Three vectors `vecA,vecB` and `vecC` are such that `vecA=vecB+vecC` and their magnitudes are in ratio 5:4:3 respectively. Find angle between vector `vecA` and `vecC`

Explanation

Given that : `vecA= vecB+ vecC rArr vecA-vecC = vecB`
By taking self dot product on both sides `(vecA- vecC)*( vecA-vecC) = vecB*vecB rArr A^(2) + C^(2) - 2vecA*vecC= B^(2) `
Now let angle between `vecA and vecC` be `theta ` then `A^(2) + C^(2) - 2AC cos theta = B^(2)`
`therefore cos theta = (A^(2) + C^(2) -B^(2))/(2AC) = ((5)^(2)+ (3)^(2) - (4)^(2))/(2(5)(3))= (18)/(30) = (3)/(5) rArr theta = cos ^(-1)((3)/(5))= 53^(@)`
`" "OR" "`
Since `5^(2) = 4^(2) + 3^(2) ` the vectors `vecA, vecB and vecC` with `vecA= vecB+ vecC`, make a triangle with angle between `vecB and vecC` as `90^(@)`. If `theta` is the angle between `vecA and vecC`, then `costheta= (3)/(5)" "therefore theta = 53^(@)`