Let L_1: frac{x+2}{5}=frac{y-3}{2}=frac{z-6}{ | vedclass

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(B) Lines $L_1$ and $L_2$ are parallel to the vectors $\vec{b}_1 = 5\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b}_2 = 4\hat{i} + 3\hat{j} + 5\hat{k}$ res

Vedclass · Accepted Answer

$\frac{\hat{i}-3 \hat{j}+\hat{k}}{\sqrt{11}}$

Vedclass · Answer

(B) Lines $L_1$ and $L_2$ are parallel to the vectors $\vec{b}_1 = 5\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b}_2 = 4\hat{i} + 3\hat{j} + 5\hat{k}$ respectively.The unit vector perpendicular to both $L_1$ and $L_2$ is given by $\hat{n} = \pm \frac{\vec{b}_1 	imes \vec{b}_2}{|\vec{b}_1 	imes \vec{b}_2|}$.Calculating the cross product: $\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 5 & 2 & 1 \ 4 & 3 & 5 \end{vmatrix} = \hat{i}(10-3) - \hat{j}(25-4) + \hat{k}(15-8) = 7\hat{i} - 21\hat{j} + 7\hat{k}$.The magnitude is $|\vec{b}_1 	imes \vec{b}_2| = \sqrt{7^2 + (-21)^2 + 7^2} = \sqrt{49 + 441 + 49} = \sqrt{539} = 7\sqrt{11}$.Thus,$\hat{n} = \pm \frac{7(\hat{i} - 3\hat{j} + \hat{k})}{7\sqrt{11}} = \pm \frac{\hat{i} - 3\hat{j} + \hat{k}}{\sqrt{11}}$.Comparing with the given options,option $B$ is correct.

Let $L_1: \frac{x+2}{5}=\frac{y-3}{2}=\frac{z-6}{1}$ and $L_2: \frac{x-3}{4}=\frac{y+2}{3}=\frac{z-3}{5}$ be the given lines. Then the unit vector perpendicular to both $L_1$ and $L_2$ is

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