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(B) The direction vector of line $L$ is perpendicular to the direction vectors of the two given lines,$\vec{v_1} = (2, 1, -2)$ and $\vec{v_2} = (1, 3,

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) The direction vector of line $L$ is perpendicular to the direction vectors of the two given lines,$\vec{v_1} = (2, 1, -2)$ and $\vec{v_2} = (1, 3, 4)$.Thus,the direction vector $\vec{v}$ of $L$ is $\vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -2 \ 1 & 3 & 4 \end{vmatrix} = \hat{i}(4 - (-6)) - \hat{j}(8 - (-2)) + \hat{k}(6 - 1) = 10\hat{i} - 10\hat{j} + 5\hat{k}$.We can take the direction vector as $\vec{d} = (2, -2, 1)$.The equation of line $L$ passing through $P(2, -1, 3)$ is $\frac{x-2}{2} = \frac{y+1}{-2} = \frac{z-3}{1} = \lambda$.Any point $Q$ on line $L$ is of the form $(2\lambda + 2, -2\lambda - 1, \lambda + 3)$.Since $Q$ lies on the $yz$-plane,its $x$-coordinate must be $0$.$2\lambda + 2 = 0 \implies \lambda = -1$.Substituting $\lambda = -1$ into the coordinates of $Q$,we get $Q(0, -2(-1) - 1, -1 + 3) = Q(0, 1, 2)$.The distance $PQ = \sqrt{(2-0)^2 + (-1-1)^2 + (3-2)^2} = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.

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