A unit vector which is in the opposite direct | vedclass

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(A) Let the given vector be $\vec{a} = (2, -2, 1)$.First,find the magnitude of $\vec{a}$:$|\vec{a}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \

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$\left(\frac{-2}{3}, \frac{2}{3}, \frac{-1}{3}\right)$

Vedclass · Answer

(A) Let the given vector be $\vec{a} = (2, -2, 1)$.First,find the magnitude of $\vec{a}$:$|\vec{a}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.The unit vector in the direction of $\vec{a}$ is $\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \left(\frac{2}{3}, \frac{-2}{3}, \frac{1}{3}\right)$.The unit vector in the opposite direction of $\vec{a}$ is $-\hat{a} = -\left(\frac{2}{3}, \frac{-2}{3}, \frac{1}{3}\right) = \left(\frac{-2}{3}, \frac{2}{3}, \frac{-1}{3}\right)$.

$A$ unit vector which is in the opposite direction of $(2, -2, 1)$ is $........$

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