Question

find the angles between the pairs of vectors A = -2.00i + 6.00j and B = 2.00i - 3.00j

Answer 1

\(A.B = |A||B|\cos(\theta)\)

\(
A.B = (-2.00i + 6.00j). (2.00i - 3.00j) = -2*2 + 6*(-3) = -4 - 18 = -22\\
|A| = \sqrt{(-2)^2+(6)^2} = \sqrt{40}=2\sqrt{10} \\
|B| = \sqrt{(2)^2+(-3)^2} = \sqrt{13} \\
\cos(\theta) =\large\frac{-22}{2\sqrt{10}*\sqrt{13}} =
\frac{-11}{\sqrt{130}} = -0.9648\\
\theta = 164.8 \text{ degrees.}
\)

Answer 2

Thank you so so so so so so mcuh!! ;D

Answer 3

You are welcome.