Answer
(a) $\theta = 124.68$ degrees
(b) $\theta = 67.38$ degrees
Work Step by Step
(a) $\vec A =3\vec i + 4 \vec j $
$\vec B =2 \vec i - 6 \vec j $
$ \vec A. \vec B = |A||B|cos\theta $
$cos\theta = \frac { \vec A. \vec B }{ |A||B|}$
$|A| =\sqrt {{3^2}+{ 4^2}} = \sqrt{25} =5$
$|B| = \sqrt{{2^2} + {6^2}} = \sqrt{40} = 2\sqrt{10}$
$ \vec A. \vec B = (3\vec i + 4 \vec j ).(2 \vec i - 6 \vec j )$
$ \vec A. \vec B = 6 -24 = - 18$
$cos\theta = \frac { \vec A. \vec B }{ |A||B|}$
$cos\theta = \frac { -18}{5×2 \sqrt{10}}$
$cos\theta= -0.569$
$\theta =cos^{-1}(-0.569)$
$\theta = 124.68 degree$
(b) $\vec A =3\vec i -2 \vec j $
$\vec B =6 \vec i + 4 \vec j $
$ \vec A. \vec B = |A||B|cos\theta $
$cos\theta = \frac { \vec A. \vec B }{ |A||B|}$
$|A| =\sqrt {{3^2}+{ 2^2}} = \sqrt{13} $
$|B| = \sqrt{{6^2} + {4^2}} = \sqrt{52} = 2\sqrt{13}$
$ \vec A. \vec B = (3\vec i -2 \vec j ).(6 \vec i +4 \vec j )$
$ \vec A. \vec B = 18- 8 = 10$
$cos\theta = \frac { \vec A. \vec B }{ |A||B|}$
$cos\theta = \frac { 10}{\sqrt{13}×2 \sqrt{13}} $
$cos\theta= \frac{5}{13}$
$\theta =cos^{-1}(\frac{5}{13})$
$\theta = 67.38$ degree
