Question

The vector equation of the plane passing through the intersection of the planes $\vec{r}.(\hat{i}+\hat{j}+\hat{k})=1$ and $\vec{r}.(\hat{i}-2\hat{j})=-2,$ and the point (1,0,2) is :

• A. $\vec{r}.(\hat{i}+7\hat{j}+3\hat{k})=\frac{7}{3}$
• B. $\vec{r}\cdot (3\hat{i}+7\hat{j}+3\hat{k})=7$
• C. $\vec{r}\cdot (\hat{i}+7\hat{j}+3\hat{k})=7$
• D. $\vec{r}.(\hat{i}-7\hat{j}+3\hat{k})=\frac{7}{3}$

Answer 1

Answer: (C)

$\vec{r}\cdot (\hat{i}+\hat{j}+\hat{k})=1$
$\vec{r}\cdot (\hat{i}-2\hat{j})=-2$
point (1,0,2) Eq ${}^n$ of plane
$\vec{r}\cdot (\hat{i}+\hat{j}+\hat{k})-1+\lambda \{r.(\hat{i}-2\hat{j})+2\}=0$
$\vec{r}\cdot \{\hat{i}(1+\lambda )+\hat{j}(1-2\lambda )+\hat{k}(1)\}-1+2\lambda =0$
Point $\hat{i}+0\hat{j}+2\hat{k}=\vec{r}$
$\therefore (\hat{i}+2\hat{k})\cdot \{\hat{i}(1+\lambda )+\hat{j}(1-2\lambda )+\hat{k}(1)\}-1+2\lambda =0$
$1+\lambda +2-1+2\lambda =0$
$\lambda =-\frac{2}{3}$
$\therefore \vec{r}\cdot \left[\hat{i}\left(\frac{1}{3}\right)+\hat{j}\left(\frac{7}{3}\right)+\hat{k}\right]$
$=\frac{7}{3}$
$r\cdot [\hat{i}+7\hat{j}+3\hat{k}]=7$