Question

If the angle bwteen a line $x=\frac{y-1}{2}=\frac{z-3}{\lambda }$ and normal to the plane $x+2y+3z=4$ is ${\cos }^{-1}\sqrt{\frac{5}{14}}$, then possible value(s) of $\lambda$ is/are

• A. $\frac{5}{2}$
• B. $\frac{2}{5}$
• C. 00
• D. $\frac{2}{3}$

Answer 1

The correct option is A $\frac{5}{2}$

Consider the given line equation $x=\frac{y-1}{2}=\frac{z-3}{\lambda }$ and plane equation$x+2y+3z=4$.

Let $\theta$ be the angle between the line and normal to plane converting the given equations into normal form, we have

$\overrightarrow{r}=0.\hat{i}+\hat{j}+3\hat{k}+\beta (\hat{i}+2\hat{j}+\lambda \hat{k})$

$\overrightarrow{r}=\hat{i}+2\hat{j}+3.\hat{k}=3$

Now,

$\overrightarrow{b}=(\hat{i}+2\hat{j}+\lambda \hat{k})$

$\overrightarrow{n}=\hat{i}+2\hat{j}+3.\hat{k}=3$

We know that,

$\cos \theta =\left|\frac{\hat{b}.\hat{n}}{\left|\hat{b}\right|\left|\hat{n}\right|}\right|$

$=\left|\frac{(\hat{i}+2\hat{j}+\lambda \hat{k}).(\hat{i}+2\hat{j}+3.\hat{k})}{|\hat{i}+2\hat{j}+\lambda \hat{k}||\hat{i}+2\hat{j}+3.\hat{k}|}\right|$

$\cos \theta =\left|\frac{1+4+3\lambda }{\sqrt{1^2+2^2+{\lambda }^2}\sqrt{1^2+2^2+3^2}}\right|=\left|\frac{5+\lambda }{\sqrt{5+{\lambda }^2}\sqrt{14}}\right|$

But given that $\theta ={\cos }^{-1}\left(\sqrt{\frac{5}{14}}\right)$ ,so

$\cos {\cos }^{-1}\sqrt{\frac{5}{14}}=\left|\frac{5+\lambda }{\sqrt{5+{\lambda }^2}\sqrt{14}}\right|$

$\sqrt{\frac{5}{14}}=\left|\frac{5+\lambda }{\sqrt{5+{\lambda }^2}\sqrt{14}}\right|$

Taking square both sides ,we get

$\frac{5}{14}={\left|\frac{5+\lambda }{\sqrt{5+{\lambda }^2}\sqrt{14}}\right|}^2=\frac{{(5+\lambda )}^2}{(5+{\lambda }^2)\left(14\right)}$

$\frac{{(5+\lambda )}^2}{(5+{\lambda }^2)}=5$

$25+{\lambda }^2+10\lambda =25+5{\lambda }^2$

$4{\lambda }^2-10\lambda =0$

$2\lambda (2\lambda -5)=0$

$\lambda (2\lambda -5)=0$

$\lambda =0,\lambda =\frac{5}{2}$

Ignore $\lambda =0$ as it is in denominator. Therefore,

$\lambda =\frac{5}{2}$

This is the answer .