Find the angle between the planes whose vector equations are

Question:

Find the angle between the planes whose vector equations are

$\vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=5$ and $\vec{r} \cdot(3 \hat{i}-3 \hat{j}+5 \hat{k})=3$.

Solution:

The equations of the given planes are $\vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=5$ and $\vec{r} \cdot(3 \hat{i}-3 \hat{j}+5 \hat{k})=3$

It is known that if $\vec{n}_{1}$ and $\vec{n}_{2}$ are normal to the planes, $\vec{r} \cdot \vec{n}_{1}=d_{1}$ and $\vec{r} \cdot \vec{n}_{2}=d_{2}$, then the angle between them, $Q$, is given by,

$\cos Q=\left|\frac{\vec{n}_{1} \cdot \vec{n}_{2}}{\left|\vec{n}_{1}\right|\left|\vec{n}_{2}\right|}\right|$                  $\ldots(1)$

Here, $\vec{n}_{1}=2 \hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{n}_{2}=3 \hat{i}-3 \hat{j}+5 \hat{k}$

$\therefore \vec{n}_{1} \cdot \vec{n}_{2}=(2 \hat{i}+2 \hat{j}-3 \hat{k})(3 \hat{i}-3 \hat{j}+5 \hat{k})=2.3+2 .(-3)+(-3) .5=-15$

$\left|\vec{n}_{1}\right|=\sqrt{(2)^{2}+(2)^{2}+(-3)^{2}}=\sqrt{17}$

$\left|\vec{n}_{2}\right|=\sqrt{(3)^{2}+(-3)^{2}+(5)^{2}}=\sqrt{43}$

Substituting the value of $\vec{n} \cdot \vec{n}_{2},\left|\vec{n}_{1}\right|$ and $\left|\vec{n}_{2}\right|$ in equation (1), we obtain

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

$\Rightarrow \cos Q=\frac{15}{\sqrt{731}}$

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

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