Question

A, B, C, D are the points i − k, − i + 2 j, 2i − 3k, 3i − 2 j − k respectively. Show that the
projection of AB on CD is equal to that of CD on AB . Also find the cosine of their inclination.

Answer 1

1.

A(1;0;-1)

B(-1;2;0)

C(2;0;-3)

D(3;-2;-1)

Then

$\overrightarrow{AB}=(-1-1;2-0;0+1)=(-2;2;1)$

$$\overrightarrow{CD}=(3-2;-2-0;-1+3)=(1;-2;2)$$

The projection AB on CD is:

$$\overrightarrow{AB}\cdot\overrightarrow{CD}/|CD|=(-2-4+2)/\sqrt{1²+(-2)²+2²}=-4/3$$

The projection CD on AB is:

$$\overrightarrow{CD}\cdot\overrightarrow{AB}/|AB|=(-2-4+2)/\sqrt{(-2)²+2²+1²}=-4/3$$

Then the projection AB on CD is equal to the projection CD on AB

2.

$$cos(\overrightarrow{AB},\overrightarrow{CD})=\overrightarrow{AB}\cdot\overrightarrow{CD}/(|AB|\cdot|CD|)=$$

$$=(-2-4+2)/\sqrt{1²+(-2)²+2²}\cdot\sqrt{(-2)²+2²+1²}=-4/9$$