The Cartesian equation of the line passing th | vedclass

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(C) Let the given vectors be $\vec{a} = \hat{i}-2\hat{j}+\hat{k}$ and $\vec{b} = 2\hat{i}+\hat{j}-\hat{k}$.Since the line is perpendicular to both vec

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$\frac{x+3}{1}=\frac{y}{3}=\frac{z-1}{5}$

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(C) Let the given vectors be $\vec{a} = \hat{i}-2\hat{j}+\hat{k}$ and $\vec{b} = 2\hat{i}+\hat{j}-\hat{k}$.Since the line is perpendicular to both vectors,its direction vector $\vec{v}$ is given by the cross product $\vec{a} 	imes \vec{b}$.$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 1 \ 2 & 1 & -1 \end{vmatrix} = \hat{i}(2-1) - \hat{j}(-1-2) + \hat{k}(1+4) = 1\hat{i} + 3\hat{j} + 5\hat{k}$.The line passes through $(-3, 0, 1)$ with direction ratios $(1, 3, 5)$.The Cartesian equation is $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$.Substituting the values,we get $\frac{x - (-3)}{1} = \frac{y - 0}{3} = \frac{z - 1}{5}$,which simplifies to $\frac{x+3}{1} = \frac{y}{3} = \frac{z-1}{5}$.

The Cartesian equation of the line passing through the point $(-3,0,1)$ and perpendicular to vectors $\hat{i}-2\hat{j}+\hat{k}$ and $2\hat{i}+\hat{j}-\hat{k}$ is

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