Question

The vector equation of the plane through the point $(1,-2,-3)$ and parallel to the vectors $(2,-1,3)$ and $(2,3,-6)$ is $\overline{r}=$

• A. $(1+2t+2s)\overline{i}-(2+t-3s)\overline{j}-(3-3t+6s)\overline{k}$
• B. $(1+2t+2s)\overline{i}+(2+t+3s)\overline{j}-(3+3t+6s)\overline{k}$
• C. $(1+2t+2s)\overline{i}+(2+t+3s)\overline{j}+(3+3t+6s)\overline{k}$
• D. $(1+2t+2s)\overline{i}+(2+t-3s)\overline{j}+(3+3t+6s)\overline{k}$

Answer 1

The correct option is A $(1+2t+2s)\overline{i}-(2+t-3s)\overline{j}-(3-3t+6s)\overline{k}$

Let $\overrightarrow{r}$ be a point in a plane of $(2,-1,3)$ and $(2,3,-6)$. then, $\overrightarrow{r}=s(2\hat{i}+3\hat{j}-6\hat{k})+t(2\hat{i}-\hat{j}+3\hat{k}),t,\vee IR$.

This above equation is the locus of points on the plane formed by two points $(2,3,-6)$ and $(2,-1,3)$. this plane passes through $(1,-2,-3)$.

$\therefore \overrightarrow{r}=1\hat{i}-2\hat{j}-3\hat{k}$ must satisfy it.

$\therefore \hat{i}-2\hat{j}-3\hat{k}=s(2\hat{i}+3\hat{j}-6\hat{k})+t(2\hat{i}-\hat{j}+3\hat{k})$

$\Rightarrow (1-2s-2t)\hat{i}+(-2-3s+t)\hat{j}+(-3+6s-3t)\hat{k}=0$

Since, $s$ and $t$ are constants, $\therefore$-s,-t$ which are also constants.

$\Rightarrow (1+2s+2t)\hat{i}-(2-3s+t)\hat{j}-(-3+6s-3t)\hat{k}=0$