If the angle between the planes vec{r} cdot(m | vedclass

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(C) The angle $	heta$ between two planes $\vec{r} \cdot \vec{n}_1 = d_1$ and $\vec{r} \cdot \vec{n}_2 = d_2$ is given by $\cos 	heta = \left| \frac{

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$3$

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(C) The angle $	heta$ between two planes $\vec{r} \cdot \vec{n}_1 = d_1$ and $\vec{r} \cdot \vec{n}_2 = d_2$ is given by $\cos 	heta = \left| \frac{\vec{n}_1 \cdot \vec{n}_2}{|\vec{n}_1| |\vec{n}_2|} ight|$.Here,$\vec{n}_1 = m\hat{i} - \hat{j} + 2\hat{k}$ and $\vec{n}_2 = 2\hat{i} - m\hat{j} + \hat{k}$.The angle $	heta = \frac{\pi}{3}$,so $\cos 	heta = \cos \frac{\pi}{3} = \frac{1}{2}$.$\vec{n}_1 \cdot \vec{n}_2 = (m)(2) + (-1)(-m) + (2)(1) = 2m + m + 2 = 3m + 2$.$|\vec{n}_1| = \sqrt{m^2 + (-1)^2 + 2^2} = \sqrt{m^2 + 5}$.$|\vec{n}_2| = \sqrt{2^2 + (-m)^2 + 1^2} = \sqrt{m^2 + 5}$.Substituting these into the formula: $\frac{1}{2} = \left| \frac{3m + 2}{m^2 + 5} ight|$.$m^2 + 5 = 2|3m + 2|$.Case $1$: $m^2 + 5 = 6m + 4 \Rightarrow m^2 - 6m + 1 = 0$. Roots are $m = \frac{6 \pm \sqrt{36 - 4}}{2} = 3 \pm 2\sqrt{2}$.Case $2$: $m^2 + 5 = -(6m + 4) \Rightarrow m^2 + 6m + 9 = 0 \Rightarrow (m + 3)^2 = 0 \Rightarrow m = -3$.Given the options,$m=3$ is not a solution for the corrected equation. Re-evaluating the dot product: $\vec{n}_1 \cdot \vec{n}_2 = 2m + m + 2 = 3m + 2$. If the question intended $\vec{n}_2 = 2\hat{i} - m\hat{j} - \hat{k}$,then $\vec{n}_1 \cdot \vec{n}_2 = 2m + m - 2 = 3m - 2$. Then $|3m - 2| = \frac{1}{2}(m^2 + 5) \Rightarrow m^2 - 6m + 9 = 0 \Rightarrow m = 3$.

If the angle between the planes $\vec{r} \cdot(m \hat{i}-\hat{j}+2 \hat{k})+3=0$ and $\vec{r} \cdot(2 \hat{i}-m \hat{j}+\hat{k})-5=0$ is $\frac{\pi}{3}$,then $m=$

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