Question

Vector equation of the plane $\overrightarrow{r}=\hat{i}-\hat{j}+\lambda (\hat{i}+\hat{j}+\hat{k})+\mu (\hat{i}-2\hat{j}+3\hat{k})$ in the scalar product from is

• A. $\overrightarrow{r}.(5\hat{i}-2\hat{j}+3\hat{k})=7$
• B. $\overrightarrow{r}.(5\hat{i}+2\hat{j}-3\hat{k})=7$
• C. $\overrightarrow{r}.(5\hat{i}-2\hat{j}-3\hat{k})=7$
• D. $\overrightarrow{r}.(5\hat{i}+2\hat{j}+3\hat{k})=7$

Answer 1

Answer: (C)

$\overrightarrow{r}.(5\hat{i}-2\hat{j}-3\hat{k})=7$

Given equation of plane,

$\overrightarrow{r}=\hat{i}-\hat{j}+\lambda (\hat{i}+\hat{j}+\hat{k})+\mu (\hat{i}-2\hat{j}+3\hat{k})\overrightarrow{r}-(\hat{i}-\hat{j})=\lambda (\hat{i}+\hat{j}+\hat{k})+\mu (\hat{i}-2\hat{j}+3\hat{k})-\left(i\right)$

If two non-collinear vector are given by $\overrightarrow{b_1}$ and $\overrightarrow{b_2}$

The equation of the plane formed by them can be given by

$\overrightarrow{r}=\lambda \overrightarrow{b_1}+\mu \overrightarrow{b_2}$

Similarly,

$\overrightarrow{r}-(\hat{i}-\hat{j})=\lambda (\hat{i}+\hat{j}+\hat{k})+\mu (\hat{i}-2\hat{j}+3\hat{k})$ is also a plane with $\hat{i}+\hat{j}+\hat{k}$ & $\hat{i}-2\hat{j}+3\hat{k}$ as two directions in the plane.

$\therefore$ direction of normal of plane will be $(\hat{i}+\hat{j}+\hat{k})\times (\hat{i}-2\hat{j}+3\hat{k})$

$\overrightarrow{n}=(\hat{i}+\hat{j}+\hat{k}).(\hat{i}-2\hat{j}+3\hat{k})=5\hat{i}-2\hat{j}-3\hat{k}$

equation of plane,

$\overrightarrow{r}.\overrightarrow{n}=p\overrightarrow{r}.(5\hat{i}-2\hat{j}-3\hat{k})=p$

In $\left(i\right)$ when $\mu =\lambda =0,\overrightarrow{r}=\hat{i}-\hat{j}$ is a point on plane .

$\therefore p=(\hat{i}-\hat{j}).(5\hat{i}-2\hat{j}-3\hat{k})p=7\therefore \overrightarrow{r}.(5\hat{i}-2\hat{j}-3\hat{k})=7$