Question
Find the difference of the vectors $ \vec{v_1} = \left(\dfrac{ 3 }{ 4 },~2\right) $ and $ \vec{v_2} = \left(3,~-2\right) $ .
Answer
$$ v_1 - v_2 = \left(-\dfrac{ 9 }{ 4 },~4\right) $$
Explanation
To find the difference $ v $ of the vectors $ v_1=(a_1, b_1) $ and $ v_2=(a_2, b_2) $ we use formula
$$ v = (a_1 - a_2, b_1 - b_2) $$In this example $ a_1 = \frac{ 3 }{ 4 } $ , $ b_1 = 2 $, $ a_2 = 3 $ and $ b_2 = -2 $, therefore:
$$ \begin{aligned}v &= (a_1 - a_2, b_1 - b_2) \\v &= \left(\frac{ 3 }{ 4 } - 3, 2 +2 \right) \\v &= \left(-\dfrac{ 9 }{ 4 },~4\right) \end{aligned} $$This page was created using
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