Question

Find the equation of the plane through the intersection of the planes ${\displaystyle \overrightarrow{r}\cdot (\hat{i}+3\hat{j})-6}$ = 0 and ${\displaystyle \overrightarrow{r}\cdot (3\hat{i}+\hat{j}+4\hat{k})}$ = 0, whose perpendicular distance from origin is unity.

Answer 1

The equation of family of planes passing through the intersection of two given planes 

${\displaystyle \overrightarrow{r}\cdot (\hat{i}+3\hat{j})-6}$ = 0 and ${\displaystyle \overrightarrow{r}\cdot (3\hat{i}+\hat{j}+4\hat{k})}$ = 0 is given by

${\displaystyle [\overrightarrow{r}\cdot (\hat{i}+3\hat{j})-6]+\lambda [\overrightarrow{r}\cdot (3\hat{i}-\hat{j}-4\hat{k})]}$ = 0

⇒ ${\displaystyle \overrightarrow{r}\cdot [(\hat{i}+3\hat{j})+\lambda (3\hat{i}-\hat{j}-4\hat{k})]-6}$ = 0

⇒ ${\displaystyle \overrightarrow{r}\cdot [(1+3\lambda )\hat{i}+(3-\lambda )\hat{j}-4\lambda \hat{k}]-6}$ = 0.   ......(1)

Given that the perpendicular distance of the origin from the required plane is unity.

Now, the position vector of origin is ${\displaystyle \overrightarrow{a}=0\hat{i}+0\hat{j}+0\hat{k}}$

The normal vector to the required plane is 

${\displaystyle \overrightarrow{N}=(1+3\lambda )\hat{i}+(3-\lambda )\hat{j}-4\lambda \hat{k}}$ and d = 6.

(Because equation of required plane is ${\displaystyle \overrightarrow{r}[(1+3\lambda )\hat{i}+(3-\lambda )\hat{j}-4\lambda \hat{k}]}$ = 6 and the equation of plane in normal form is given by ${\displaystyle \overrightarrow{N}=(1+3\lambda )\hat{i}+(3-\lambda )\hat{j}-4\lambda \hat{k}}$ and d = 6)

We know that the perpendicular distance of a point whose position vector is ${\displaystyle \overrightarrow{a}}$ from the plane

${\displaystyle \overrightarrow{r}\cdot \overrightarrow{n}}$ = d is given by ${\displaystyle \left|\frac{\overrightarrow{a}.\overrightarrow{N}-d}{\left|\overrightarrow{N}\right|}\right|}$

Therefore, the perpendicular distance of the origin from the required plane is ${\displaystyle \left|\frac{\overrightarrow{a}.\overrightarrow{N}-d}{\left|\overrightarrow{N}\right|}\right|}$ = 1.

⇒ ${\displaystyle \left|\frac{(0\hat{i}+0\hat{j}+0\hat{k})\cdot ((1+3\lambda )\hat{i}+(3-\lambda )\hat{j}-4\lambda \hat{k})-6}{|(1+3\lambda )\hat{i}+(3-\lambda )\hat{j}+-4\lambda \hat{k}|}\right|}$ = 1    .....${\displaystyle (\because \hat{i}\cdot \hat{i}=\hat{j}\cdot \hat{j}=\hat{k}\cdot \hat{k}=1\text{and}\hat{i}\cdot \hat{j}=\hat{j}\cdot \hat{k}\cdot \hat{i}=0)}$

⇒ ${\displaystyle \left|\frac{-6}{\sqrt{{(1+3\lambda )}^2+{(3-\lambda )}^2+{(-4\lambda )}^2}}\right|}$ = 1

⇒ ${\displaystyle \sqrt{{(1+3\lambda )}^2+{(3-\lambda )}^2+{(-4k=\lambda )}^2}}$ = 6

⇒ ${\displaystyle {(1+3\lambda )}^2+{(3-\lambda )}^2+{(-4\lambda )}^2}$ = 36

⇒ ${\displaystyle 1+9{\lambda }^2+6\lambda +9+{\lambda }^2-6\lambda +16{\lambda }^2}$ = 36

⇒ ${\displaystyle 26{\lambda }^2}$ = 26

⇒ ${\displaystyle {\lambda }^2}$ = 1

⇒ ${\displaystyle \lambda =\pm 1}$.

Therefore, the equation of required plane is 

${\displaystyle \overrightarrow{r}\cdot [(1+3\left(1\right)\hat{i}+(3-1)\hat{j}-4\left(1\right)\hat{k}]-6}$ = 0

or 

${\displaystyle \overrightarrow{r}\cdot [(1+3(-1))\hat{i}+(3-(-1))\hat{j}-4(-1)\hat{k}]-6}$ = 0.

Hence, the equation of required plane is

${\displaystyle \overrightarrow{r}\cdot [4\hat{i}+2\hat{j}-4\hat{k}]}$ = 6 or ${\displaystyle \overrightarrow{r}\cdot [-2\hat{i}+4\hat{j}+4\hat{k}]}$ = 6.

And the Cartesian equation of required plane is 4x + 2y − 4z = or – 2x + 4y + 4z = 6.