Question

The equation of the plane $\overrightarrow{r}=\hat{i}-\hat{j}+\lambda \left(\hat{i}+\hat{j}+\hat{k}\right)+\mu \left(\hat{i}-2\hat{j}+3\hat{k}\right)$ in scalar product form is
(a) $\overrightarrow{r}·\left(5\hat{i}-2\hat{j}-3\hat{k}\right)=7$
(b) $\overrightarrow{r}·\left(5\hat{i}+2\hat{j}-3\hat{k}\right)=7$
(c) $\overrightarrow{r}·\left(5\hat{i}-2\hat{j}+3\hat{k}\right)=7$
(d) None of these

Answer 1

(a) $\overrightarrow{r}·\left(5\hat{i}-2\hat{j}-3\hat{k}\right)=7$

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