Three Dimensional Geometry Question 10
Question 10 - 2024 (27 Jan Shift 2)
The lines $\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16}$ and $\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}$ intersect at the point $P$. If the distance of $P$ from the line $\frac{x+1}{2}=\frac{y-1}{3}=\frac{z-1}{1}$ is $l$, then $14 l^{2}$ is equal to
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Answer (108)
Solution
Sol. $\frac{x-2}{1}=\frac{y}{-1}=\frac{z-7}{8}=\lambda$
$\frac{\mathrm{x}+3}{4}=\frac{\mathrm{y}+2}{3}=\frac{\mathrm{z}+2}{1}=\mathrm{k}$
$\Rightarrow \lambda+2=4 \mathrm{k}-3$
$-\lambda=3 \mathrm{k}-2$
$\Rightarrow \mathrm{k}=1, \lambda=-1$
$8 \lambda+7=\mathrm{k}-2$
$\therefore \mathrm{P}=(1,1,-1)$
Projection of $2 \hat{i}-2 \hat{k}$ on $2 \hat{i}+3 \hat{j}+\hat{k}$ is
$=\frac{4-2}{\sqrt{4+9+1}}=\frac{2}{\sqrt{14}}$
$\therefore l^{2}=8-\frac{4}{14}=\frac{108}{14}$
$\Rightarrow 14 l^{2}=108$
