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The foot of the perpendicular from $(0,2,3)$ to the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$ is
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Verified Answer
The correct answer is:
$(2,3,-1)$
Let $N$ be the foot of the perpendicular from the point $P(0,2,3)$ on the given line
$\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=r \quad$ (say) ... (i)
Any point on the line (i) is
$(5 r-3,2 r+1,3 r-4)$
If this point is $N$, The direction ratio's of $N P$ are
$5 r-3,2 r+1-2,3 r-4-3$
i.e.
$5 r-3,2 r-1,3 r-7$
Since, $N P$ is perpendicular to the given line, then
$\begin{aligned} & \qquad 5(5 r-3)+2(2 r-1)+3(3 r-7)=0 \\ & \Rightarrow \quad 38 r-38=0 \Rightarrow r=1 \\ & \text { The point } N \text { is } \quad(5-3,2+1,3-4) \\ & \text { i.e., } \quad(2,3,-1) .\end{aligned}$
$\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=r \quad$ (say) ... (i)
Any point on the line (i) is
$(5 r-3,2 r+1,3 r-4)$
If this point is $N$, The direction ratio's of $N P$ are
$5 r-3,2 r+1-2,3 r-4-3$
i.e.
$5 r-3,2 r-1,3 r-7$
Since, $N P$ is perpendicular to the given line, then
$\begin{aligned} & \qquad 5(5 r-3)+2(2 r-1)+3(3 r-7)=0 \\ & \Rightarrow \quad 38 r-38=0 \Rightarrow r=1 \\ & \text { The point } N \text { is } \quad(5-3,2+1,3-4) \\ & \text { i.e., } \quad(2,3,-1) .\end{aligned}$
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