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Find the acute angle between the planes \vec{r}\cdot \left ( \hat{i}-2\hat{j}-2\hat{k} \right )= 1  and \vec{r}\cdot \left ( 3\hat{i}-6\hat{j}+2\hat{k} \right )= 0\cdot

 

 

 

 
 
 
 
 

Answers (1)
R Ravindra Pindel

The angle between the two plane is same as the angle between their normal vectors. The normals of the given plane are \hat{i}-2\hat{j}-2\hat{k}\: \: \& \: \: 3\hat{i}-6\hat{j}+2\hat{k}
So, \cos \theta =\tfrac{\left | \left ( \hat{i}-2\hat{j}-2\hat{k}\right )\cdot \left ( 3\hat{i}-6\hat{j}+2\hat{k} \right ) \right |}{\sqrt{1+4+4}\sqrt{9+36+4}} 
where  \theta is the required acute angle.
\therefore \theta = \cos^{-1}\frac{11}{21}
 

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