Question

The vector equation of a plane passing through the line of intersection of the planes $\overrightarrow{r}.\overrightarrow{n_1}=q_1$ and $\overrightarrow{r}.\overrightarrow{n_2}=q_2$ is:

• A. $\overrightarrow{r}\times \overrightarrow{n_1}+\overrightarrow{r}\times \overrightarrow{n_2}=q_1+q_2$
• B. $(\overrightarrow{r}.\overrightarrow{n_1}-q_1)+\lambda (\overrightarrow{r}.\overrightarrow{n_2}-q_2)=0$
• C. $\overrightarrow{r}\times \overrightarrow{n_1}-\overrightarrow{r}\times \overrightarrow{n_2}=q_1+q_2$
• D. $\overrightarrow{r}.\overrightarrow{n_1}+\overrightarrow{r}.\overrightarrow{n_2}=q_1+\lambda q_2$

Answer 1

The correct option is D $\overrightarrow{r}.\overrightarrow{n_1}+\overrightarrow{r}.\overrightarrow{n_2}=q_1+\lambda q_2$

Given:Suppose we are given two non-parallel planes whose vector equation are $\overrightarrow{r}.\overrightarrow{n_1}=\overrightarrow{q_1}$

and $\overrightarrow{r}.\overrightarrow{n_2}=\overrightarrow{q_2}$

Their line of intersection will be perpendicular to both $\overrightarrow{n_1}$ and $\overrightarrow{n_2}$, since these are the normals to the two planes. The line of intersection will thus be parallel to $\overrightarrow{n_1}\times \overrightarrow{n_2}$, and all it remains to do, to obtain the vector equation of this line, is to determine any point on it. For convenience, we may take the point (common to both planes) for which one of $x,y$ or $z$

Thus, $\overrightarrow{r}.\overrightarrow{n_1}+\overrightarrow{r}.\overrightarrow{n_2}=q_1+\lambda q_2$