Let a, b and c be three coplanar unit vectors | vedclass

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(A) Since $a, b, c$ are coplanar unit vectors and $d$ is perpendicular to them, we have $d$ as a unit vector perpendicular to the plane of $a, b, c$.

Vedclass · Accepted Answer

$\frac{3}{13} i - \frac{4}{13} j + \frac{12}{13} k$

Vedclass · Answer

(A) Since $a, b, c$ are coplanar unit vectors and $d$ is perpendicular to them, we have $d$ as a unit vector perpendicular to the plane of $a, b, c$. Thus, $(a 	imes b) = |a||b| \sin(30^\circ) d = (1)(1)(1/2) d = \frac{1}{2} d$.Using the vector triple product identity $(a 	imes b) 	imes (c 	imes d) = ((a 	imes b) \cdot d) c - ((a 	imes b) \cdot c) d$.Since $(a 	imes b) = \frac{1}{2} d$, we have $(a 	imes b) \cdot d = \frac{1}{2} (d \cdot d) = \frac{1}{2}$ and $(a 	imes b) \cdot c = \frac{1}{2} (d \cdot c) = 0$ (as $d \perp c$).Substituting these into the identity: $\frac{1}{2} c - 0 = \frac{3}{26} i - \frac{2}{13} j + \frac{6}{13} k$.Therefore, $c = 2(\frac{3}{26} i - \frac{2}{13} j + \frac{6}{13} k) = \frac{3}{13} i - \frac{4}{13} j + \frac{12}{13} k$.

Let $a, b$ and $c$ be three coplanar unit vectors. $A$ unit vector $d$ is perpendicular to them. If $(a \times b) \times (c \times d) = \frac{3}{26} i - \frac{2}{13} j + \frac{6}{13} k$ and the angle between $a$ and $b$ is $30^\circ$, then $c$ is equal to...

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