The foot of the perpendicular fromP5,7,3  to the line which is passing through the points A12,21,10,B9,13,15 is

# The foot of the perpendicular from$P\left(5,7,3\right)$  to the line which is passing through the points $A\left(12,21,10\right)$,$B\left(9,13,15\right)$ is

1. A

$\left(2,19,7\right)$

2. B

$\left(-2,19,7\right)$

3. C

$\left(5,7,3\right)$

4. D

$\left(9,13,15\right)$

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### Solution:

The equation of the line passing through the points$A\left(12,21,10\right)$  and$B\left(9,13,15\right)$  is

$\frac{x-12}{3}=\frac{y-21}{8}=\frac{z-10}{-5}$

General point on the line is$Q\left(3k+12,8k+21,-5k+10\right)$

Suppose that $Q$be the foot of the perpendicular to the $P\left(5,7,3\right)$point   on the above line$\overline{PQ}$

Hence, the line  is perpendicular to the above line

It implies that

$\begin{array}{c}\left(3k+7\right)\cdot 3+\left(8k+14\right)\cdot 8+\left(-5k+7\right)\cdot \left(-5\right)=0\\ 9k+21+64k+112+25k-35=0\\ 98k+98=0\\ k=-1\end{array}$

Therefore, the foot of the perpendicular is   $Q\overline{)\left(9,13,15\right)}$  +91

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