QJEEAdv15Madda4: JEE Advanced 2015 Mathematics - Class 12 JEEAdv15Madda4 Solution: In let be a straight lines passing through the origin. Suppose…

Let the equation of the line passing through the origin be $\mathrm{L}: \frac{x}{l}=\frac{y}{m}=\frac{z}{n}\quad \text{... (i)}$

Given, planes

$\begin{aligned}& \mathrm{P}_1: x+2 y-z+1=0 \quad \text{... (ii)}\\& \mathrm{P}_2: 2 x-y+z-1=0 \quad \text{... (iii)}\end{aligned}$

since, all the points on $\mathrm{L}$ are at a constant distance from the planes $\mathrm{P}_1$ and $\mathrm{P}_2$

$\therefore$ Line $\mathrm{L}$ is perpendicular to normal of plane $\mathrm{P}_1$ and $\mathrm{P}_2$

As we know if two lines are perpendicular then the sum of products of their respective direction ratios is zero.

$\begin{aligned}& \Rightarrow l+2 m-n=0 \ldots \ldots . \text { (iv) } \quad \left\{\because L \perp P_1\right\}\\& \Rightarrow 2 l-m+n=0 \ldots \ldots . \text { (v) } \quad \left\{\because L \perp P_2\right\}\end{aligned}$

On solving equation (iv) and equation (v), we get

$\frac{l}{1}=\frac{m}{-3}=\frac{n}{-5}$

$\therefore \quad$ Equation of line L is $\frac{x}{1}=\frac{y}{-3}=\frac{z}{-5}$

Let any point on L is $\mathrm{A}(k,-3 k,-5 k)$

Now foot of perpendicular from $A$ to plane $P_1$ is given by

$\begin{aligned}& \frac{x-k}{1}=\frac{y+3 k}{2}=\frac{z+5 k}{-1}=-\frac{(k-6 k+5 k+1)}{1^2+2^2+1^2} \\\Rightarrow \quad & \frac{x-k}{1}=\frac{y+3 k}{2}=\frac{z+5 k}{-1}=-\frac{1}{6} \\\Rightarrow \quad & x=k-\frac{1}{6}, y=-3 k-\frac{1}{3}, z=-5 k+\frac{1}{6}\end{aligned}$

So, coordinates of foot of perpendicular is $\left(k-\frac{1}{6},-3 k-\frac{1}{3},-5 k+\frac{1}{6}\right)$

For $k=0$ , foot of perpendicular $\left(-\frac{1}{6},-\frac{1}{3}, \frac{1}{6}\right)$

For $k=\frac{1}{6}$ , foot of perpendicular $\left(0,-\frac{5}{6},-\frac{2}{3}\right)$

Hint :

(i) since all points on L are at constant distance from the planes $\mathrm{P}_1$ and $\mathrm{P}_2$ . So line $L$ is perpendicular to normal of $P_1$ and $P_2$ .

(ii) Find equation of line L using above condition and assume any point on L .

(iii) Use coordinates of foot of perpendicular drawn from a point $\left(x_1 y_1 z_1\right)$ to the plane $a x+b y+c z+d=0$ is given by

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=-\frac{a x_1+b y_1+c z_1+d}{a^2+b^2+c^2}$