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(D) The direction of motion of particle $P$ is normal to the plane containing $\vec{A}$ and $\vec{B}$. Thus,the unit vector $\hat{v}_1$ is given by $\

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) The direction of motion of particle $P$ is normal to the plane containing $\vec{A}$ and $\vec{B}$. Thus,the unit vector $\hat{v}_1$ is given by $\hat{v}_1 = \pm \frac{\vec{A} 	imes \vec{B}}{|\vec{A} 	imes \vec{B}|}$.Calculating the cross product: $\vec{A} 	imes \vec{B} = (\hat{i} + \hat{j}) 	imes (\hat{j} + \hat{k}) = \hat{i} 	imes \hat{j} + \hat{i} 	imes \hat{k} + \hat{j} 	imes \hat{j} + \hat{j} 	imes \hat{k} = \hat{k} - \hat{j} + 0 + \hat{i} = \hat{i} - \hat{j} + \hat{k}$.The magnitude is $|\vec{A} 	imes \vec{B}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$.So,$\hat{v}_1 = \pm \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}}$.The direction of motion of particle $Q$ is normal to the plane containing $\vec{A}$ and $\vec{C}$. Thus,$\hat{v}_2 = \pm \frac{\vec{A} 	imes \vec{C}}{|\vec{A} 	imes \vec{C}|}$.Calculating the cross product: $\vec{A} 	imes \vec{C} = (\hat{i} + \hat{j}) 	imes (-\hat{i} + \hat{j}) = \hat{i} 	imes (-\hat{i}) + \hat{i} 	imes \hat{j} + \hat{j} 	imes (-\hat{i}) + \hat{j} 	imes \hat{j} = 0 + \hat{k} + \hat{k} + 0 = 2\hat{k}$.The magnitude is $|\vec{A} 	imes \vec{C}| = 2$.So,$\hat{v}_2 = \pm \frac{2\hat{k}}{2} = \pm \hat{k}$.The angle $	heta$ between $\hat{v}_1$ and $\hat{v}_2$ is given by $\cos 	heta = |\hat{v}_1 \cdot \hat{v}_2| = |\pm \frac{1}{\sqrt{3}} \hat{k} \cdot \hat{k}| = \frac{1}{\sqrt{3}}$.Comparing this with $\cos 	heta = \frac{1}{\sqrt{x}}$,we get $x = 3$.

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