Question

If `vec"a", vec"b", vec"c"` are three vectors such that `vec"a" + 2vec"b" + vec"c"` = 0 and `|vec"a"| = 3, |vec"b"| = 4, |vec"c"| = 7`, find the angle between `vec"a"` and `vec"b"`

Answer 1

`vec"a" + 2vec"b" + vec"c"` = 0

 `|vec"a"| = 3, |vec"b"| = 4, |vec"c"| = 7`

`vec"a" + 2vec"b" = -vec"c"`

Squaring on both sides,

`(vec"a" + 2vec"b")^2 = (- vec"c")^2`

`vec"a"^2 + 4vec"b"^2 + 2vec"a" * 2vec"b" = vec"c"^2`

`|vec"a"|^2 + 4|vec"b"|^2 + 4vec"a" * vec"b" = |vec"c"|^2`

`3^2 + 4 xx 4^2 + 4 |vec"a"||vec"b"| cos theta = 7^2`

`9 + 64 + 4 xx 3 xx 4 cos theta` = 49

48 cos θ = 49 – 73

48 cos θ = – 24

cos θ = `- 24/48`

cos θ = `- 1/2`

cos θ = = `- cos  pi/3`

cos θ = `cos(pi - pi/3)`

= `cos((3pi- pi)/3)`

cos θ = `cos  ((2pi)/3)`

θ = `(2pi)/3`