Answer to Question #87784 in Analytic Geometry for Shivam Nishad

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

$$\overrightarrow{AB} =(-1-1)i+(2-0)j+(0+1)k=-2i+2j+k$$

$$\overrightarrow{CD} =(3-2)i+(-2-0)j+(-1+3)k=1i-2j+2k$$

Then projection AB on CD is

$$\overrightarrow{AB}\cdot\overrightarrow{CD} /|CD|=(-2-4+2)/\sqrt{(-2)²+2²+1}$$

$$=-4/3$$

And projection CD on AB is

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

$$=-4/3$$

Then projection AB on CD is equal CD on AB

2.

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

$$=(-2-4+2)/(\sqrt{1+(-2)²+2²}\cdot\sqrt{1+(-2)²+2²})=$$

$$(−2−4+2)/((1+(−2)²+2²)\cdot(1+(−2)²+2²))$$ $$=-4/9$$