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(A) The equations of the two planes are $x+2y=0$ and $3y-z=0$.Let $\vec{n}_{1}$ and $\vec{n}_{2}$ be the normals to the two planes,respectively.$\vec{

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The equations of the two planes are $x+2y=0$ and $3y-z=0$.Let $\vec{n}_{1}$ and $\vec{n}_{2}$ be the normals to the two planes,respectively.$\vec{n}_{1} = \hat{i} + 2\hat{j} + 0\hat{k}$ and $\vec{n}_{2} = 0\hat{i} + 3\hat{j} - \hat{k}$.Since the required line is parallel to both planes,it must be parallel to the cross product of their normals,$\vec{b} = \vec{n}_{1} 	imes \vec{n}_{2}$.$\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 0 \ 0 & 3 & -1 \end{vmatrix} = \hat{i}(-2 - 0) - \hat{j}(-1 - 0) + \hat{k}(3 - 0) = -2\hat{i} + \hat{j} + 3\hat{k}$.The line passes through the point $(3, 0, 1)$.The equation of the line in symmetric form is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.Substituting the point $(3, 0, 1)$ and the direction vector $(-2, 1, 3)$,we get $\frac{x-3}{-2} = \frac{y-0}{1} = \frac{z-1}{3}$.

Find the equation of the line passing through the point $(3, 0, 1)$ and parallel to the planes $x+2y=0$ and $3y-z=0$.

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