If a is perpendicular to b and c,|a| = 2,|b| | vedclass

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(C) The scalar triple product is defined as $[a \; b \; c] = a \cdot (b 	imes c)$.Since $a$ is perpendicular to both $b$ and $c$,$a$ must be parallel

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) The scalar triple product is defined as $[a \; b \; c] = a \cdot (b 	imes c)$.Since $a$ is perpendicular to both $b$ and $c$,$a$ must be parallel to the vector $b 	imes c$.Let $n$ be the unit vector perpendicular to the plane containing $b$ and $c$,such that $b 	imes c = |b||c| \sin(	heta) n$,where $	heta = \frac{2\pi}{3}$.Then $b 	imes c = 3 	imes 4 	imes \sin(\frac{2\pi}{3}) n = 12 	imes \frac{\sqrt{3}}{2} n = 6\sqrt{3} n$.Since $a$ is perpendicular to $b$ and $c$,$a$ is parallel to $n$. Thus,$a = \pm 2n$.Therefore,$[a \; b \; c] = a \cdot (b 	imes c) = (\pm 2n) \cdot (6\sqrt{3} n) = \pm 12\sqrt{3} (n \cdot n) = \pm 12\sqrt{3}$.Considering the magnitude or the standard convention for such problems,the value is $12\sqrt{3}$.

If $a$ is perpendicular to $b$ and $c$,$|a| = 2$,$|b| = 3$,$|c| = 4$ and the angle between $b$ and $c$ is $\frac{2\pi}{3}$,then $[a \; b \; c]$ is equal to (in $\sqrt{3}$):

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