Question

Find the vector equation of the plane passing through the intersection of the planes $\overrightarrow{r}·\left(\hat{i}+\hat{j}+\hat{k}\right)=6\mathrm{and}\overrightarrow{r}·\left(2\hat{i}+3\hat{j}+4\hat{k}\right)=-5$ and the point (1, 1, 1).

Answer 1

$$\begin{gathered} \text{The equation of the plane passing through the line of intersection of the given planes is} \\ \overrightarrow{r}.\left(\hat{i}+\hat{j}+\hat{k}\right)-6+\lambda \left(\overrightarrow{r}.\left(2\hat{i}+3\hat{j}+4\hat{k}\right)+5\right)=0 \\ \overrightarrow{r}.\left[\left(1+2\lambda \right)\hat{i}+\left(1+3\lambda \right)\hat{j}+\left(1+4\lambda \right)\hat{k}\right]-6+5\lambda =0...\left(1\right) \\ \text{This passes through}\hat{i}\text{+}\hat{j}\text{+}\hat{k}\text{. So,} \\ \left(\hat{i}+\hat{j}+\hat{k}\right).\left[\left(1+2\lambda \right)\hat{i}+\left(1+3\lambda \right)\hat{j}+\left(1+4\lambda \right)\hat{k}\right]-6+5\lambda =0 \\ \Rightarrow 1+2\lambda +1+3\lambda +1+4\lambda -6+5\lambda =0 \\ \Rightarrow 14\lambda -3=0 \\ \Rightarrow \lambda =\frac{3}{14} \\ \text{Substituting this in (1), we get} \\ \overrightarrow{r}.\left[\left(1+2\left(\frac{3}{14}\right)\right)\hat{i}+\left(1+3\left(\frac{3}{14}\right)\right)\hat{j}+\left(1+4\left(\frac{3}{14}\right)\right)\hat{k}\right]-6+5\left(\frac{3}{14}\right)=0 \\ \Rightarrow \overrightarrow{r}.\left(20\hat{i}+23\hat{j}+26\hat{k}\right)=69 \end{gathered}$$