The Cartesian equation of a line passing thro | vedclass

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(B) Let the direction ratios of the line be $(a, b, c)$. Since the line is parallel to the planes $x - y + 2z = 5$ and $3x + y + z = 6$,its direction

Vedclass · Accepted Answer

$\frac{x-1}{-3} = \frac{y-2}{5} = \frac{z-3}{4}$

Vedclass · Answer

(B) Let the direction ratios of the line be $(a, b, c)$. Since the line is parallel to the planes $x - y + 2z = 5$ and $3x + y + z = 6$,its direction vector is perpendicular to the normals of both planes. The normals are $\vec{n_1} = (1, -1, 2)$ and $\vec{n_2} = (3, 1, 1)$. The direction vector $\vec{v}$ is given by the cross product $\vec{n_1} 	imes \vec{n_2}$: $\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 2 \ 3 & 1 & 1 \end{vmatrix} = \hat{i}(-1 - 2) - \hat{j}(1 - 6) + \hat{k}(1 - (-3)) = -3\hat{i} + 5\hat{j} + 4\hat{k}$. Thus,the direction ratios are $(-3, 5, 4)$. The line passes through $(1, 2, 3)$,so the Cartesian equation is $\frac{x-1}{-3} = \frac{y-2}{5} = \frac{z-3}{4}$.

The Cartesian equation of a line passing through $(1, 2, 3)$ and parallel to the planes $x - y + 2z = 5$ and $3x + y + z = 6$ is:

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