Question

Find the vector equation of the plane passing through the intersection of the planes $r.(2\hat{i}+2\hat{j}-3\hat{k})=7,r.(2\hat{i}+5\hat{j}+3\hat{k})=9$ and through the point (2,1,3).

Answer 1

Given planes are $r.(2\hat{i}+2\hat{j}-3\hat{k})=7,r.(2\hat{i}+5\hat{j}+3\hat{k})=9$

These equation can be written as

$r.(2\hat{i}+2\hat{j}-3\hat{k})-7=0$

and $r.(2\hat{i}+5\hat{j}+3\hat{k})-9=0$

The equation of any plane through the intersection of the planes given in Eqs.(i) and (ii) can be written as

$[r.(2\hat{i}+2\hat{j}-3\hat{k})-7]+\lambda [r.(2\hat{i}+5\hat{j}+3\hat{k})-9]=0\Rightarrow r.\left[\right(2\hat{i}+2\hat{j}-3\hat{k}\left)\right]+\lambda (2\hat{i}+5\hat{j}+3\hat{k})]=9\lambda +7\Rightarrow r.[(2+2\lambda )\hat{i}+(2+5\lambda )\hat{j}+(3\lambda -3)\hat{k}]=9\lambda +7$

The plane passes through the point (2,1,3). Therefore, its position vector is given by $r.2\hat{i}+\hat{j}+3\hat{k}$

Substituting in Eq.(iii), we obtain

$(2\hat{i}+\hat{j}+3\hat{k}).\left[\right(2+2\lambda )\hat{i}+(2+5\lambda )\hat{j}+(3\lambda -3\left)\hat{k}\right]=9\lambda +7\Rightarrow 2(2+2\lambda )+2+5\lambda +3(3\lambda -3)=9\lambda +7\Rightarrow (4+4\lambda )+(2+5\lambda )+(9\lambda -9)=9\lambda +7\Rightarrow -3+18\lambda =9\lambda +7\Rightarrow 9\lambda =10\Rightarrow \lambda =\frac{10}{9}$

Substituting this value of $\lambda$ in Eq.(iii), we obtain the required plane as

$[r.(2\hat{i}+2\hat{j}-3\hat{k})-7]+\frac{10}{9}[r.(2\hat{i}+5\hat{j}+3\hat{k})-9]=0\Rightarrow r.[18\hat{i}+18\hat{j}-27\hat{k}-63+20\hat{i}+50\hat{j}+30\hat{k}-90]=0\Rightarrow r.[38\hat{i}+68\hat{j}+3\hat{k}-153]=0\Rightarrow r.(38\hat{i}+68\hat{j}+3\hat{k})=153$

This is the vector equation of the required plane.