Question
Find the magnitude of the vector $ \| \vec{v} \| = \left(4,~2,~-\dfrac{ 3 }{ 2 }\right) $ .
Answer
The magnitude of vector $ \vec{v} $ is $ \| \vec{v} \| = \dfrac{\sqrt{ 89 }}{ 2 } $ .
Explanation
To find magnitude of a vector $ v=(a,b,c) $ we use formula $ ||v|| = \sqrt{a^2+b^2+c^2} $
In this example $ a = 4 $ , $ b = 2 $ and $ c = -\frac{ 3 }{ 2 } $ so:
$$ ||v|| = \sqrt{ 4^2 + 2^2 + \left(-\frac{ 3 }{ 2 }\right)^2} = \sqrt{ 16 + 4 + \frac{ 9 }{ 4 }} = \sqrt{ \frac{ 89 }{ 4 } } = \frac{\sqrt{ 89 }}{ 2 }$$This page was created using
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