Question

The 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 scalar product form is

 

 

 

 

 

 

 

पर्याय

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

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

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

•  None of these

Answer 1

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

$$\text{ We know that the equation }\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}+\mu \overrightarrow{c}\text{ represents a plane passing through a point whose position vector is }\overrightarrow{a}\text{ and parallel to the vectors }\overrightarrow{b}\text{ and }\overrightarrow{c}.$$
$$\text{ Here },\overrightarrow{a}=\hat{i}-\hat{j}+0\hat{k};\overrightarrow{b}=\hat{i}+\hat{j}+\hat{k};\overrightarrow{c}=\hat{i}-2\hat{j}+3\hat{k}$$
$$\text{ Normal vector,}\overrightarrow{n}=\overrightarrow{b}\times \overrightarrow{c}$$
$$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 1& -2& 3\end{array}\right|$$
$$=5\hat{i}-2\hat{j}-3\hat{k}$$
$$\text{ The vector equation of the plane in scalar product form is }$$
$$\overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n}$$
$$\Rightarrow \overrightarrow{r}.(5\hat{i}-2\hat{j}-3\hat{k})=(\hat{i}-\hat{j}+0\hat{k}).(5\hat{i}-2\hat{j}-3\hat{k})$$
$$\Rightarrow \overrightarrow{r}.(5\hat{i}-2\hat{j}-3\hat{k})=5+2+0$$
$$\Rightarrow \overrightarrow{r}.(5\hat{i}-2\hat{j}-3\hat{k})=7$$
$$\Rightarrow \overrightarrow{r}.(5\hat{i}-2\hat{j}-3\hat{k})=7$$