The vector having magnitude of 2 sqrt{29} uni | vedclass

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(B) Step $1$: Find the magnitude of vector $\vec{a} = 4 \hat{i} + 3 \hat{j} - 2 \hat{k}$.$|\vec{a}| = \sqrt{4^2 + 3^2 + (-2)^2} = \sqrt{16 + 9 + 4} =

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$8 \hat{i} + 6 \hat{j} - 4 \hat{k}$

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(B) Step $1$: Find the magnitude of vector $\vec{a} = 4 \hat{i} + 3 \hat{j} - 2 \hat{k}$.$|\vec{a}| = \sqrt{4^2 + 3^2 + (-2)^2} = \sqrt{16 + 9 + 4} = \sqrt{29}$.Step $2$: Find the unit vector in the direction of $\vec{a}$,denoted by $\hat{a}$.$\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{4 \hat{i} + 3 \hat{j} - 2 \hat{k}}{\sqrt{29}}$.Step $3$: The required vector with magnitude $2 \sqrt{29}$ in the direction of $\vec{a}$ is given by $2 \sqrt{29} 	imes \hat{a}$.Vector $= 2 \sqrt{29} 	imes \left( \frac{4 \hat{i} + 3 \hat{j} - 2 \hat{k}}{\sqrt{29}} ight) = 2(4 \hat{i} + 3 \hat{j} - 2 \hat{k}) = 8 \hat{i} + 6 \hat{j} - 4 \hat{k}$.

The vector having magnitude of $2 \sqrt{29}$ units in the direction of vector $\vec{a} = 4 \hat{i} + 3 \hat{j} - 2 \hat{k}$ is . . . . . . .

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