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

A
$(2, 19, 7)$
B
$(- 2, 19, 7)$
C
$(5, 7, 3)$
D
$(9, 13, 15)$

Solution:

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

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

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

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

Hence, the line is perpendicular to the above line

It implies that

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

Therefore, the foot of the perpendicular is $Q (9, 13, 15)$

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

Solution:

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