Question

approximately what is the angle between two position vectors if their terminal points are (1,5) and (6,3)​

Answer 1

The angle between them = 52.1°

Explanation:

∵ The position vector of the first point is
`\left[\begin{array}{ccc}1\\5\end{array}\right]`

∵ The position vector of the second point is
`\left[\begin{array}{ccc}6\\3\\\end{array}\right]`

∵ The magnitude of the first = √(1²+5²) = √26

∵ The magnitude of the second = √(6²+3²) = √45

∵ The scalar product of them = (1 × 6) + (5 × 3) = 6 + 15 = 21

∵ cosФ = scalar product/(magnitude 1st × magnitude 2nd)

∴ cosФ = 21/(√26 × √45) = 0.61394

∴ Ф = 52.1°