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

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