JEE Main 2018 — Three Dimensional Geometry Question with Solution

Given lines are $\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ and $\frac{5-x}{-2}=\frac{7y-14}{p}=\frac{z-3}{4}$

$\Rightarrow \frac{x-5}{2}=\frac{y-2}{\left({\displaystyle \frac{P}{7}}\right)}=\frac{z-3}{4}$

We know that the angle between the lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ is given by $\theta ={\cos }^{-1}\left|\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\times \sqrt{a_2^2+b_2^2+c_2^2}}\right|$

Given, the angle between both lines is ${\cos }^{-1}\left(\frac{2}{3}\right)$

$\Rightarrow {\cos }^{-1}\left(\frac{2}{3}\right)={\cos }^{-1}\left(\frac{2\times 2+{\displaystyle 2\times \frac{P}{7}}+1\times 4}{\sqrt{2^2+2^2+1^2}\times \sqrt{2^2+{\left({\displaystyle \frac{P}{7}}\right)}^2+4^2}}\right)$

$\Rightarrow \frac{2}{3}=\frac{4+{\displaystyle \frac{2P}{7}}+4}{3\times \sqrt{4+{\displaystyle \frac{P^2}{49}}+16}}$

$\Rightarrow \frac{2}{3}=\frac{56+2P}{3\times \sqrt{P^2+980}}$

$\Rightarrow \sqrt{P^2+980}=P+28$

$\Rightarrow P^2+980={\left(P+28\right)}^2$

$\Rightarrow P^2+980=P^2+56P+784$

$\Rightarrow 56P=196$

$\Rightarrow P=\frac{7}{2}.$