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Equation of a Plane : Point Normal Form

The equation ...

Question

The equation of the plane passing through the origin and perpendicular to the line x=2 y=3 z is

A

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B

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C

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D

None of these

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Solution

The correct option is A

Obviously the line is $\frac{x}{6} = \frac{y}{3} = \frac{z}{2}$ So required plane is 6(x-0)+3(y-0)+2(z-0)=0.

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Similar questions

The equation of a straight line passing through the origin and perpendicular to the straight line 2x +3y -7 = 0 is

Q. The equation of the plane parallel to the lines x − 1 = 2y − 5 = 2z and 3x = 4y − 11 = 3z − 4 and passing through the point (2, 3, 3) is
(a) x − 4y + 2z + 4 = 0
(b) x + 4y + 2z + 4 = 0
(c) x − 4y + 2z − 4 = 0
(d) None of these

Q. The equation of the plane through the intersection of the planes x + 2y + 3z = 4 and 2x + y − z = −5 and perpendicular to the plane 5x + 3y + 6z + 8 = 0 is

(a) 7x − 2y + 3z + 81 = 0
(b) 23x + 14y − 9z + 48 = 0
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Q. The equation of the plane passing through the point

(- 1, 2, 0)

and parallel to the lines

L_{1} : \frac{x}{3} = \frac{y + 1}{0} = \frac{z - 2}{- 1}

L_{2} : \frac{x - 1}{1} = \frac{2 y + 1}{2} = \frac{z + 1}{- 1}

Q. Equation of the plane perpendicular to the line

\frac{x - 1}{3} = \frac{y - 5}{2} = \frac{z + 3}{- 4}

and passing through the point

(1, - 2, 3)

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