Question

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

Answer 1

$n_1=2i+j-3k$

$n_2=2i+5j+3k$

$p_1=7$

$p_2=9$

$r.(n_1+\lambda n_2)=p_1+\lambda p_2$

$r[2i+j-3k+\lambda (2j+5j+3k\left)\right]=7+\lambda 9$

$r\left[i\right(2+2\lambda )+j(1+5\lambda )+k(-3+3\lambda \left)\right]=7+9\lambda$

Taking $r=xi+yj+zj$

We get,

$(2+2\lambda )x+(1+5\lambda )y+(-3+3\lambda )z=7+9\lambda$

$(2x+y-3z-7)+\lambda (2x+5y+3z-9)=0$

Given, it passes through $(2,1,3)$

$(4+1-9-7)+\lambda (4+5+9-9)=0$

$\lambda =\frac{11}{9}$

Putting in above equation,

$r.(\frac{40}{9}i+\frac{64}{9}j+\frac{6}{9}k)=18$....Ans.