First slide

Introduction to 3-D Geometry

Question

If $P Q$ be the projections of $(3, - 2, 4)$ on the coordinate planes $x y, y z$ respectively then $P Q$ is

Very Easy

A

2
B

$\sqrt{20}$
C

5
D

$\sqrt{29}$

Solution

The projection $(3, - 2, 4)$ of on $x y$ plane is $P (3, - 2, 4)$
The projection $(3, - 2, 4)$ of on $y z$ plane is $Q (0, - 2, 4)$
The distance between the points $P Q = \sqrt{3^{2} + 0 + 4^{2}} = 5$

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