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A point $P$ lies on a line through $Q(1,-2,3)$ and is parallel to the line $\frac{x}{1}=\frac{y}{4}=\frac{z}{5},$ If $P$ lies on the plane $2 x+3 y-4 z+22=0,$ then segment PQ equals
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Verified Answer
The correct answer is:
$\sqrt{42}$ units
Equation of line through $Q(1,-2,3)$ and
parallel to the line $\frac{x}{1}=\frac{y}{4}=\frac{z}{5}$ is $\frac{x-1}{1}=\frac{y+2}{4}=\frac{z-3}{5}=\lambda \quad$ (say)
since, point $P$ lies on above line.
$\therefore$
$$
P(\lambda+1,4 \lambda-2,5 \lambda+3)
$$
since, $P$ lies on the given plane. $\therefore \quad 2(\lambda+1)+3(4 \lambda-2)-4(5 \lambda+3)+22=0$
$\Rightarrow \quad 2 \lambda+2+12 \lambda-6-20 \lambda-12+22=0$
$$
\Rightarrow \quad-6 \lambda+6=0
$$
$\Rightarrow$
$\lambda=1$
$\therefore$
$P(2,2,8)$
$\therefore \quad P Q=\sqrt{(2-1)^{2}+(2+2)^{2}+(3-8)^{2}}$
$\Rightarrow \quad P Q=\sqrt{1+16+25}=\sqrt{42}$
parallel to the line $\frac{x}{1}=\frac{y}{4}=\frac{z}{5}$ is $\frac{x-1}{1}=\frac{y+2}{4}=\frac{z-3}{5}=\lambda \quad$ (say)
since, point $P$ lies on above line.
$\therefore$
$$
P(\lambda+1,4 \lambda-2,5 \lambda+3)
$$
since, $P$ lies on the given plane. $\therefore \quad 2(\lambda+1)+3(4 \lambda-2)-4(5 \lambda+3)+22=0$
$\Rightarrow \quad 2 \lambda+2+12 \lambda-6-20 \lambda-12+22=0$
$$
\Rightarrow \quad-6 \lambda+6=0
$$
$\Rightarrow$
$\lambda=1$
$\therefore$
$P(2,2,8)$
$\therefore \quad P Q=\sqrt{(2-1)^{2}+(2+2)^{2}+(3-8)^{2}}$
$\Rightarrow \quad P Q=\sqrt{1+16+25}=\sqrt{42}$
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