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.