Question

A parallelogram is determined by the vectors a = (-2,5) and b = (3,2). Determined the angles between the diagonals of the parallelogram?

Answer 1

Using the Law of Cosines

Explanation:

As Stefan indicated, the quick method is to use the Law of Cosines or Dot Product.
`||<-2, 5>|| = sqrt(29)`
`||<3, 2>|| = sqrt(13)`

Find the two diagonals first:
<-2, 5> + <3, 2> = <1, 7>
`||<1, 7>|| = sqrt(50) = 5sqrt(2)`

<-2, 5> - <3, 2> = <-5, 3>
`||<-5, 3>|| = sqrt(34)`

A triangle is formed by half of the lengths of the two diagonals, and either of the lengths of the original vectors.

Using the Law of Cosines...
`a^2 = b^2 + c^2 - 2abcos(alpha)`

In this case...
`13 = 34/4 + 50/4 - 2(sqrt(34)/2)(sqrt(50)/2)cos(alpha)`

`13 = 21 - (sqrt(34))(5sqrt(2)/2)cos(alpha)`

`13 = 21 - 5sqrt(17)cos(alpha)`

`cos(alpha) = 8/(5sqrt(17))`

`alpha = arccos(8/(5sqrt(17)))`

In radians `alpha ~~ 1.7227`.

The other angle is `pi - alpha ~~ 1.96932`

If you prefer degrees, multiply the radian angles by `180/pi`.