The equation of the perpendicular from the po | vedclass

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Question

(B) The perpendicular line from a point $(\alpha, \beta, \gamma)$ to a plane $ax + by + cz + d = 0$ passes through the point $(\alpha, \beta, \gamma)$

Vedclass · Accepted Answer

$\frac{x - \alpha}{a} = \frac{y - \beta}{b} = \frac{z - \gamma}{c}$

Vedclass · Answer

(B) The perpendicular line from a point $(\alpha, \beta, \gamma)$ to a plane $ax + by + cz + d = 0$ passes through the point $(\alpha, \beta, \gamma)$.Since the line is perpendicular to the plane,its direction ratios are the same as the normal vector of the plane,which are $(a, b, c)$.The equation of a line passing through $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is given by $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$.Substituting the given point,the equation is $\frac{x - \alpha}{a} = \frac{y - \beta}{b} = \frac{z - \gamma}{c}$.

The equation of the perpendicular from the point $(\alpha, \beta, \gamma)$ to the plane $ax + by + cz + d = 0$ is

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