Find the angle between the planes whose vecto | vedclass

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(A) 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

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

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(A) The equations of the given planes are $\vec{r} \cdot (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) = 5$ and $\vec{r} \cdot (3 \hat{i} - 3 \hat{j} + 5 \hat{k}) = 3$.The normal vectors to the planes are $\vec{n}_{1} = 2 \hat{i} + 2 \hat{j} - 3 \hat{k}$ and $\vec{n}_{2} = 3 \hat{i} - 3 \hat{j} + 5 \hat{k}$.The angle $Q$ between the two planes is given by the formula:$\cos Q = \left| \frac{\vec{n}_{1} \cdot \vec{n}_{2}}{|\vec{n}_{1}| |\vec{n}_{2}|} \right|$First,calculate the dot product $\vec{n}_{1} \cdot \vec{n}_{2}$:$\vec{n}_{1} \cdot \vec{n}_{2} = (2)(3) + (2)(-3) + (-3)(5) = 6 - 6 - 15 = -15$.Next,calculate the magnitudes $|\vec{n}_{1}|$ and $|\vec{n}_{2}|$:$|\vec{n}_{1}| = \sqrt{2^2 + 2^2 + (-3)^2} = \sqrt{4 + 4 + 9} = \sqrt{17}$.$|\vec{n}_{2}| = \sqrt{3^2 + (-3)^2 + 5^2} = \sqrt{9 + 9 + 25} = \sqrt{43}$.Now,substitute these values into the formula:$\cos Q = \left| \frac{-15}{\sqrt{17} \cdot \sqrt{43}} \right| = \frac{15}{\sqrt{731}}$.Thus,the angle $Q$ is $\cos^{-1} \left( \frac{15}{\sqrt{731}} \right)$.

Find the angle between the planes whose vector equations are $\vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=5$ and $\vec{r} \cdot(3 \hat{i}-3 \hat{j}+5 \hat{k})=3$.

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