If planes vec{r} cdot (p hat{i} - hat{j} + 2 | vedclass

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(D) The normal vectors to the planes are $\vec{n}_1 = p \hat{i} - \hat{j} + 2 \hat{k}$ and $\vec{n}_2 = 2 \hat{i} - p \hat{j} - \hat{k}$.Given that th

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$3$

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(D) The normal vectors to the planes are $\vec{n}_1 = p \hat{i} - \hat{j} + 2 \hat{k}$ and $\vec{n}_2 = 2 \hat{i} - p \hat{j} - \hat{k}$.Given that the angle between the planes is $	heta = \frac{\pi}{3}$,the cosine of the angle between their normals is given by $\cos 	heta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| |\vec{n}_2|}$.$\vec{n}_1 \cdot \vec{n}_2 = (p)(2) + (-1)(-p) + (2)(-1) = 2p + p - 2 = 3p - 2$.$|\vec{n}_1| = \sqrt{p^2 + (-1)^2 + 2^2} = \sqrt{p^2 + 5}$.$|\vec{n}_2| = \sqrt{2^2 + (-p)^2 + (-1)^2} = \sqrt{p^2 + 5}$.Thus,$\cos \left( \frac{\pi}{3} ight) = \frac{|3p - 2|}{\sqrt{p^2 + 5} \sqrt{p^2 + 5}}$.$\frac{1}{2} = \frac{|3p - 2|}{p^2 + 5}$.Case $1$: $3p - 2 = \frac{1}{2}(p^2 + 5) \implies 6p - 4 = p^2 + 5 \implies p^2 - 6p + 9 = 0 \implies (p - 3)^2 = 0 \implies p = 3$.Case $2$: $-(3p - 2) = \frac{1}{2}(p^2 + 5) \implies -6p + 4 = p^2 + 5 \implies p^2 + 6p + 1 = 0 \implies p = \frac{-6 \pm \sqrt{36 - 4}}{2} = -3 \pm 2\sqrt{2}$.Since $3$ is the only option provided in the choices,the value of $p$ is $3$.

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

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