The unit vector in the direction of the vecto | vedclass

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Question

(A) Let $\vec{a} = (2, 2, -1)$.The magnitude of the vector $\vec{a}$ is given by $|\vec{a}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9}

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $\vec{a} = (2, 2, -1)$.The magnitude of the vector $\vec{a}$ is given by $|\vec{a}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.The unit vector in the direction of $\vec{a}$ is denoted by $\hat{a}$ and is calculated as $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$.Substituting the values,we get $\hat{a} = \frac{1}{3}(2, 2, -1) = \left(\frac{2}{3}, \frac{2}{3}, \frac{-1}{3}\right)$.

The unit vector in the direction of the vector $\vec{a} = (2, 2, -1)$ is $......$

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