Question

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

Answer 1

Given planes are
$r.(\hat{i}+2\hat{j}+3\hat{k})-4=0\cdots \left(i\right)r.(2\hat{i}+\hat{j}-\hat{k})+5=0\cdots \left(ii\right)$
and $r.(5\hat{i}+3\hat{j}-6\hat{k})+8=0$ . . . (iii)
Any plane through the line of intersection of Eqs. (i) and (ii) can be written as
$[r.(\hat{i}+2\hat{j}+3\hat{k})-4]+\lambda [r.(2\hat{i}+\hat{j}-\hat{k}\left)\right]+5=0$
$\Rightarrow [r.(1+2\lambda )\hat{i}+(2+\lambda )\hat{j}+(3-\lambda \left)\hat{k}\right]+(5\lambda -4)=0$ . . . (iv)
This plane is at right angle to Eq. (iii),
$\therefore 5(1+2\lambda )+3(2+\lambda )-6(3-\lambda )=0\Rightarrow 19\lambda =7\Rightarrow \lambda =\frac{7}{19}$
Substituting this value of $\lambda$ in Eq. (iv), we get the required plane as
$r.[(1+\frac{14}{19})\hat{i}+(2+\frac{7}{19})\hat{j}+(3-\frac{7}{19})\hat{k}]+\frac{35}{19}-4=0\Rightarrow r[\left(\frac{33}{19}\right)\hat{i}+\left(\frac{45}{19}\right)\hat{j}+\left(\frac{50}{19}\right)\hat{k}]-\frac{41}{19}=0\Rightarrow r.(33\hat{i}+45\hat{j}+50\hat{k})-41=0\cdots \left(v\right)$
This is the vector equation of the required plane.
Now, converting the equation into Cartesian equation of this plane can be obtained by substituting $r=x\overrightarrow{i}+y\hat{j}+z\hat{k}$ in Eq. (v), we get
$(x\hat{i}+y\hat{j}+z\hat{k}).(33\hat{i}+45\hat{j}+50\hat{k})-41=0\Rightarrow 33x+45y+50z-41=0$