The direction cosines of the line which is pe | vedclass

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(B) Let the direction ratios of the two lines be $\vec{v_1} = (2, -3, 1)$ and $\vec{v_2} = (1, 2, -2)$.Since the required line is perpendicular to bot

Vedclass · Accepted Answer

$\pm \frac{4}{\sqrt{90}}, \pm \frac{5}{\sqrt{90}}, \pm \frac{7}{\sqrt{90}}$

Vedclass · Answer

(B) Let the direction ratios of the two lines be $\vec{v_1} = (2, -3, 1)$ and $\vec{v_2} = (1, 2, -2)$.Since the required line is perpendicular to both lines,its direction ratios $(a, b, c)$ are given by the cross product $\vec{v_1} 	imes \vec{v_2}$.$\vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -3 & 1 \ 1 & 2 & -2 \end{vmatrix} = \hat{i}(6-2) - \hat{j}(-4-1) + \hat{k}(4+3) = 4\hat{i} + 5\hat{j} + 7\hat{k}$.Thus,the direction ratios are $(4, 5, 7)$.The magnitude is $\sqrt{4^2 + 5^2 + 7^2} = \sqrt{16 + 25 + 49} = \sqrt{90}$.The direction cosines are $\pm \frac{4}{\sqrt{90}}, \pm \frac{5}{\sqrt{90}}, \pm \frac{7}{\sqrt{90}}$.

The direction cosines of the line which is perpendicular to the lines $\frac{x-7}{2}=\frac{y+17}{-3}=\frac{z-6}{1}$ and $\frac{x+5}{1}=\frac{y+3}{2}=\frac{z-6}{-2}$ are

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