First slide

Planes in 3D

Question

$The distance of the plane passing through the point P (1, 1, 1) and perpendicular to the line$ $\frac{x - 1}{3} = \frac{y - 1}{0} = \frac{z - 1}{4} from the origin is$

Moderate

A

$\frac{3}{4}$
B

$\frac{4}{3}$
C

$\frac{7}{5}$
D

$1$

Solution

$Equation of the plane is a (x - 1) + b (y - 1) + c (z - 1) = 0 \dots \dots . (1)$
Since the line is perpendicular to the plane (1)
$\begin{array}{l} ∴ 3 (x - 1) + 0 (y - 1) + 4 (z - 1) = 0 \\ \Rightarrow 3 x + 0 y + 4 z - 7 = 0 \dots \dots (2) \end{array}$
Distance from origin (0, 0, 0) to equation (2) is
$d = \frac{|a x_{1} + b y_{1} + c z_{1} + d|}{\sqrt{a^{2} + b^{2} + c^{2}}}$
$\begin{array}{l} ∴ d = |\frac{- 7}{5}| = \frac{7}{5} \\ Therefore, the correct answer is (3). \end{array}$

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