Question

A plane passing through the intersection of 2 planes $\overline{r}.\overline{n_1}=d_1$ and $\overline{r}.\overline{n_2}=d_2$ can be given by
$\overline{r}.(\overline{n_1}+\lambda \overline{n_1})=d_1+\lambda d_2$

• A. True
• B. False

Answer 1

The correct option is A True

Let take the case of 2 planes meeting in a line. But we will see it will satisfy for other cases too.
Plane 1 will be $\overline{r}.\overline{n_1}=d_1$
Plane 2 will be $\overline{r}.\overline{n_2}=d_2$
We can write the plane through intersection as $(\overline{r}.\overline{n_1}-d_1)+\lambda (\overline{r}.\overline{n_2}-d_2)=0$. If we simplify this the form it takes is $\overline{r}.(\overline{n_1}+\lambda \overline{n_2})=d_1+\lambda d_2$. This can be viewed as $\overline{r}.\overline{n_3}=d_3$, which represents the equation of a plane and satisfies for the points on the given planes only when they exist on the line of intersection unless $\lambda =0$. Therefore given statement is true.