First slide

Introduction to 3-D Geometry

Question

If the projections of the line segment $\bar{A B}$ on the $y z -$ plane, $z x -$ plane, $x y -$ plane are $\sqrt{160}, \sqrt{153} and 5$ respectively, then the projection of $\bar{A B}$ on the $z -$ axis is

Easy

A

22
B

24
C

12
D

14

Solution

The projection of the line segment $\bar{A B}$ on the $y z -$ plane is $\sqrt{{(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}} = \sqrt{160}$
The projection of the line segment $\bar{A B}$ on the $z x -$ plane is $\sqrt{{(z_{2} - z_{1})}^{2} + {(x_{2} - x_{1})}^{2}} = \sqrt{153}$
The projection of the line segment $\bar{A B}$ on the $x y -$ plane is $\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = 5$
It implies that
${(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2} = 169$
Hence, ${(z_{2} - z_{1})}^{2} = 169 - 25 = 144$
Apply square root on both sides $|z_{2} - z_{1}| = 12$ , this is the length of the projection of $\bar{A B}$ on $z -$ axis.

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