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.