Find the equation of planes passing through the intersection of the planes \vec{r}\cdot \left ( 2\hat{i} +6\hat{j}\right )+12= 0\; and\;\vec{r}\cdot \left ( 3\hat{i} -\hat{j}+4\hat{k}\right ) = 0  and are at  a unit distance from origin.

 

 

 

 
 
 
 
 

Answers (1)

Required plane is \vec{r}\cdot \left ( 2\hat{i}+6\hat{j} \right )+12+\lambda \left [\vec{r}\cdot \left ( 3\hat{i}-\hat{j}+4\hat{k} \right ) \right ]= 0
ie \vec{r}\cdot \left \{ \left ( 2\hat{i} +6\hat{j}\right )+\lambda \left ( 3\hat{i}-\hat{j} +4\hat{k}\right )\right \}+12= 0---(i)
writing (i) in cartesian form \left ( 2+3\lambda \right )x+\left ( 6-\lambda \right )y+4\lambda z+12= 0
Distance of this plane from origin is 
                               = \frac{\left | {\left ( 2+3\lambda \right )\cdot 0+\left ( 6-\lambda \right )0+4\lambda \cdot 0+12} \right |}{\sqrt{\left ( 2+3\lambda \right )^{2}+\left ( 6-\lambda \right )^{2}+\left ( 4\lambda \right )^{2}}}= 1
\Rightarrow \frac{144}{26\lambda ^{2}+40}= 1\; \; \Rightarrow \lambda ^{2}= 4\; \; \therefore \lambda = 2,-2
so, the equations of the plane are 2x+y+2z+3= 0\; \; and\; \; x-2y+2z-3= 0
That is \vec{r}\cdot \left ( 2\hat{i}+\hat{j}+2\hat{k} \right )+3= 0\; \; and\; \; \vec{r}\cdot \left ( \hat{i}-2\hat{j}+2\hat{k} \right )-3= 0
 

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