Find the equation of the plane through the intersection of the planes
3x-y+2z-4 =0 and x+y+z-2=0 and the point (2, 2, 1).
Find the equation of the plane through the intersection of the planes
3x-y+2z-4 =0 and x+y+z-2=0 and the point (2, 2, 1).
Solution and Explanation
The equation of any plane through the intersection of the planes,
3x-y+2z-4=0 and x+y+z-2=0, is
(3x-y+2z-4)+ \(\alpha\) (x+y+z-2)=0, where \(\alpha \in\) R...(1)
The plane passes through the point (2 ,2,1).
Therefore, this point will satisfy equation(1).
∴ (3×2-2+2×1-4)+α(2+2+1-2)=0
\(\Rightarrow \) 2+3\(\alpha\) =0
\(\Rightarrow \alpha=-\frac{2}{3}\)
Substituting \(\alpha=-\frac{2}{3}\) in equation(1), we obtain
(3x-y+2z-4)-\(\frac{2}{3}\) (x+y+z-2)=0
\(\Rightarrow 3\) (3x-y+2z-4)-2(x+y+z-2)=0
\(\Rightarrow\) (9x-3y+6z-12)-2(x+y+z-2)=0
\(\Rightarrow \) 7x-5y+4z-8=0
This is the required equation of the plane.
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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.