How do you find a) u+v, b) u-v, c) 2u-3v given u=3j, v=2i?

Feb 5, 2017

a)" "u+v=sqrt(13)" , "tan alpha=3/2
b)" "u-v=sqrt(13)" , "tan beta=-3/2
c)" "2u-3v=6sqrt(2)" , "tan theta=-1

Explanation:

$\text{the vector u is shown in figure below}$

$\text{the vector v is shown in the diagram below}$

$\text{a)the vector w=u+v is shown in the diagram below}$

$w = 2. i + 3. j$

$\text{magnitude of w can be calculated :}$
$w = \sqrt{{2}^{2} + {3}^{2}}$
$w = \sqrt{4 + 9} \text{ , } w = \sqrt{13}$
$\text{direction of the vector w is}$
$\tan \alpha = \frac{3}{2}$

$\text{a)the vector z=u-v is shown in the diagram below}$

$z = - 2. i + 3. j$
$\text{magnitude of z can be calculated :}$
$w = \sqrt{{\left(- 2\right)}^{2} + {3}^{2}}$
$w = \sqrt{4 + 9} \text{ , } w = \sqrt{13}$
$\text{direction of the vector w is}$
$\tan \beta = - \frac{3}{2}$

c)"The vector 2u is shown in the figure below"

$\text{The vector 3v is shown in the figure below}$

$\text{p=2u-3v is shown in the figure below}$

$\text{magnitude of p can be calculated :}$
$p = \sqrt{{\left(- 6\right)}^{2} + {6}^{2}}$
$p = \sqrt{36 + 36} \text{ , } p = \sqrt{72} = 6 \sqrt{2}$
$\text{direction of the vector p is}$
$\tan \theta = - \frac{6}{6}$=-1