 # The vector equation of the plane passing Question:

The vector equation of the plane passing through the intersection of the planes

$\overrightarrow{\mathrm{r}} \cdot(\hat{i}+\hat{j}+\hat{k})=1$ and $\overrightarrow{\mathrm{r}} \cdot(\hat{i}-2 \hat{j})=-2$, and the point $(1,0,2)$ is :

1. (1) $\overrightarrow{\mathbf{r}} \cdot(\hat{\mathbf{i}}-7 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})=\frac{7}{3}$

2. (2) $\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+7 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=7$

3. (3) $\overrightarrow{\mathrm{r}} \cdot(3 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=7$

4. (4) $\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+7 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=\frac{7}{3}$

Correct Option: , 2

Solution:

Plane passing through intersection of plane is

$\{\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})-1\}+\lambda\{\vec{r} \cdot(\hat{i}-2 \hat{j})+2\}=0$

Passes through $\hat{\mathrm{i}}+2 \hat{\mathrm{k}}$, we get

$(3-1)+\lambda(1+2)=0 \Rightarrow \lambda=-\frac{2}{3}$

Hence, equation of plane is $3\{\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})-1\}-2\{\vec{r} \cdot(\hat{i}-2 \hat{j})+2\}=0$

$\Rightarrow \overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+7 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=7$