A particle is moving with speed 6 , m/s along | vedclass

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(B) The velocity vector $\vec{v}$ is given by the product of its speed and the unit vector in the direction of motion.First,find the unit vector $\hat

Vedclass · Accepted Answer

$(4\hat{i} + 4\hat{j} - 2\hat{k}) \, m/s$

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(B) The velocity vector $\vec{v}$ is given by the product of its speed and the unit vector in the direction of motion.First,find the unit vector $\hat{A}$ in the direction of $\vec{A} = 2\hat{i} + 2\hat{j} - \hat{k}$:$|\vec{A}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.Thus,$\hat{A} = \frac{\vec{A}}{|\vec{A}|} = \frac{2\hat{i} + 2\hat{j} - \hat{k}}{3}$.The velocity vector is $\vec{v} = |\vec{v}| \hat{A} = 6 \left( \frac{2\hat{i} + 2\hat{j} - \hat{k}}{3} \right)$.Simplifying the expression:$\vec{v} = 2(2\hat{i} + 2\hat{j} - \hat{k}) = 4\hat{i} + 4\hat{j} - 2\hat{k} \, m/s$.

Column $I$	Column $II$
$(A)$ Component along $x$-axis	$(p)$ $5 \text{ unit}$
$(B)$ Component along vector $(2 \hat{i} + \hat{j} + 2 \hat{k})$	$(q)$ $4 \text{ unit}$
$(C)$ Component along $(6 \hat{i} + 8 \hat{j} - 10 \hat{k})$	$(r)$ $0$
$(D)$ Component along $(-3 \hat{i} - 4 \hat{j} + 5 \hat{k})$	$(s)$ None

$A$ particle is moving with speed $6 \, m/s$ along the direction of $\vec{A} = 2\hat{i} + 2\hat{j} - \hat{k}$. What is its velocity vector?

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