Question

Find the Cartesian equation of the plane, passing through the line of intersection of the planes: $\overrightarrow{r}.(2\hat{i}+3\hat{j}-4\hat{k})+5=0$ and $\overrightarrow{r}.(\hat{i}-5\hat{j}+7\hat{k})+2=0$ and intersecting $y$-axis at $(0,3)$

Answer 1

Any plane passing through the intersection of the given planes is

$[\overrightarrow{r}.(2\hat{i}+3\hat{j}-4\hat{k})]+\lambda [\overrightarrow{r}.(\hat{i}-5\hat{j}+7\hat{k})+2]=0$

$\overrightarrow{r}\left[\right(2+\lambda )\hat{i}+(3-5\lambda \hat{j})+(-4+7\lambda \left)\hat{k}\right]=-5-2\lambda .....\left(i\right)$

If equation $\left(i\right)$ intersects the y-axis at $(0,3)$

$\therefore$ $\overrightarrow{r}=0\hat{i}+0\hat{j}+3\hat{k}$

$(0\hat{i}+0\hat{j}+3\hat{k}).\left[\right(2+\lambda )\hat{i}+(3-5\lambda \hat{j}+(-4+7\lambda )\hat{k}]=0$

$0(2+\lambda )+0(3-\lambda )+3(-4+7\lambda )=0$

$\lambda =\frac{4}{7}$

$\therefore$ The required plane is
$(2+\frac{4}{7})x+(3-5\times \frac{4}{7})y+(-4+7\times \frac{4}{7})z=-5-2\times \frac{4}{7}$

$\left(\frac{14+4}{7}\right)x+\left(\frac{21-20}{7}\right)y+\left(\frac{-28+28}{7}\right)z=\frac{-35-8}{7}$

$18x+y=-43$

or $18x+y+43=0$ which is the required equation.