Question

Find the vector equation of the plane passing through the intersection of the planes

$\overrightarrow r.(2\hat i+2\hat j-3\hat k)=7$,$\overrightarrow r.(2\hat i+5\hat j+3\hat k)=9$ and through the point ( 2, 1, 3 ).

Answer 1

The equations of the planes are $\overrightarrow r.(2\hat i+2\hat j-3\hat k)=7$ and $\overrightarrow r.(2\hat i+5\hat j+3\hat k)=9$

$\Rightarrow \overrightarrow r.(2\hat i+2\hat j-3\hat k)-7=0$...(1)

$\Rightarrow\overrightarrow r.(2\hat i+5\hat j+3\hat k)-9=0$...(2)

The equation of any plane through the intersection of the planes given in equation (1) and (2) is given by,

[ $\overrightarrow r.(2\hat i+2\hat j-3\hat k)=7$ ]+$\lambda$ [ $\overrightarrow r.(2\hat i+5\hat j+3\hat k)=9$ ]=0, where $\lambda\in$ R

$\overrightarrow r.[(2\hat i+2\hat j-3\hat k)$+$\lambda(2\hat i+5\hat j+3\hat k)]=9\lambda+7$

$\overrightarrow r[(2+2\lambda)\hat i+(2+5\lambda)\hat j+(3\lambda-3)\hat k]=9\lambda+7$ ...(3)

The plane passes through the point (2,1, 3).

Therefore, its position vector is given by, $\overrightarrow r.2\hat i+2\hat j+3\hat k$

Substituting in equation (3), we obtain

(2\hat i+2\hat j-3\hat k).$$[(2+2\lambda)\hat i+(2+5\lambda)\hat j+(3\lambda-3)\hat k]=9\lambda+7

$\Rightarrow[(2+2\lambda)+(2+5\lambda)+(3\lambda-3)]=9\lambda+7$

$\Rightarrow$ 18λ-3=9λ+7

$\Rightarrow$ 9λ=10

$\Rightarrow$ λ=$\frac{10}{9}$

Substituting $\lambda=\frac{10}{9}$ in equation(3), we obtain

$\overrightarrow r\bigg(\frac{38}{9}\hat i+\frac{68}{9}\hat j+\frac{3}{9}\hat k\bigg)=17$

$\Rightarrow \overrightarrow r (38\hat i+68\hat j+3\hat k)=153$

This is the vector equation of the required plane.