Question

Let a line $L$ passing through the point $P(1,1,1)$ be perpendicular to the lines \[ \frac{x-4}{4}=\frac{y-1}{1}=\frac{z-1}{1} \quad \text{and} \quad \frac{x-17}{1}=\frac{y-71}{1}=\frac{z}{0}. \] Let the line $L$ intersect the $yz$-plane at the point $Q$.
Another line parallel to $L$ and passing through the point $S(1,0,-1)$ intersects the $yz$-plane at the point $R$.
Then the square of the area of the parallelogram $PQRS$ is equal to

Hint:

For parallelogram area, always use the magnitude of cross product of adjacent sides.

Answer 1

Correct Answer: 72

Step 1: Finding direction ratios of line $L$.
Direction ratios of the given lines are \[ \vec d_1=(4,1,1),\quad \vec d_2=(1,1,0) \] Since $L$ is perpendicular to both, its direction vector is \[ \vec d=\vec d_1\times\vec d_2 =\begin{vmatrix} \hat i & \hat j & \hat k \\ 4 & 1 & 1 \\ 1 & 1 & 0 \end{vmatrix} =(-1,1,3) \] Step 2: Coordinates of points $Q$ and $R$.
Equation of line $L$: \[ (x,y,z)=(1,1,1)+\lambda(-1,1,3) \] At $yz$-plane, $x=0 \Rightarrow \lambda=1$, \[ Q=(0,2,4) \] Similarly, line through $S(1,0,-1)$ parallel to $L$: \[ (x,y,z)=(1,0,-1)+\mu(-1,1,3) \] At $x=0 \Rightarrow \mu=1$, \[ R=(0,1,2) \] Step 3: Area of parallelogram.
Adjacent sides are \[ \vec{PQ}=(-1,1,3),\quad \vec{PS}=(0,-1,-2) \] \[ \text{Area}^2=|\vec{PQ}\times\vec{PS}|^2 \] \[ =72 \]