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

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Answer 1 · 2017-09-19T09:41:14

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This is the sum or difference of $2$ $2$ vectors and multiplication by a scalar

$\to u = - 2 ˆ i + ˆ j$ $vecu=-2hati+hatj$

$\to v = - ˆ i + 2 ˆ j$ $vecv=-hati+2hatj$

Therefore,

$\to u + \to v = (- 2 ˆ i + ˆ j) + (- ˆ i + 2 ˆ j) = - 3 ˆ i + 3 ˆ j$ $vecu + vecv =(-2hati+hatj)+(-hati+2hatj)=-3hati+3hatj$

$\to u - \to v = (- 2 ˆ i + ˆ j) - (- ˆ i + 2 ˆ j) = - ˆ i - ˆ j$ $vecu - vecv =(-2hati+hatj)-(-hati+2hatj)=-hati-hatj$

$2 \to u - 3 \to v = 2 (- 2 ˆ i + ˆ j) - 3 (- ˆ i + 2 ˆ j) = - ˆ i - 4 ˆ j$ $2vecu - 3vecv =2(-2hati+hatj)-3(-hati+2hatj)= -hati-4hatj$

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