Question

Find the angle between the vectors. (first find an exact expression and then approximate to the nearest degree.)a = 5, −1, 6, b = −2, 6, 3

Answer 1

the points given in the question is [5,-1,6] and [-2,6,3]

the position vectors associated with these points are given as

a=5i^-j^+6k^,b= -2i^+6j^+3k^

the dot product of these vectors are given as -
`a.b=ab*cos\alpha`

where α is the angle between them .

hence we have cosα=
`(a.b)/(ab)`

here
`a=√(5^2+[-1]^2+6^2)`

and
`b=√([-2]^2+6^2+3^2)`

hence a=
`√(62)` and b=
`√(49)`

hence cosα=
`(a.b)/(ab)`

=5×[-2]+[-1]×6+6×3
`(1)/(√(62) *7)`

=
`(2)/(√(62)*7 )`

=0.0362857507

hence α=2.07947 degree