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Question

(A) The equation of a line passing through the point $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is given by $\frac{x - x_1}{a} = \frac{y - y_

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The equation of a line passing through the point $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is given by $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$.The given point is $(1, 2, 3)$,so the equation is $\frac{x - 1}{a} = \frac{y - 2}{b} = \frac{z - 3}{c}$.$A$ line is perpendicular to a plane if its direction ratios are proportional to the normal vector of the plane.The equation of the plane is $x + 2y - 5z + 9 = 0$,so the normal vector is $\vec{n} = (1, 2, -5)$.Thus,the direction ratios of the line are $(1, 2, -5)$.Substituting these into the line equation,we get $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{-5}$.

The equation of the straight line passing through $(1, 2, 3)$ and perpendicular to the plane $x + 2y - 5z + 9 = 0$ is

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