrow{\text{c}}\text{and} \overrightarrow{\text{c}} + \overrightarrow{\text{a}}$ are coplanar.
AnswerGiven that $\overrightarrow{\text{a}} +\overrightarrow{\text{b}}, \overrightarrow{\text{b}} + \overrightarrow{\text{c}}, \overrightarrow{\text{c}} + \overrightarrow{\text{a}}$ are coplanar
$\therefore[\overrightarrow{\text{a}} +\overrightarrow{\text{b}}, \overrightarrow{\text{b}} + \overrightarrow{\text{c}}, \overrightarrow{\text{c}} + \overrightarrow{\text{a}}] = 0$
$\text{i.e} \overrightarrow{\text{(a}} +\overrightarrow{\text{b})}, \big\{\overrightarrow{\text{(b}} + \overrightarrow{\text{c})} \times\overrightarrow{\text{(c}} + \overrightarrow{\text{a})}\big\} = 0$
$\overrightarrow{\text{(a}} +\overrightarrow{\text{b})}. \big\{\overrightarrow{\text{(b}} \times \overrightarrow{\text{c}}+ \overrightarrow{\text{b}} \times\overrightarrow{\text{a}}\times\overrightarrow{\text{c}} \times \overrightarrow{\text{a})}\big\} = 0$
$\Rightarrow\overrightarrow{\text{a}}.(\overrightarrow{\text{b}}\times\overrightarrow{\text{c})} + \overrightarrow{\text{a}}.(\overrightarrow{\text{b}}\times\overrightarrow{\text{a})} + \overrightarrow{\text{a}}(\overrightarrow{\text{c}}\times\overrightarrow{\text{a})}+\overrightarrow{\text{b}}. (\overrightarrow{\text{b}}\times\overrightarrow{\text{c})} + \overrightarrow{\text{b}}.(\overrightarrow{\text{b}}\times\overrightarrow{\text{a})} + \overrightarrow{\text{b}}.(\overrightarrow{\text{c}}\times\overrightarrow{\text{a})} = 0$
$\Rightarrow 2 [\overrightarrow{\text{a}}, \overrightarrow{\text{b}}, \overrightarrow{\text{c}}] = \text{0 or} [\overrightarrow{\text{a}},\overrightarrow{\text{b}}, \overrightarrow{\text{c] }} = 0$
$\Rightarrow\overrightarrow{\text{a}}, \overrightarrow{\text{b}}, \overrightarrow{\text{c}}\text{are coplanar}.$
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