If a non-zero vector vec{a} is parallel to th | vedclass

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(D) The normal $\vec{n}_1$ to the plane $P_1$ determined by $\hat{j}-\hat{k}$ and $3\hat{j}-2\hat{k}$ is given by the cross product: $\vec{n}_1 = (\ha

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$\cos^{-1}\left(\pm\frac{1}{\sqrt{3}}\right)$

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(D) The normal $\vec{n}_1$ to the plane $P_1$ determined by $\hat{j}-\hat{k}$ and $3\hat{j}-2\hat{k}$ is given by the cross product: $\vec{n}_1 = (\hat{j}-\hat{k}) 	imes (3\hat{j}-2\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 1 & -1 \ 0 & 3 & -2 \end{vmatrix} = \hat{i}(-2+3) = \hat{i}$. The normal $\vec{n}_2$ to the plane $P_2$ determined by $2\hat{i}+3\hat{j}$ and $\hat{i}-3\hat{j}$ is given by: $\vec{n}_2 = (2\hat{i}+3\hat{j}) 	imes (\hat{i}-3\hat{j}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 0 \ 1 & -3 & 0 \end{vmatrix} = \hat{k}(-6-3) = -9\hat{k}$. Since $\vec{a}$ is parallel to the line of intersection of planes $P_1$ and $P_2$,it is parallel to $\vec{n}_1 	imes \vec{n}_2$: $\vec{a} = k(\vec{n}_1 	imes \vec{n}_2) = k(\hat{i} 	imes -9\hat{k}) = k(9\hat{j}) = 9k\hat{j}$. We can take $\vec{a} = \pm\hat{j}$. The angle $	heta$ between $\vec{a} = \hat{j}$ and $\vec{b} = \hat{i}+\hat{j}+\hat{k}$ is given by: $\cos 	heta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{1}{1 \cdot \sqrt{1^2+1^2+1^2}} = \frac{1}{\sqrt{3}}$. Considering the direction $\vec{a} = -\hat{j}$,we get $\cos 	heta = -\frac{1}{\sqrt{3}}$. Thus,$	heta = \cos^{-1}\left(\pm\frac{1}{\sqrt{3}}ight)$.

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