Login/Sign Up
Home » Find the vector and Cartesian equations of the planes (a) that passes through the point and the normal to the plane is (b) that passes through the point and the normal vector to the plane is
Find the vector and Cartesian equations 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}
CBSE | Class 12 | Exercise 11.3 | Maths | NCERT | Three Dimensional Geometry
Find the vector and Cartesian equations 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:

(a) That passes through the point (1,0,-2) and the normal to the plane is \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}
Let’s say that the position vector of the point (1,0,-2) be
\overrightarrow{\mathrm{a}}=(1 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})
It is known that Normal \vec{N} \perp to the plane is given as:
\overrightarrow{\mathrm{N}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathrm{k}}
Vector eq. of the plane is given as:
(\vec{r}-\vec{a}) \cdot \vec{N}=0
Now,
\begin{array}{l} x-1-2 y+8+z-6=0 \\ x-2 y+z+1=0 \\ x-2 y+z=-1 \end{array}
\therefore The required Cartesian equation of the plane is x-2 y+z=-1
(\vec{r}-(\hat{i}-2 \hat{k})) \cdot \hat{i}+\hat{j}-\hat{k}=0 \ldots(1)
Since,
\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}
So eq.(1) becomes,
\begin{array}{l} (\mathrm{x} \hat{\mathrm{i}}+\mathrm{yj}+z \hat{\mathrm{k}}-\hat{\mathrm{i}}+2 \hat{\mathrm{k}}) \cdot \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}=0 \\ {[(\mathrm{x}-1) \hat{\mathrm{i}}+\mathrm{yj}+(z+2) \hat{\mathrm{k}}] \cdot \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}=0} \\ \mathrm{x}-1+\mathrm{y}-\mathrm{z}-2=0 \\ \mathrm{x}+\mathrm{y}-\mathrm{z}-3=0 \end{array}
As a result, the required Cartesian equation of the plane is x+y-z=3

(b) That passes through the point (1,4,6) and the normal vector to the plane is \hat{i}-2 \hat{j}+\hat{k}
Let’s say that the position vector of the point (1,0,-2) be
\overrightarrow{\mathrm{a}}=(1 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})
It is known that Normal \vec{N} \perp to the plane is given as:
\overrightarrow{\mathrm{N}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}
Vector eq. of the plane is given as:
(\vec{r}-\vec{a}) \cdot \vec{N}=0
Now,
(\vec{r}-(\hat{i}+4 \hat{j}+6 \hat{k})) \cdot \hat{i}-2 \hat{j}+\hat{k}=0
Since,
\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}
So eq.(1) becomes,
(x \hat{i}+\hat{y}+z \hat{k}-\hat{i}-4 \hat{j}-6 \hat{k}) \cdot \hat{i}-2 \hat{j}+\hat{k}=0
\begin{array}{l} \quad[(x-1) \hat{i}+(y-4) \hat{j}+(z-6) \hat{k}] \hat{i}-2 \hat{j}+\hat{k}=0 \\ x-1-2 y+8+z-6=0 \\ x-2 y+z+1=0 \\ x-2 y+z=-1 \end{array}
As a result, the required Cartesian equation of the plane is x-2 y+z=-1

i

More Material