JEE 2022 — Mathematics PYQ

To find the square of the distance between the feet of the perpendiculars, we can use the concept of the projection of the vector $\overrightarrow{PQ}$ onto the plane.

1. Find the vector $\overrightarrow{PQ}$

The vector connecting points $P(1,2,-1)$ and $Q(2,-1,3)$ is:

2. Identify the normal vector to the plane

The given plane is $-x+y+z=1$. The normal vector $\vec{n}$ to this plane is:

3. Concept

The distance $d$ between the feet of the perpendiculars is the length of the projection of vector $\overrightarrow{PQ}$ parallel to the plane.

We can find this by using the formula:

4. Calculations

• Magnitude squared of $\overrightarrow{PQ}$:

  $$|\overrightarrow{PQ}{|}^2=(1{)}^2+(-3{)}^2+(4{)}^2=1+9+16=26$$

• Projection of $\overrightarrow{PQ}$ on $\vec{n}$:

  $${\text{Proj}}_n=\frac{|\overrightarrow{PQ}\cdot \vec{n}|}{|\vec{n}|}$$
  $$\overrightarrow{PQ}\cdot \vec{n}=(1)(-1)+(-3)(1)+(4)(1)=-1-3+4=0$$
  $$|\vec{n}|=\sqrt{(-1{)}^2+1^2+1^2}=\sqrt{3}$$
  $${\text{Proj}}_n=\frac{0}{\sqrt{3}}=0$$

5. Final Result

Correct Option: (A) 26