For the components of a vector vec{A} = (3 ha | vedclass

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(A) The component of a vector $\vec{A}$ along a direction defined by unit vector $\hat{n}$ is given by $\vec{A} \cdot \hat{n}$.$(A)$ Component along $

Vedclass · Accepted Answer

$(A \rightarrow q, B \rightarrow r, C \rightarrow s, D \rightarrow s)$

Vedclass · Answer

(A) The component of a vector $\vec{A}$ along a direction defined by unit vector $\hat{n}$ is given by $\vec{A} \cdot \hat{n}$.$(A)$ Component along $x$-axis: $\vec{A} \cdot \hat{i} = (3 \hat{i} + 4 \hat{j} - 5 \hat{k}) \cdot \hat{i} = 3$. This is not in the options,so $(A \rightarrow s)$.$(B)$ Component along $\vec{B} = (2 \hat{i} + \hat{j} + 2 \hat{k})$: $\vec{A} \cdot \frac{\vec{B}}{|B|} = \frac{(3)(2) + (4)(1) + (-5)(2)}{\sqrt{2^2 + 1^2 + 2^2}} = \frac{6 + 4 - 10}{3} = 0$. Thus,$(B \rightarrow r)$.$(C)$ Component along $\vec{C} = (6 \hat{i} + 8 \hat{j} - 10 \hat{k}) = 2\vec{A}$. Since it is parallel,the component is the magnitude $|\vec{A}| = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{50} = 5\sqrt{2}$. This is not in the options,so $(C \rightarrow s)$.$(D)$ Component along $\vec{D} = (-3 \hat{i} - 4 \hat{j} + 5 \hat{k}) = -\vec{A}$. Since it is anti-parallel,the component is $-|\vec{A}| = -5\sqrt{2}$. This is not in the options,so $(D \rightarrow s)$.Therefore,the correct matching is $(A \rightarrow s, B \rightarrow r, C \rightarrow s, D \rightarrow s)$.

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

Column $I$	Column $II$
$(A)$ $\theta=60^{\circ}$	$(p)$ $n=\sqrt{3}$
$(B)$ $\theta=90^{\circ}$	$(q)$ $n=1$
$(C)$ $\theta=120^{\circ}$	$(r)$ $n=\sqrt{2}$
$(D)$ $\theta=180^{\circ}$	$(s)$ $n=2$

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