A plane passes through $(2, 3, - 1)$ and is perpendicular to the line having direction ratios $3, - 4, 7$ . The perpandicular distance from the origin to this plane is

\frac{3}{\sqrt{74}}

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\frac{5}{\sqrt{74}}

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\frac{6}{\sqrt{74}}

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\frac{13}{\sqrt{74}}

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Solution

The correct option is C $\frac{13}{\sqrt{74}}$
Equation of plane :
$\begin{matrix} 3 (x - 2) - 4 (y - 3) + 7 (z + 1) = 0 3 x - 4 y + 7 z + 7 + 12 - 6 = 0 3 x - 4 y + 7 z + 13 = 0 \end{matrix}$
Perpendicular distance from origin $d = \frac{13}{\sqrt{9 + 16 + 49}}$
$= \frac{13}{\sqrt{74}}$
Then,
Option $D$ is correct answer.

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A plane passes through (2,3, -1) and is perpendicular to the line having direction ratios 3, -4, 7 The perpendicular distance from the origin to this plane is

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and

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A plane passes through (2,3,−1) and is perpendicular to the line having direction ratios 3,−4,7. The perpandicular distance from the origin to this plane is

The correct option is C 13√74Equation of plane :3(x−2)−4(y−3)+7(z+1)=03x−4y+7z+7+12−6=03x−4y+7z+13=0Perpendicular distance from origin d=13√9+16+49=13√74Then,Option D is correct answer.

A plane passes through $(2, 3, - 1)$ and is perpendicular to the line having direction ratios $3, - 4, 7$ . The perpandicular distance from the origin to this plane is