First slide

Introduction to 3-D Geometry

Question

If a parallelepiped is formed by planes drawn through the points (5, 8, 10) and (3, 6, 8) parallel to
the coordinate planes, then the length of diagonal of the parallelepiped is

Easy

A

$2 \sqrt{3}$
B

$3 \sqrt{2}$
C

$\sqrt{2}$
D

$\sqrt{3}$

Solution

The edge of a parallelepiped = (3 - 5,6 - 8,8 - 10)
= (2, -2, -2)
Length of diagonal
$\begin{matrix} = \sqrt{(- 2)^{2} + (- 2)^{2} + (- 2)^{2}} \\ = \sqrt{4 + 4 + 4} = 2 \sqrt{3} \end{matrix}$

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