The foot of the perpendicular from 2,−4,0to the join of 0,0,0 , 0,3,0 is

# The foot of the perpendicular from $\left(2,-4,0\right)$to the join of $\left(0,0,0\right)\text{\hspace{0.17em}},\text{\hspace{0.17em}}\left(0,3,0\right)$ is

1. A

$\left(\frac{4}{3},0,3\right)$

2. B

$\left(-\frac{4}{3},0,0\right)$

3. C

$\left(0,-4,0\right)$

4. D

$\left(0,-\frac{4}{3},0\right)$

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### Solution:

Equation of the line joining $\left(0,0,0\right)\text{\hspace{0.17em}},\text{\hspace{0.17em}}\left(0,3,0\right)$ is$\frac{x}{0}=\frac{y}{3}=\frac{z}{0}$

General point on the above line is$Q\left(0,3{y}_{1},0\right)$

The direction ratios of the line joining  $P\left(2,-4,0\right)$ and $Q\left(0,3{y}_{1},0\right)$is$〈-2,3{y}_{1}+4,0〉$

Since the line $PQ$ and $\frac{x}{0}=\frac{y}{3}=\frac{z}{0}$are perpendicular to each other.

$\begin{array}{c}3\left(3{y}_{1}+4\right)=0\\ {y}_{1}=-\frac{4}{3}\end{array}$

Therefore, the foot of the perpendicular of$\left(2,-4,0\right)$ to the join of $\left(0,0,0\right)\text{\hspace{0.17em}},\text{\hspace{0.17em}}\left(0,3,0\right)$ is $Q\overline{)\left(0,-4,0\right)}$

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