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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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