The equation of the line passing through (-4, | vedclass

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(C) Let the required line $L$ pass through $P(-4, 1, 3)$ with direction ratios $(a, b, c)$.The equation of the line is $\frac{x + 4}{a} = \frac{y - 1}

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the required line $L$ pass through $P(-4, 1, 3)$ with direction ratios $(a, b, c)$.The equation of the line is $\frac{x + 4}{a} = \frac{y - 1}{b} = \frac{z - 3}{c}$.Since $L$ is parallel to the plane $x + 2y - z - 5 = 0$,the normal to the plane $(1, 2, -1)$ is perpendicular to the line $L$. Thus,$a + 2b - c = 0$.Let the line $L$ intersect the given line $L_1: \frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z - 2}{-1}$ at point $Q$. The vector $\vec{PQ}$ must be perpendicular to the normal of the plane and the direction of $L_1$. The vector $\vec{PQ}$ is $(x_Q - (-4), y_Q - 1, z_Q - 3)$.Alternatively,for the line to intersect $L_1$ and be parallel to the plane,the direction vector $(a, b, c)$ must satisfy the condition that the scalar triple product of the direction of $L_1$ $(-3, 2, -1)$,the normal to the plane $(1, 2, -1)$,and the vector connecting the points $(-4, 1, 3)$ and $(-1, 3, 2)$ is zero.The vector connecting the points is $(-1 - (-4), 3 - 1, 2 - 3) = (3, 2, -1)$.The condition for intersection is $\begin{vmatrix} 3 & 2 & -1 \ -3 & 2 & -1 \ 1 & 2 & -1 \end{vmatrix} = 0$. Since this is not zero,we solve for $(a, b, c)$ using $a + 2b - c = 0$ and the condition that the line lies in the plane containing $L_1$ and parallel to the given plane.Solving the system,we get the direction ratios $(3, -1, 1)$.Thus,the equation is $\frac{x + 4}{3} = \frac{y - 1}{-1} = \frac{z - 3}{1}$.

The equation of the line passing through $(-4, 1, 3)$,parallel to the plane $x + 2y - z - 5 = 0$ and intersecting the line $\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z - 2}{-1}$ is

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