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

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

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Answer 1 · 2017-02-05T14:59:19

$a) u + v = \sqrt{13}, tan α = \frac{3}{2}$ $a)" "u+v=sqrt(13)" , "tan alpha=3/2$
$b) u - v = \sqrt{13}, tan β = - \frac{3}{2}$ $b)" "u-v=sqrt(13)" , "tan beta=-3/2$
$c) 2 u - 3 v = 6 \sqrt{2}, tan θ = - 1$ $c)" "2u-3v=6sqrt(2)" , "tan theta=-1$

Explanation:

$the vector u is shown in figure below$ $"the vector u is shown in figure below"$

enter image source here

$the vector v is shown in the diagram below$ $"the vector v is shown in the diagram below"$

enter image source here

$a)the vector w=u+v is shown in the diagram below$ $"a)the vector w=u+v is shown in the diagram below"$

$w = 2 . i + 3 . j$ $w=2.i+3.j$

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

enter image source here

$a)the vector z=u-v is shown in the diagram below$ $"a)the vector z=u-v is shown in the diagram below"$

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

enter image source here

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

enter image source here

$The vector 3v is shown in the figure below$ $"The vector 3v is shown in the figure below"$

enter image source here

$p=2u-3v is shown in the figure below$ $"p=2u-3v is shown in the figure below"$

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

enter image source here

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

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

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