# 12. Find the angle between the planes whose vector equations are $\overrightarrow{r}.(2\widehat{i}+2\widehat{j}-3\widehat{k})= 5$ and $\overrightarrow{r}.(3\widehat{i}-3\widehat{j}+5\widehat{k})= 3$.

Given two vector equations of plane

$\overrightarrow{r}.(2\widehat{i}+2\widehat{j}-3\widehat{k})= 5$ and $\overrightarrow{r}.(3\widehat{i}-3\widehat{j}+5\widehat{k})= 3$.

Here, $\vec{n_{1}} = 2\widehat{i}+2\widehat{j}-3\widehat{k}$  and  $\vec{n_{2}} = 3\widehat{i}-3\widehat{j}+5\widehat{k}$

The formula for finding the angle between two planes,

$\cos A = \left | \frac{\vec{n_{1}}.\vec{n_{2}}}{|\vec{n_{1}}||\vec{n_{2}}|} \right |$                                         .............................(1)

$\vec{n_{1}}.\vec{n_{2}} = (2\widehat{i}+2\widehat{j}-3\widehat{k})(3\widehat{i}-3\widehat{j}+5\widehat{k}) = 2(3)+2(-3)-3(5) = -15$

$|\vec{n_{1}}| =\sqrt{(2)^2+(2)^2+(-3)^2} =\sqrt{17}$

and       $|\vec{n_{2}}| =\sqrt{(3)^2+(-3)^2+(5)^2} =\sqrt{43}$

Now, we can substitute the values in the angle formula (1) to get,

$\cos A = \left | \frac{-15}{\sqrt{17}\sqrt{43}} \right |$

or  $\cos A =\frac{15}{\sqrt{731}}$

or  $A = \cos^{-1}\left ( \frac{15}{\sqrt{731}} \right )$

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