Question

Prove that equation of the plane passing through (1, 2, 3) and perpendicular to planes $\overrightarrow{r}.(\hat{i}+\hat{j}+\hat{k})=3$ and $\overrightarrow{r}.(2\hat{i}+3\hat{j}+4\hat{k})=0$ is $\overrightarrow{r}.(\hat{i}-2\hat{j}-\hat{k})=0$

Answer 1

It is given that point $P(1,2,3)$ lies on the required plane.

Also, planes $\overrightarrow{r}.(\hat{i}+\hat{j}+\hat{k})=3$ and $\overrightarrow{r}.(2\hat{i}+3\hat{j}+4\hat{k})=0$ are perpendicular to required plane. Hence, their perpendiculars $\overrightarrow{n_1}=\hat{i}+\hat{j}+\hat{k}$ and $\overrightarrow{n_2}=2\hat{i}+3\hat{j}+4\hat{k}$ are also perpendicular to required plane.

Now, the required plane is also perpendicular to its normal vector $\overrightarrow{n}$. Hence, using perpendicularity theorem,

$\overrightarrow{n}=\overrightarrow{n_1}\times \overrightarrow{n_2}$

$\therefore \overrightarrow{n}=(\hat{i}+\hat{j}+\hat{k})\times (2\hat{i}+3\hat{j}+4\hat{k})$

$\therefore \overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 2& 3& 4\end{array}\right|$

$\therefore \overrightarrow{n}=(4-3)\hat{i}-(4-2)\hat{j}+(3-2)\hat{k}$

$\therefore \overrightarrow{n}=\hat{i}-2\hat{j}+\hat{k}$

Now, equation of plane with normal vector $\overrightarrow{n}$ and passing through $P$ is

$(\overrightarrow{r}-\overrightarrow{p}).\overrightarrow{n}=0$

$\therefore (\overrightarrow{r}-(\hat{i}+2\hat{j}+3\hat{k}\left)\right).(\hat{i}-2\hat{j}+\hat{k})=0$

$\therefore \left(\overrightarrow{r}\right).(\hat{i}-2\hat{j}+\hat{k})-(\hat{i}+2\hat{j}+3\hat{k}).(\hat{i}-2\hat{j}+\hat{k})=0$

$\therefore \left(\overrightarrow{r}\right).(\hat{i}-2\hat{j}+\hat{k})-1+4-3=0$

$\therefore \left(\overrightarrow{r}\right).(\hat{i}-2\hat{j}+\hat{k})=0$

Hence, proved.