The equation of the line,passing through (1, | vedclass

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(A) The line passes through the point $(1, 2, 3)$ and is parallel to the planes $x - y + 2z = 5$ and $3x + y + z = 6$. Let the normal vectors to the p

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The line passes through the point $(1, 2, 3)$ and is parallel to the planes $x - y + 2z = 5$ and $3x + y + z = 6$. Let the normal vectors to the planes be $\vec{n_1} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{n_2} = 3\hat{i} + \hat{j} + \hat{k}$. The direction vector $\vec{v}$ of the line is parallel to the cross product of the normals: $\vec{v} = \vec{n_1} 	imes \vec{n_2}$. $\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 2 \ 3 & 1 & 1 \end{vmatrix} = \hat{i}(-1 - 2) - \hat{j}(1 - 6) + \hat{k}(1 + 3) = -3\hat{i} + 5\hat{j} + 4\hat{k}$. The equation of the line passing through $(x_1, y_1, z_1)$ with direction vector $(a, b, c)$ is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$. Substituting the values,we get $\frac{x-1}{-3} = \frac{y-2}{5} = \frac{z-3}{4}$.

The equation of the 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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