Unit vectors a, b, c are coplanar. A unit vec | vedclass

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(A) Since $a, b, c$ are coplanar,$[a, b, c] = 0$. Given $(a 	imes b) 	imes (c 	imes d) = \frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k$. Using the vec

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$\frac{i - 2j + 2k}{3}$

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(A) Since $a, b, c$ are coplanar,$[a, b, c] = 0$. Given $(a 	imes b) 	imes (c 	imes d) = \frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k$. Using the vector triple product formula $(A 	imes B) 	imes C = (A \cdot C)B - (B \cdot C)A$,we have: $[(a 	imes b) \cdot d]c - [(a 	imes b) \cdot c]d = \frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k$. Since $a, b, c$ are coplanar,$[a, b, c] = (a 	imes b) \cdot c = 0$. Thus,$[(a 	imes b) \cdot d]c = \frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k$. $|a 	imes b| = |a||b| \sin 30^{\circ} = (1)(1)(\frac{1}{2}) = \frac{1}{2}$. Let $\hat{n}$ be the unit vector perpendicular to the plane of $a$ and $b$. Then $a 	imes b = \frac{1}{2} \hat{n}$. Since $d$ is perpendicular to $a, b, c$,$d$ must be parallel to $\hat{n}$,so $d = \pm \hat{n}$. Then $[(a 	imes b) \cdot d] = (\frac{1}{2} \hat{n}) \cdot (\pm \hat{n}) = \pm \frac{1}{2}$. So,$\pm \frac{1}{2} c = \frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k$. If we take the positive sign,$c = 2(\frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k) = \frac{1}{3}i - \frac{2}{3}j + \frac{2}{3}k = \frac{i - 2j + 2k}{3}$.

Unit vectors $a, b, c$ are coplanar. $A$ unit vector $d$ is perpendicular to the given vectors. If $(a \times b) \times (c \times d) = \frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k$ and the angle between $a$ and $b$ is $30^{\circ}$,then $c = ....$

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