Question

If `A = <2 ,1 ,6 >`, `B = <5 ,1 ,3 >` and `C=A-B`, what is the angle between A and C?

Answer 1

<`theta = 63.8^@` ( 1 decimal place )

Explanation:

`vecC = vecA - vecB =( 2 , 1 ,6 ) - (5 , 1 ,3 ) =( - 3 , 0 , 3 )`

To calculate the angle`(theta) between vecA and vecC`

use the following formula :

`costheta =( vecA . vecC)/ (|vecA| |vecC|)`

The 'dot product' `vecA . vecC =(2 , 1 , 6 ) . ( - 3 , 0 , 3 )`

= -6 + 0 + 18 = 12

and`|vecA| = sqrt(2^2 +1^2 + 6^2) = sqrt(4 + 1 + 36 ) = sqrt41`

`|vecC| = sqrt( (- 3 )^2 +0^2 + 3^2) = sqrt(9 + 0 + 9) = sqrt18`

substituting these values into the formula :

`costheta = 12/(sqrt41 xx sqrt18) = 0.4417...`

`rArr theta = 63.8^@`