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 $\hat i-2\hat j+\hat k$ .

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  $\hat i-2\hat j+\hat k$ .

cdquestions Admin · Accepted Answer

(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.

List - I		List - II
(P)	γ equals	(1)	\(-\hat{i}-\hat{j}+\hat{k}\)
(Q)	A possible choice for \(\hat{n}\) is	(2)	\(\sqrt{\frac{3}{2}}\)
(R)	\(\overrightarrow{OR_1}\) equals	(3)	1
(S)	A possible value of \(\overrightarrow{OR_1}.\hat{n}\) is	(4)	\(\frac{1}{\sqrt6}\hat{i}-\frac{2}{\sqrt6}\hat{j}+\frac{1}{\sqrt6}\hat{k}\)
		(5)	\(\sqrt{\frac{2}{3}}\)

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

Top Questions on Three Dimensional Geometry

Questions Asked in CBSE CLASS XII exam

Notes on Three Dimensional Geometry

Concepts Used:

Plane

Properties of a Plane:

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 \(\hat i-2\hat j+\hat k\).