If a plane passes through the point 1- 1- 1 and is perpendicular to the line x-13-y-10-z-14 then its perpendicular distance from the origin is

The correct option is C 75 [ Let eq u^ n of plane be : a( x 1 ) +b( y 1 ) +( z 1 ) =0; 3( x 1 ) +0( y 1 ) ...

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Byju's Answer

Standard XII

Mathematics

Equation of a Plane : Vector Form

If a plane pa...

Question

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

\frac{3}{4}

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\frac{4}{3}

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\frac{7}{5}

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Solution

The correct option is C $\frac{7}{5}$
$\begin{matrix} L e t e q u^{n} o f p l a n e b e : a (x - 1) + b (y - 1) + (z - 1) = 0 3 (x - 1) + 0 (y - 1) + 4 (z - 1) = 0 3 x + 4 z = 7 p e r p e n d i c u l a r d i s tan c e f r o m o r i g i n d = \frac{7}{5} H e n c e, t h e o p t i o n C i s c o r r e c t a n s w e r . \end{matrix}$

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Similar questions

Q. 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

Q. 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

Q. 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
(a) 3/4
(b) 4/3
(c) 7/5
(d) 1

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

Q. The distance of the plane passing through (1, 1, 1) and perpendicular to the line

\frac{x - 1}{3} = \frac{y - 1}{0} = \frac{z - 1}{4}

from the origin is

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If a plane passes through the point (1,1,1) and is perpendicular to the line x−13=y−10=z−14 then its perpendicular distance from the origin is

The correct option is C 75Letequnofplanebe:a(x−1)+b(y−1)+(z−1)=03(x−1)+0(y−1)+4(z−1)=03x+4z=7perpendiculardistancefromorigind=75Hence,theoptionCiscorrectanswer.

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