Question

Find the equation of the plane which contains the line of intersection of the planes $\overrightarrow{r}.(\hat{i}+2\hat{j}+3\hat{k})-4=0$ and $\overrightarrow{r}.(2\hat{i}+\hat{j}-\hat{k})+5=0$ and which is perpendicular to the plane $\overrightarrow{r}.(5\hat{i}+3\hat{j}-6\hat{k})+8=0$.

Answer 1

$P_1:\overline{r}\cdot (i+2j+3k)-4=0$

$P_2:\overline{r}\cdot (2i+j-k)+5=0$

$P_3:\overline{r}\cdot (5i+3j-6k)+8=0$

The plane passing through the line of intersection of $P_1,P_2$ is of the form $P_1+\lambda P_2=0$

$\overline{r}\cdot \left(\right(1+2\lambda )i+(2+\lambda \left)j\right(3-\lambda \left)k\right)+5\lambda -4=0$

It is perpendicular to $P_3:\overline{r}\cdot (5i+3j-6k)+8=0$

So $(1+2\lambda )5+(2+\lambda )3+(3-\lambda )(-6)=0$

Dot product of normals of cosines is zero.

$5+10\lambda +6+3\lambda -18+6\lambda =0$

$19\lambda =7$

Hence $\lambda =\frac{7}{19}$

So the plane is $\overline{r}\cdot (1+2\times \frac{7}{19})i+(2+\frac{7}{19})j+(3-\frac{7}{19})k)+(5\times \frac{7}{19}-4)=0$

$\overline{r}\cdot \left[\right(\frac{33}{19})i+(\frac{45}{19})j+(\frac{50}{19}\left)k\right]=\frac{41}{19}$

$\Rightarrow \overline{r}\cdot (33i+45j+50k)=41$