(a) Find the angle between the vectors (a = begin{pmatrix} -3 \ 4 end{pmatrix}) and (b = begin{pmatrix} -8 \ -15 end{pmatrix}).
(b) Given that (a = (4N, 060°)) and (b = (3N, 120°)), find, in component form, the unit vector along (a – b).
Explanation
(a) (a = (-3i + 4j) ; b = (-8i – 15j))
Let the angle between them be (theta).
(a cdot b = |a||b| cos theta)
(24 – 60 = (sqrt{(-3)^{2} + 4^{2}})(sqrt{(-8)^{2} + (-15)^{2}}) cos theta)
(-36 = 5 times 17 times cos theta)
(cos theta = frac{-36}{85} = -0.4235 implies theta = cos^{-1} (-0.4235))
(theta = 115.06°)
(b) (a = 4 cos 60° \ 4 sin 60° end{pmatrix} = begin{pmatrix} 2 \ 3.464 end{pmatrix})
(b = begin{pmatrix} 3 cos 120° \ 3 sin 120° end{pmatrix} = begin{pmatrix} -1.5 \ 2.598 end{pmatrix})
(a – b = begin{pmatrix} 2 – (-1.5) \ 3.464 – 2.598 end{pmatrix} = begin{pmatrix} 3.5 \ 0.866 end{pmatrix})
(|a – b| = sqrt{(3.5)^{2} + (0.866)^{2}} = sqrt{12.25 + 0.75})
= (sqrt{13})
Unit vector along a – b = (frac{3.5i + 0.866j}{sqrt{13}}).