Question

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 \(\hat i-2\hat j+\hat k\).

Answer 1

(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.

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(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.