Find the vector and cartesian equation of the planes
(a) that passes through the point (1,0,-2)and the normal to the plane is \(\hat i+\hat j-\hat k\).
(b) that passes through the point(1,4,6)and the normal vector to the plane is \(\hat i-2\hat j+\hat k\).
Find the vector and cartesian equation of the planes
(a) that passes through the point (1,0,-2)and the normal to the plane is \(\hat i+\hat j-\hat k\).
(b) that passes through the point(1,4,6)and the normal vector to the plane is \(\hat i-2\hat j+\hat k\).
Solution and Explanation
(a) The position vector of point (1,0,-2) is \(\overrightarrow a=\hat i-2\hat k\)
The normal vector N→ perpendicular to the plane is \(\overrightarrow N=\hat i+\hat j-\hat k\)
The vector equation of the plane is given by (\(\overrightarrow r-\overrightarrow a\)).\(\overrightarrow N\)=0
\(\Rightarrow [\overrightarrow r-(\hat i-2\hat k)].(\hat i+\hat j-\hat k)=0\)...(1)
\(\overrightarrow r\) is the position vector of any point P(x,y,z) in the plane.
∴ \(\overrightarrow r=x\hat i+y\hat j+z\hat k\)
Therefore, equation(1) becomes
[\((x\hat i+y\hat j+z\hat k)\)-\((\hat i-2\hat k)].(\hat i+\hat j-\hat k)=0\)
\(\Rightarrow [(x-1)\hat i+y\hat j+(z+2)\hat k].(\hat i+\hat j-\hat k)=0\)
\(\Rightarrow\) (x-1)+y-(z+2)=0
\(\Rightarrow\) x+y-z-3=0
\(\Rightarrow\) x+y-z=3
This is the cartesian equation of the required plane.
(b) The position vector of the point (1,4,6) is \(\overrightarrow a=\hat i+4\hat j+6\hat k\)
The normal vector \(\overrightarrow N\) perpendicular to the plane is \(\overrightarrow N=\hat i-2\hat j+\hat k\)
The vector equation of the plane is given by \((\overrightarrow r-\overrightarrow a).\overrightarrow N\) =0
\(\Rightarrow [ \overrightarrow r\)-(\(\hat i+4\hat j+6\hat k\))].(\(\hat i-2\hat j+\hat k\))=0...(1)
\(\overrightarrow r\) is the position vector of of any point P(x,y,z)in the plane.
∴\(\overrightarrow r\) =\(x\hat i+y\hat j+z\hat k\)
Therefore, equation(1) becomes
[(\(x\hat i+y\hat j+z\hat k\))-(\(\hat i+4\hat j+6\hat k\))].(\(\hat i-2\hat j+\hat k\))=0
\(\Rightarrow [(x-1)\hat i+(y-4)\hat j+(z-6)\hat k].(\hat i-2\hat j+\hat k)=0\)
\(\Rightarrow\) (x-1)-2(y-4)+(z-6)=0
\(\Rightarrow\) x-2y+z+1=0
This is the cartesian equation of the required 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.