The Cartesian equation of the line passing th | vedclass

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(D) Let the direction ratios of the required line be $(a, b, c)$.The line passes through the point $P(-1, 3, -2)$.The direction ratios of the given li

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$\frac{x+1}{2} = \frac{y-3}{-7} = \frac{z+2}{4}$

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(D) Let the direction ratios of the required line be $(a, b, c)$.The line passes through the point $P(-1, 3, -2)$.The direction ratios of the given lines are $\vec{v_1} = (1, 2, 3)$ and $\vec{v_2} = (-3, 2, 5)$.Since the required line is perpendicular to both given lines,its direction vector $\vec{v} = (a, b, c)$ must be parallel to the cross product $\vec{v_1} 	imes \vec{v_2}$.$\vec{v} = \vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ -3 & 2 & 5 \end{vmatrix} = \hat{i}(10 - 6) - \hat{j}(5 - (-9)) + \hat{k}(2 - (-6)) = 4\hat{i} - 14\hat{j} + 8\hat{k}$.Dividing by $2$,we get the direction ratios as $(2, -7, 4)$.The equation of the line passing through $(-1, 3, -2)$ with direction ratios $(2, -7, 4)$ is $\frac{x - (-1)}{2} = \frac{y - 3}{-7} = \frac{z - (-2)}{4}$,which simplifies to $\frac{x+1}{2} = \frac{y-3}{-7} = \frac{z+2}{4}$.

The Cartesian equation of the line passing through the point $(-1, 3, -2)$ and perpendicular to the lines $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ and $\frac{x+2}{-3} = \frac{y-1}{2} = \frac{z+1}{5}$ is

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