Find the equation of the plane passing through the intersection of the planes

Find the equation of the plane passing through the intersection of the planes $\vec{r}.\left ( \hat{i}+\hat{j} +\hat{k}\right )= 1$ and $\vec{r}.\left ( 2\hat{i}+3\hat{j} -\hat{k}\right )+4= 0$ and parallel to x-axis. Hence .find the distance of the plane from x-axis.

Answers (1)

Equation of plane passing through the intersection of the planes
$\vec{r}\cdot \left ( \hat{i}+\hat{j}+\hat{k} \right )-1= 0\:\: \S\: \:\vec{r}\cdot \left ( 2\hat{i}+3\hat{j}-\hat{k} \right ) +4= 0$
$\left [ \vec{r} \cdot \left ( \hat{i}+\hat{j} +\hat{k}\right )-1\right ]+\lambda \left [ \vec{r}\cdot \left ( 2\hat{i}+3\hat{j} -\hat{k} \right )+4 \right ]= 0$
$\vec{r}\cdot \left [ \left ( 1+2\lambda \right )\hat{i}+\left ( 1+3\lambda \right )\hat{j}+\left ( 1-\lambda \right )\hat{k} \right ] = 1+4\lambda = 0\: ---(1)$
Direction ration (1,0,0)
$\left ( 1+2\lambda \right )\cdot 1+0+0= 0$
$1+2\lambda = 0$
$\lambda = \frac{-1}{2}$
Equation (1) $\vec{r}\cdot \left [ \left ( \hat{i}+3\left ( \frac{-1}{2} \right )\hat{j} +\hat{k}\left ( 1-\left ( \frac{-1}{2} \right ) \right )\right ) \right ]$
$\Rightarrow\vec{r}\cdot \left [ -\frac{1}{2}\hat{j}+\frac{3}{2}\hat{k} \right ]= 3$
$\Rightarrow\vec{r}\cdot \left [ -\hat{j}+3\hat{k} \right ]= 6$
$-y+3z= 6\: \: \Rightarrow y-3z+6= 0$
Distance of the point from x axis: x(x,0,0)
$d= \frac{\left | Ax+By+Cz+D \right |}{\sqrt{A^{2}+B^{2}+C^{2}}} \: \: x= x$
$y= 0$
$z= 0$
$= \frac{\left | 0\times x+1\times 0+\: \: \times 0+6 \right |}{\sqrt{0+1+\left ( 3 \right )^{2}}}$
$\Rightarrow \frac{6}{\sqrt{10}}= \frac{6}{\sqrt{10}} \: units$