Three Dimensional Geometry Question 24
Question 24 - 30 January - Shift 2
Let a line $L$ pass through the point $P(2,3,1)$ and be parallel to the line $x+3 y-2 z-2=0=x-y+2 z$. If the distance of $L$ from the point $(5,3,8)$ is $\alpha$, then $3 \alpha^{2}$ is equal to __________
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Answer: 158
Solution:
Formula: Angle Between Two Line Segments
$ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & -2 \\ 1 & -1 & 2\end{vmatrix} =4 \hat{i}-4 \hat{j}-4 \hat{k}$
$\therefore$ Equation of line is $\frac{x-2}{1}=\frac{y-3}{-1}=\frac{z-1}{-1}$
Let $Q$ be $(5,3,8)$ and foot of $\perp$ from $Q$ on this line be $R$.
Now, $R \equiv(k+2,-k+3,-k+1)$
$DR$ of $QR$ are $(k-3,-k,-k-7)$
$\therefore(1)(k-3)+(-1)(-k)+(-1)(-k-7)=0$
$\Rightarrow k=-\frac{4}{3}$
$\therefore \alpha^{2}=(\frac{13}{3})^{2}+(\frac{4}{3})^{2}+(\frac{17}{3})^{2}=\frac{474}{9}$
$\therefore 3 \alpha^{2}=158$
