Exercise 6.10 - Chapter 6 Applications of Vector Algebra 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $6.10$
Choose the correct or the most suitable answer from the given four alternatives:
Question 1.
If $\vec{a}$ and $\vec{b}$ are parallel vector, then $[\vec{a}, \vec{c}, \vec{b}]$ is equal to
(a) 2
(b) $-1$
(c) 1
(d) 0
Solution:
(d) 0
Hint:
$\vec{a}$ and $\vec{b}$ are parallel vectors, so $\vec{a} \times \vec{b}=0$
then $[\vec{a} \vec{c} \vec{b}]=-[\vec{a} \vec{b} \vec{c}]=-(\vec{a} \times \vec{b}) \cdot \vec{c}=0$
Question 2.
If a vector $\vec{\alpha}$ lies in the plane of $\vec{\beta}$ and $\vec{\gamma}$, then .
(a) $[\vec{\alpha}, \vec{\beta}, \vec{\gamma}]=1$
(b) $[\vec{\alpha}, \vec{\beta}, \vec{\gamma}]=-1$
(c) $[\vec{\alpha}, \vec{\beta}, \vec{\gamma}]=0$
(d) $[\vec{\alpha}, \vec{\beta}, \vec{\gamma}]=2$
Solution:
(c) $[\vec{\alpha}, \vec{\beta}, \vec{\gamma}]=0$
Hint:
Since $[\vec{\alpha}, \vec{\beta}, \vec{\gamma}]$ are lie in the same plane
so $[\vec{\alpha}, \vec{\beta}, \vec{\gamma}]=0$
Question 3.
If $\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}=0$, then the value of $|[\vec{a}, \vec{b}, \vec{c}]|$ is
(a) $|\vec{a}||\vec{b}||\vec{c}|$
(b) $\frac{1}{3}|\vec{a}||\vec{b}||\vec{c}|$
(c) 1
(d) $-1$
Solution:
(a) $|\vec{a}||\vec{b}||\vec{c}|$

Hint:
$
\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}=0
$
$\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular vectors
$|[\vec{a}, \vec{b}, \vec{c}]|=|\vec{a}||\vec{b}||\vec{c}|$ is volume of cuboid where $\vec{a}, \vec{b}, \vec{c}$ are coterminus edges.
Question 4.
If $\vec{a}, \vec{b}, \vec{c}$ are three unit vectors such that $\vec{a}$ is perpendicular to $\vec{b}$, and is parallel to $\vec{c}$ then $\vec{a} \times(\vec{b} \times \vec{c})$ is equal to .
(a) $\vec{a}$
(b) $\vec{b}$
(c) $\vec{c}$
(d) $\overrightarrow{0}$
Solution:
(b) $\vec{b}$
Hint:
$\vec{a}$ is perpendicular to $\vec{b} \Rightarrow \vec{a} \cdot \vec{b}=0$
$\vec{a}$ is parallel to $\vec{c} \Rightarrow \vec{a} \times \vec{c}=0$
then $\quad \vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}$
$
\begin{aligned}
& =(1) \vec{b}-0 \vec{c} \\
& =\vec{b}
\end{aligned}
$
Question 5.
If $[\vec{a}, \vec{b}, \vec{c}]=1$, then the value of $\frac{\vec{a} \cdot(\vec{b} \times \vec{c})}{(\vec{c} \times \vec{a}) \cdot \vec{b}}+\frac{\vec{b} \cdot(\vec{c} \times \vec{a})}{(\vec{a} \times \vec{b}) \cdot \vec{c}}+\frac{\vec{c} \cdot(\vec{a} \times \vec{b})}{(\vec{c} \times \vec{b}) \cdot \vec{a}}$ is
(a) 1
(b) $-1$
(c) 2
(d) 3
Solution:
(a) 1

Hint:
$
\begin{gathered}
{[\vec{a}, \vec{b}, \vec{c}]=1(\text { Given })} \\
\frac{\vec{a} \cdot(\vec{b} \times \vec{c})}{(\vec{c} \times \vec{a}) \cdot \vec{b}}+\frac{\vec{b} \cdot(\vec{c} \times \vec{a})}{(\vec{a} \times \vec{b}) \cdot \vec{c}}+\frac{\vec{c} \cdot(\vec{a} \times \vec{b})}{(\vec{c} \times \vec{b}) \cdot \vec{a}}=\frac{[\vec{a} \vec{b} \vec{c}]}{[\vec{a} \vec{b} \vec{c}]}+\frac{[\vec{a} \vec{b} \vec{c}]}{[\vec{a} \vec{b} \vec{c}]}+\frac{[\vec{a} \vec{b} \vec{c}]}{-[\vec{a} \vec{b} \vec{c}]}=1+1-1=1
\end{gathered}
$
Question 6.
The volume of the parallelepiped with its edges represented by the vectors $\hat{i}+\hat{j}, i+2 \hat{j}, \hat{i}+\hat{j}+\pi \hat{k}$ is

(a) $\frac{\pi}{2}$
(b) $\frac{\pi}{3}$
(c) $\pi$
(d) $\frac{\pi}{4}$
Solution:
(c) $\pi$
Hint:
$
\begin{aligned}
& \text { Volume }=[\vec{a} \vec{b} \vec{c}]=\left|\begin{array}{ccc}
1 & 1 & 0 \\
1 & 2 & 0 \\
1 & 1 & \pi
\end{array}\right| \\
& =\pi(2-1)=\pi \text { cubic units }
\end{aligned}
$

Question 7.
If $\vec{a}$ and $\vec{b}$ are unit vectors such that $[\vec{a}, \vec{b}, \vec{a} \times \vec{b}]=\frac{\pi}{4}$ then the angle between $\vec{a}$ and $\vec{b}$ is
(a) $\frac{\pi}{6}$
(b) $\frac{\pi}{4}$
(c) $\frac{\pi}{3}$
(d) $\frac{\pi}{2}$
Solution:
(a) $\frac{\pi}{6}$
Hint:

Question 8.
If $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}+\hat{j}, \vec{c}=\hat{i}$ and $(\vec{a} \times \vec{b}) \times \vec{c}=\lambda \vec{a}+\mu \vec{b}$, then the value of $\lambda+\mu$ is
(a) 0
(b) 1
(c) 6
(d) 3
Solution:
(a) 0
Hint:
But $\quad \begin{aligned}(\vec{a} \times \vec{b}) \times \vec{c} & =(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{b} \cdot \vec{c}) \vec{a}=\vec{b}-\vec{a} \\ (\vec{a} \times \vec{b}) \times \vec{c} & =\lambda \vec{a}+\mu \vec{b} \\ \vec{b}-\vec{a} & =\lambda \vec{a}+\mu \vec{b} \\ -\vec{k} & =\lambda \vec{i}+\lambda \vec{j}+\lambda \vec{k}+\mu \vec{i}+\mu \vec{j}\end{aligned}$
Equate corresponding coefficients on both sides
$\lambda+\mu=0$ and $\lambda=-1$ this gives $\mu=1$
$\therefore$ Then the value of $\lambda+\mu=0$.

Question 9.
If $\vec{a}, \vec{b}, \vec{c}$ are three non-coplanar vectors such that $\vec{a} \times(\vec{b} \times \vec{c})=\frac{\vec{b}+\vec{c}}{\sqrt{2}}$, then the angle between $\vec{a}$ and $\vec{b}$ is
(a) 81
(b) 9
(c) 27
(d) 18
Solution:
(a) 81
Hint:
$
\left\{\left[\begin{array}{lll}
\vec{a} \times \vec{b} & \vec{b} \times \vec{c} & \vec{c} \times \vec{a}
\end{array}\right]\right\}^2=\left\{\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]^2\right\}^2=\left\{(3)^2\right\}^2=\{9\}^2=81
$
Question 10.
If $\vec{a}, \vec{b}, \vec{c}$ are three non-coplanar vector such that $\vec{a} \times(\vec{b} \times \vec{c})=\frac{\vec{b}+\vec{c}}{\sqrt{2}}$, then the angle between $\vec{a}$ and $\vec{b}$ is
(a) $\frac{\pi}{2}$
(b) $\frac{3 \pi}{4}$
(c) $\frac{\pi}{4}$
(d) $\pi$
Solution:
(b) $\frac{3 \pi}{4}$

Hint:
$
\begin{aligned}
& \vec{a} \times(\vec{b} \times \vec{c})=\frac{\vec{b}+\vec{c}}{\sqrt{2}} \\
& (\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}=\frac{\vec{b}+\vec{c}}{\sqrt{2}} \\
& \therefore \quad \vec{a} \cdot \vec{c}=\frac{1}{\sqrt{2}} \\
& \vec{a} \cdot \vec{b}=-\frac{1}{\sqrt{2}} \\
& |\vec{a}||\vec{b}| \cos \theta=-\frac{1}{\sqrt{2}} \\
& \cos \theta=-\frac{1}{\sqrt{2}} \\
& \theta=\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\frac{3 \pi}{4} \\
&
\end{aligned}
$
Question 11.
If the volume of the parallelepiped with $\vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a}$ as coterminous edges is 8 cubic units, then the volume of the parallelepiped with $(\vec{a} \times \vec{b}) \times(\vec{b} \times \vec{c}),(\vec{b} \times \vec{c}) \times(\vec{c} \times \vec{a})$ and $(\vec{c} \times \vec{a}) \times(\vec{a} \times \vec{b})$ as coterminous edges is
(a) 8 cubic units
(b) 512 cubic units
(c) 64 cubic units
(d) 24 cubic units
Solution:
(c) 64 cubic units
Hint:
$
\begin{aligned}
{[\vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a}] } & =8 \text { (Given) } \\
{[(\vec{a} \times \vec{b}) \times(\vec{b} \times \vec{c}),(\vec{b} \times \vec{c}) \times(\vec{c} \times \vec{a}),(\vec{c} \times \vec{a}) \times(\vec{a} \times \vec{b})] } & =[(\vec{a} \times \vec{b}),(\vec{b} \times \vec{c}),(\vec{c} \times \vec{a})]^2 \\
& =(8)^2=64
\end{aligned}
$

Question 12.
Consider the vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ such that $(\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})=\overrightarrow{0}$. Let $\mathrm{P}_1$ and $\mathrm{P}_2$ be the planes determined by the pairs of vectors $\vec{a}, \vec{b}$ and $\vec{c}, \vec{d}$ respectively. Then the angle between $\mathrm{P}_1$ and $\mathrm{P}_2$ is
(a) $0^{\circ}$
(b) $45^{\circ}$
(c) $60^{\circ}$
(d) $90^{\circ}$
Solution:
(a) $0^{\circ}$
Hint:
$
\begin{aligned}
& \begin{array}{l|l}
\mathrm{P}_1 \rightarrow \vec{a}, \vec{b} & \mathrm{P}_2 \rightarrow \vec{c}, \vec{d}
\end{array} \\
& \vec{n}_1 \rightarrow \vec{a} \times \vec{b} \quad \vec{n}_2 \rightarrow \vec{c} \times \vec{d} \\
& \text { Given. } \quad \vec{n}_1 \times \vec{n}_2=0 \quad\left(\vec{n}_1, \vec{n}_2 \text { are normal unit vectors }\right) \\
& \left|\vec{n}_1\right|\left|\vec{n}_2\right| \sin \theta=0 \\
& \sin \theta=0 \quad \Rightarrow \theta=0^{\circ} \\
&
\end{aligned}
$

Question 13.
If $\vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \times \vec{b}) \times \vec{c}$, where $\vec{a}, \vec{b}, \vec{c}$ are any three vectors such that $\vec{b} \cdot \vec{c} \neq 0$ and $\vec{a} \cdot \vec{b} \neq 0$, then $\vec{a}$ and $\vec{c}$ are
(a) perpendicular
(b) parallel
(c) inclined at an angle $\frac{\pi}{3}$
(d) inclined at an angle $\frac{\pi}{6}$
Solution:
(b) parallel
Hint:
$
\begin{aligned}
\vec{a} \times(\vec{b} \times \vec{c}) & =(\vec{a} \times \vec{b}) \times \vec{c} \\
(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c} & =(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{b} \cdot \vec{c}) \vec{a} \\
\vec{a} & =\left(\frac{\vec{a} \cdot \vec{b}}{\vec{b} \cdot \vec{c}}\right) \vec{c} \\
\vec{a} & =\lambda \vec{c}
\end{aligned}
$
$\therefore \vec{a}$ and $\vec{c}$ are parallel.

Question 14.
If $\vec{a}=2 \hat{i}+3 \hat{j}-\hat{k}, \vec{b}=\hat{i}+2 \hat{j}-5 \hat{k}, \vec{c}=3 \hat{i}+5 \hat{j}-\hat{k}$, then a vector perpendicular to $\vec{a}$ and lies in the plane containing $\vec{b}$ and $\vec{c}$ is
(a) $-17 \hat{i}+21 \hat{j}-97 \hat{k}$
(b) $-17 \hat{i}+21 \hat{j}-123 \hat{k}$
(c) $-17 \hat{i}-21 \hat{j}+97 \hat{k}$
(d) $-17 \hat{i}-21 \hat{j}-97 \hat{k}$
Solution:
(d) $-17 \hat{i}-21 \hat{j}-97 \hat{k}$
Hint:
A vector $\perp \mathrm{r}$ to $\vec{a}$ and lies in the plane containing $\vec{b}$ and $\vec{c}$ is $\vec{a} \times(\vec{b} \times \vec{c})$
$
\begin{aligned}
\vec{b} \times \vec{c} & =\left|\begin{array}{rrr}
\vec{i} & \vec{j} & \vec{k} \\
1 & 2 & -5 \\
3 & 5 & -1
\end{array}\right|=23 \vec{i}-14 \vec{j}-\vec{k} \\
\vec{a} \times(\vec{b} \times \vec{c}) & =\left|\begin{array}{rrr}
\vec{i} & \vec{j} & \vec{k} \\
2 & 3 & -1 \\
23 & -14 & -1
\end{array}\right|=-17 \vec{i}-21 \vec{j}-97 \vec{k}
\end{aligned}
$

Question 15.
The angle lines $\frac{x-2}{3}=\frac{y+1}{-2}, z=2$ and $\frac{x-1}{1}=\frac{2 y+3}{3}=\frac{z+5}{2}$ is .................

(a) $\frac{\pi}{6}$
(b) $\frac{\pi}{4}$
(c) $\frac{\pi}{3}$
(d) $\frac{\pi}{2}$
Solution:
(d) $\frac{\pi}{2}$
Hint: $\frac{x-2}{3}=\frac{y+1}{-2}, z=2$ and $\frac{x-1}{1}=\frac{2 y+3}{3}=\frac{z+5}{2}$ $\frac{x-1}{1}=\frac{y+\frac{3}{2}}{\frac{3}{2}}=\frac{z+5}{2}$
d.r.s : $(3,-2,0)$ and d.c.s $\left(1, \frac{3}{2}, 2\right)$
$
\begin{aligned}
\cos \theta & =\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}} \\
\cos \theta & =\frac{3-3+0}{\sqrt{(3)^2+(-2)^2+(0)^2} \sqrt{(1)^3+\left(\frac{3}{2}\right)^2+(2)^2}} \\
\cos \theta & =0 \\
\theta & =\frac{\pi}{2}
\end{aligned}
$

Question 16.
If the line $
\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}
$
lies in the plane $x+3+\alpha z+\beta=0$, then $(\alpha, \beta)$ is
(a) $(-5,5)$
(b) $(-6,7)$
(c) $(5,-5)$
(d) $(6,-7)$
Solution:
(d) $(-6,7)$
Hint:
d.c.s of the first line $=(3,-5,2)$
d.c.s of the line perpendicular to plane $=(1,3,-\alpha)$
$\mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2=0$
$3-15-2 \alpha=0=>-12-2 \alpha=0$
$-2 \alpha=12 \Rightarrow \alpha=-6$
Plane passes through the point $(2,1,-2)$ so
$
2+3+6(-2)+\beta=0 \Rightarrow \beta=7
$
$(\alpha, \beta)=(-6,7)$

Question 17.
The angle between the line $\vec{r}=(\hat{i}+2 \hat{j}-3 \hat{k})+t(2 \hat{i}+\hat{j}-2 \hat{k})$ and the plane $\vec{r} \cdot(\hat{i}+\hat{j})+4=0$ is
(a) $0^{\circ}$
(b) $30^{\circ}$
(c) $45^{\circ}$
(d) $90^{\circ}$
Solution:
(c) $45^{\circ}$
Hint:
$
\begin{aligned}
\vec{b} & =2 \vec{i}+\vec{j}-2 \vec{k} ; \vec{n}=\vec{i}+\vec{j} \\
\sin \theta & =\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|} \\
\sin \theta & =\frac{2+1}{\sqrt{4+1+4} \sqrt{1+1}}=\frac{3}{3\sqrt{2}}=\frac{1}{\sqrt{2}} \\
\theta & =45^{\circ}
\end{aligned}
$

Question 18.
The coordinates of the point where the line $\vec{r}=(6 \hat{i}-\hat{j}-3 \hat{k})+t(-\hat{i}+4 \hat{k})$ meets the plane $\vec{r} \cdot(\hat{i}+\hat{j}-\hat{k})$ are
(a) $(2,1,0)$
(b) $(7,-1,-7)$
(c) $(1,2,-6)$
(c) $(5,-1,1)$
Solution:
(d) $(5,-1,1)$
Hint:
Cartesian equation of the line
$
\begin{aligned}
& \frac{x-6}{-1}=\frac{y+1}{0}+\frac{z+3}{4}=\lambda \\
& (-\lambda+6,-1,4 \lambda-3)
\end{aligned}
$
This meets the plane $x+y-z=3$
$
-\lambda+6-1-41+3=3 \Rightarrow-5 \lambda=-5
$
$
\lambda=1
$
The required point $(5,-1,1)$.

Question 19.
Distance from the origin to the plane $3 x-6 y+2 z+7=0$ is
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(b) 1
Hint:
Distance from the origin $(0,0,0)$ to the plane
$
=\left|\frac{3(0)-6(0)+2(0)+7}{\sqrt{9+36+4}}\right|=\frac{7}{\sqrt{49}}=\frac{7}{7}=1
$
Question 20.
The distance between the planes $x+2 y+3 z+7=0$ and $2 x+4 y+6 z+7=0$ is
(a) $\frac{\sqrt{7}}{2 \sqrt{2}}$
(b) $\frac{7}{2}$
(c) $\frac{\sqrt{7}}{2}$
(d) $\frac{7}{2 \sqrt{2}}$
Solution:
(a) $\frac{\sqrt{7}}{2 \sqrt{2}}$
Hint:
$
\mathrm{x}+2 \mathrm{y}+3 \mathrm{z}+1=0 ; 2 \mathrm{x}+4 \mathrm{y}+6 \mathrm{z}+7=0
$
Multiplying 2 on both sides
$
\begin{aligned}
& 2 x+4 y+6 z+14=0 \\
& a=2, b=4, c=6, d_1=14, d_2=?
\end{aligned}
$
Distance $=\left|\frac{d_1-d_2}{\sqrt{a^2+b^2+c^2}}\right|=\left|\frac{14-7}{\sqrt{4+16+36}}\right|=\left|\frac{7}{\sqrt{56}}\right|=\frac{7 \sqrt{7}}{2 \sqrt{2} \sqrt{7}}=\frac{\sqrt{7}}{2 \sqrt{2}}$
Question 21.
If the direction cosines of a line are $\frac{1}{c}, \frac{1}{c}, \frac{1}{c}$, then
(a) $\mathrm{c}=\pm 3$
(b) $c=\pm \sqrt{3}$
(c) c $>0$
(d) $0<$ c $<1$
Solution:
(b) $\mathrm{c}=\pm \sqrt{3}$
Hint:

We know that sum of the squares of direction cosines $=1$
$
\begin{aligned}
\frac{1}{c^2}+\frac{1}{c^2}+\frac{1}{c^2} & =1 & & \Rightarrow \frac{3}{c^2}=1 \\
c^2 & =3 & & \Rightarrow c=\pm \sqrt{3}
\end{aligned}
$
Question 22.
The vector equation $\vec{r}=(\hat{i}-2 \hat{j}-\hat{k})+t(6 \vec{j}-\hat{k})$ represents a straight line passing through the points and $(0,-6,1)$
(a) $(0,6,-1)$ and $(1,-2,-1)$ (b) $(0,6,-1)$ and $(-1,-4,-2)$ (c) $(1,-2,-1)$ and $(1,4,-2)$ (d) $(1,-2,-1)$

Solution:
(c) $(1,-2,-1)$ and $(1,4,-2)$
Hint:
The required vector equation is $\vec{r}=\vec{a}+t(\vec{b}-\vec{a})$
$
\begin{aligned}
\therefore \quad \vec{a} & =\vec{i}-2 \vec{j}-\vec{k} \\
\vec{b}-\vec{a} & =6 \vec{j}-\vec{k} \\
\vec{b} & =6 \vec{j}-\vec{k}+\vec{a} \quad \Rightarrow \vec{b}=6 \vec{j}-\vec{k}+\vec{i}-2 \vec{j}-\vec{k} \\
\vec{b} & =\vec{i}+4 \vec{j}-2 \vec{k}
\end{aligned}
$
From (1) and (2) The points are $(1,-2,-1)$ and $(1,4,-2)$

Question 23.
If the distance of the point $(1,1,1)$ from the origin is half of its distance from the plane $x+y+z+k=0$, then the values of $k$ are
(a) $\pm 3$ (b) $\pm 6$ (c) $-3,9$ (d) $3,-9$
Solution:
(d) $3,-9$
Hint:
$
\begin{aligned}
\sqrt{3} & =\frac{1}{2}\left[\frac{3+k}{\sqrt{3}}\right] \\
3 & =\pm \frac{1}{2}(3+k) \\
3+k & =\pm 6 \\
3+k & =6 \\
k & =3
\end{aligned} \mid \begin{aligned}
3+k & =-6 \\
& k=-9
\end{aligned}
$
Question 24.
If the planes $\vec{r} \cdot(2 \hat{i}-\lambda \hat{j}+\hat{k})=3$ and $\vec{r} \cdot(4 \hat{i}+\hat{j}-\mu \hat{k})=5$ are parallel, then the value of $\lambda$ and $\mu$ are
Solution:
(c) $-\frac{1}{2},-2$

Hint:
$
\vec{a}=2 \vec{i}-\lambda \vec{j}+\vec{k} \text { and } \vec{b}=4 \vec{i}+\vec{j}-\mu \vec{k}
$
Given $\vec{a}$ and $\vec{b}$ are parallel
$
\begin{aligned}
& \vec{a}=m \vec{b} \\
& 2 \vec{i}-\lambda \vec{j}+\vec{k}=4 m \vec{i}+m \vec{j}-m \mu \vec{k} \\
& \begin{array}{l|c|c}
4 m=2 & -\lambda=m & -m \mu=1 \\
m=1 / 2 & -\lambda=1 / 2 & \mu=-1 / m \\
& \lambda=-1 / 2 & \mu=-2
\end{array} \\
&
\end{aligned}
$
so
$
(\lambda, \mu)=(-1 / 2,-2)
$

Question 25.
If the length of the perpendicular from the origin to the plane $2 x+3 y+\lambda z=1, \lambda>0$ is $-\frac{1}{5}$, then the value of $\lambda$ is .............
(a) $2 \sqrt{3}$
(b) $3 \sqrt{2}$
(c) 0
(d) 1
Solution:
(a) $2 \sqrt{3}$
Hint:
Given length of perpendicular from origin to the plane $=-\frac{1}{5}$
$
\begin{aligned}
& \left|\frac{2(0)+3(0)+\lambda(0)-1}{\sqrt{4+9+\lambda^2}}\right|=\frac{1}{5} \\
& \frac{1}{\sqrt{13+\lambda^2}}=\frac{1}{5} \quad \Rightarrow 5=\sqrt{13+\lambda^2} \\
&
\end{aligned}
$
Squaring on both sides
$
\begin{aligned}
25 & =13+\lambda^2 & & \Rightarrow \lambda^2=12 \\
\lambda & =\sqrt{12} & & \Rightarrow \lambda=2 \sqrt{3}
\end{aligned}
$