Question

Find the vector and Cartesian equations of the planes.

that passes through the point (1,0,-2) and the normal to the plane is $\hat{i}+\hat{j}-\hat{k}.$

that passes through the point (1,4,6) and the normal to the plane is $\hat{i}-2\hat{j}+\hat{k}.$

Answer 1

The position vector of point (1,0,-2) is a= $\hat{i}-2\hat{k}$

The normal vector N perpendicular to the plane is N = $\hat{i}+\hat{j}-\hat{k}.$

The vector equation of the plane is given by (r-a).N=0

$\Rightarrow$ $[r-(\hat{i}+0\hat{j}-2\hat{k}\left)\right].(\hat{i}+\hat{j}-\hat{k})=0$ ...(i)

where, r is the position vector of any point P(x,y,z) in the plane.

i.e., $r=x\hat{i}+y\hat{j}+z\hat{k}$

Therefore, Eq.(i) becomes

$\left[\right(x\hat{i}+y\hat{j}+z\hat{k})-(\hat{i}+0\hat{j}-2\hat{k}\left)\right].(\hat{i}+\hat{j}-\hat{k})=0\Rightarrow \left[\right(x-1)\hat{i}+y\hat{j}+(z+2\left)\hat{k}\right].(\hat{i}+\hat{j}-\hat{k})=0\Rightarrow (x-1)+y-(z+2)=0\Rightarrow x+y-z-3=0\Rightarrow x+y-z=3$

This is the Cartesian equation of the required plane.

The position vector of the point (1,4,6) is a $\hat{i}+4\hat{j}+6\hat{k}.$

The normal vector N perpendicular to the plane is $N=\hat{i}-2\hat{j}+\hat{k}.$

The vector equation of the plane is given by (r-a).N=0

$\Rightarrow [r-(\hat{i}+4\hat{j}+6\hat{k}\left)\right].(\hat{i}-2\hat{j}+\hat{k})=0..\left(ii\right)\Rightarrow \left[\right(x-1)\hat{i}+(y-4)\hat{j}+(z-6\left)\hat{k}\right].(\hat{i}-2\hat{j}+\hat{k})=0(putr=x\hat{i}+y\hat{j}+z\hat{k})\Rightarrow (x-1)-2(y-4)+(z-6)=0\Rightarrow x-2y+z+1=0$

This is the Cartesian equation of the required plane.