Find the vector equation of the plane which contains the line of intersection of the planes \vec{r}\cdot \left ( \hat{i}+2\hat{j}+3\hat{k} \right )-4= 0,\vec{r}\cdot \left ( 2\hat{i}+\hat{j}-\hat{k} \right )+5= 0    and which is perpendicular to the plane \vec{r}\cdot \left ( 5\hat{i}+3\hat{j}-6\hat{k} \right )+8= 0

 

 

 

 
 
 
 
 

Answers (1)

The required plane is \vec{r}\cdot \left ( \hat{i}+2\hat{j}+3\hat{k} \right )-4+\lambda \left ( \vec{r} \cdot \left (2\hat{i}+\hat{j}-\hat{k} \right )\right +5)= 0
i.e  \vec{r}\left \{ \hat{i} \left ( 1+2\lambda \right )+\hat{j}\left ( 2+\lambda \right )+\hat{k}\left ( 3-\lambda \right )\right \}
                                          = 4-5\lambda ---(i)
As (i) is perpendicular to the plane \vec{r}\cdot \left ( 5\hat{i}+3\hat{j}-6\hat{k} \right )+8= 0
so. 5\left ( 1+2\lambda \right )+3\left ( 2+\lambda \right )-6\left ( 3-\lambda \right )= 0
                           \lambda = \frac{7}{19}
By (i) \vec{r}\cdot \left \{ \hat{i}\left ( 1+\frac{14}{19} \right )+\hat{j}\left ( 2+\frac{7}{19} \right )+\hat{k} \left ( 3-\frac{7}{19} \right )\right \}= 4-\frac{35}{19}
\therefore \vec{r}\cdot \left ( 33\hat{i}+45\hat{j}+50\hat{k} \right )= 41

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