Question

The equation of the plane through the intersection of the planes $\overrightarrow{r}.(2\hat{i}+6\hat{j})+12=0$ and $\overrightarrow{r}.(3\hat{i}-\hat{j}+$ $4\hat{k})=0$ and at a unit distance from the origin, is

• A. $\overrightarrow{r}\cdot (2\hat{i}-\hat{j}-2\hat{k})+3=0$
• B. $\overrightarrow{r}\cdot (2\hat{i}+\hat{j}+2\hat{k})+3=0$
• C. $\overrightarrow{r}\cdot (\hat{i}-2\hat{j}+2\hat{k})-3=0$
• D. $\overrightarrow{r}\cdot (\hat{i}-2\hat{j}-2\hat{k})+3=0$

Answer 1

Answer: (B), (C)

$\overrightarrow{r}\cdot (\hat{i}-2\hat{j}+2\hat{k})-3=0$

$\overrightarrow{r}.(2\hat{i}+6\hat{j})+12=0\Rightarrow 2x+6y+12=0$
$\overrightarrow{r}\cdot (3\hat{i}-\hat{j}+4\hat{k})=0\Rightarrow 3x-y+4z=0$

Intersecting plane equation is

$2x+6y+12+\lambda (3x-y+4z)=0(2+3\lambda )x+(6-\lambda )y+4\lambda z+12=0\cdots \left(i\right)$

Distance of this plane from origin is$1$
$\left|\frac{0+0+0+12}{\sqrt{(2+3\lambda {)}^2+(6-\lambda {)}^2+(4\lambda {)}^2}}\right|=1$ $\frac{12}{\sqrt{4+9{\lambda }^2+12\lambda +36+{\lambda }^2-12\lambda +16{\lambda }^2}}=\pm 1$
$144=26{\lambda }^2+4026{\lambda }^2=104{\lambda }^2=4\lambda =\pm 2$

Putting in $\left(i\right)$, $2x+6y+12\pm 2(3x-y+4z)=0$

$2x+6y+12+6x-2y+8z=0\text{OR}2x+6y+12-6x+2y-8z=08x+4y+8z+12=0OR-4x+8y-8z+12=02x+y+2z+3=0ORx-2y+2z-3=0$

In vector form

$\overrightarrow{r}\cdot (2\hat{i}+\hat{j}+2\hat{k})+3=0$ OR
$\overrightarrow{r}\cdot (\hat{i}-2\hat{j}+2\hat{k})-3=0$