Question:
The equation of plane passing through a point $A (2,-1,3)$ and parallel to the vectors $a =(3,0,-1)$ and $b =(-3,2,2)$ is:
The equation of plane passing through a point $A (2,-1,3)$ and parallel to the vectors $a =(3,0,-1)$ and $b =(-3,2,2)$ is:
Updated On: Jun 27, 2024
- 2x - 3y + 6z - 25 = 0
- 2x - 3y + 6z + 25 = 0
- 3x - 2y + 6z - 25 = 0
- 3x - 2y + 6z + 25 = 0
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The Correct Option is A
Solution and Explanation
Let the DR of the normal of required plane be $< a, b, c >$
Since plane is parallel to $(3,0,-1)$
$\therefore$ normal must be perpendicular
$\therefore 3 a +0 b - c =0 \ldots$ (1)
Also, it is parallel to $(-3,2,2)$
$\therefore-3 a +2 b +2 c =0 \ldots$ (2)
From (1) and (2)
$\frac{a}{2}=\frac{-b}{6-3}=\frac{c}{6}$
$a=2, b=-3, c=6$
It passes through $(2,-1,3)$
$2(x-2)-3(y+1)+6(z-3)=0 $
$\Rightarrow 2 x-4-3 y-3+6 z-18=0 $
$\Rightarrow 2 x-3 y+6 z-25=0$
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Concepts Used:
Plane
A surface comprising all the straight lines that join any two points lying on it is called a plane in geometry. A plane is defined through any of the following uniquely:
- Using three non-collinear points
- Using a point and a line not on that line
- Using two distinct intersecting lines
- Using two separate parallel lines
Properties of a Plane:
- In a three-dimensional space, if there are two different planes than they are either parallel to each other or intersecting in a line.
- A line could be parallel to a plane, intersects the plane at a single point or is existing in the plane.
- If there are two different lines that are perpendicular to the same plane then they must be parallel to each other.
- If there are two separate planes which are perpendicular to the same line then they must be parallel to each other.