Question

The vector equation of the plane containing the line $\overrightarrow{r}=\left(-2\hat{i}-3\hat{j}+4\hat{k}\right)+\lambda \left(3\hat{i}-2\hat{j}-\hat{k}\right)$ and the point $\hat{i}+2\hat{j}+3\hat{k}$ is

(a) $\overrightarrow{r}·\left(\hat{i}+3\hat{k}\right)=10$

(b) $\overrightarrow{r}·\left(\hat{i}-3\hat{k}\right)=10$

(c) $\overrightarrow{r}·\left(3\hat{i}+\hat{k}\right)=10$

(d) None of these

Answer 1

(a) $\overrightarrow{r}·\left(\hat{i}+3\hat{k}\right)=10$

$$\begin{gathered} \text{Let the direction ratios of the required plane be proportional to}a,b,c. \\ \text{Since the required plane contains the line}\overrightarrow{r}=\left(-2\hat{i}-3\hat{j}+4\hat{k}\right)+\lambda \left(3\hat{i}-2\hat{j}-\hat{k}\right),\text{it must pass through the point (-2, -3, 4) and it should be parallel to the line.} \\ \text{So, the equation of the plane is} \\ a\left(x+2\right)+b\left(y+3\right)+c\left(z-4\right)=0...\left(1\right)\text{and} \\ 3a-2b-c=0...\left(2\right) \\ \text{It is given that plane (1) passes through the point}\hat{i}\text{+ 2}\hat{j}\text{+ 3}\hat{k}\text{or (1, 2, 3). So,} \\ a\left(1+2\right)+b\left(2+3\right)+c\left(3-4\right)=0 \\ 3a+5b-c=0...\left(3\right) \\ \text{Solving (1), (2) and (3), we get} \\ \left|\begin{array}{ccc}x+2& y+3& z-4\\ 3& -2& -1\\ 3& 5& -1\end{array}\right|=0 \\ \Rightarrow 7\left(x+2\right)+0\left(y+3\right)+21\left(z-4\right)=0 \\ \Rightarrow x+2+3z-12=0 \\ \Rightarrow x+3z=10\text{or}\overrightarrow{r}\text{.}\left(\hat{i}+3\hat{k}\right)\text{= 10} \end{gathered}$$