First slide

Introduction to 3-D Geometry

Question

If three consecutive vertices of a parallelogram are $A (4, 3, 5),$ $B (0, 6, 0),$ $C (- 8, 1, 4)$ and D is the fourth vertex then the angle between $\overset{\leftrightarrow}{A C}$ and $\overset{\leftrightarrow}{B D}$ is

Moderate

A

$C o s^{- 1} (\frac{55}{\sqrt{149} \sqrt{161}})$
B

$C o s^{- 1} (\frac{65}{\sqrt{149} \sqrt{161}})$
C

$C o s^{- 1} (\frac{15}{\sqrt{149} \sqrt{161}})$
D

$C o s^{- 1} (\frac{3}{\sqrt{149} \sqrt{161}})$

Solution

In the figure E is mid point of AC and BD
$∴ E = (- 2, 2, \frac{9}{2})$
Since it is also midpoint BD, we have $D = (- 4, - 2, 9)$
D.r’s of AC are (12, 2, 1)
D.c’s of AC are $(\frac{12}{\sqrt{149}}, \frac{2}{\sqrt{149}}, \frac{1}{\sqrt{149}})$
D.r’s of BD are (-4, -8, 9)
D.c’s of BD are $(\frac{- 4}{\sqrt{161}}, \frac{- 8}{\sqrt{161}}, \frac{9}{\sqrt{161}})$
Angle between diagonals $θ = C o s^{- 1} (\frac{55}{\sqrt{149} \sqrt{161}})$

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