If vec a = 2sin theta hat i - hat j + 2hat k, | vedclass

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(A) Since the vectors $\vec a, \vec b$,and $\vec c$ are coplanar,their scalar triple product must be zero: $[\vec a \vec b \vec c] = 0$.This is given

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$n\pi + {(-1)^n}\frac{\pi}{6}, n \in I$

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(A) Since the vectors $\vec a, \vec b$,and $\vec c$ are coplanar,their scalar triple product must be zero: $[\vec a \vec b \vec c] = 0$.This is given by the determinant:$\begin{vmatrix} 2\sin 	heta & -1 & 2 \ 2 & 2\sin 	heta & -1 \ 4 & -1 & 4\cos^2 	heta \end{vmatrix} = 0$Expanding the determinant along the first row:$2\sin 	heta (8\sin 	heta \cos^2 	heta - 1) + 1(8\cos^2 	heta + 4) + 2(-2 - 8\sin 	heta) = 0$$16\sin^2 	heta \cos^2 	heta - 2\sin 	heta + 8\cos^2 	heta + 4 - 4 - 16\sin 	heta = 0$$4\sin^2(2	heta) + 8(1 - \sin^2 	heta) - 18\sin 	heta = 0$$4(4\sin^2 	heta \cos^2 	heta) + 8 - 8\sin^2 	heta - 18\sin 	heta = 0$Solving this equation,we find that $\sin 	heta = \frac{1}{2}$ satisfies the condition.For $\sin 	heta = \frac{1}{2}$,the general solution is $	heta = n\pi + {(-1)^n}\frac{\pi}{6}$ for $n \in I$.

If $\vec a = 2\sin \theta \hat i - \hat j + 2\hat k$,$\vec b = 2\hat i + 2\sin \theta \hat j - \hat k$ and $\vec c = 4\hat i + \hat j + 4\cos^2 \theta \hat k$ are coplanar,then $\theta$ can be equal to

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