If a plane passes through the point (1, 1, 1) | vedclass

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(C) The direction ratios of the line are $(3, 0, 4)$. Since the plane is perpendicular to this line,these direction ratios are the normal vector $(A,

Vedclass · Accepted Answer

$\frac{7}{5}$

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(C) The direction ratios of the line are $(3, 0, 4)$. Since the plane is perpendicular to this line,these direction ratios are the normal vector $(A, B, C)$ of the plane.The equation of a plane passing through $(x_1, y_1, z_1)$ with normal $(A, B, C)$ is $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$.Substituting the given values: $3(x - 1) + 0(y - 1) + 4(z - 1) = 0$.Simplifying,we get $3x - 3 + 4z - 4 = 0$,which is $3x + 4z - 7 = 0$.The perpendicular distance $d$ from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.Here,$A = 3, B = 0, C = 4, D = -7$.$d = \frac{|-7|}{\sqrt{3^2 + 0^2 + 4^2}} = \frac{7}{\sqrt{9 + 16}} = \frac{7}{\sqrt{25}} = \frac{7}{5}$.

If a plane passes through the point $(1, 1, 1)$ and is perpendicular to the line $\frac{x - 1}{3} = \frac{y - 1}{0} = \frac{z - 1}{4}$,then its perpendicular distance from the origin is

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