|(a times b) cdot c| = |a| |b| |c|,if | vedclass

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(D) The scalar triple product is defined as $|(a 	imes b) \cdot c| = |a| |b| |c| |\sin 	heta| |\cos \alpha|$,where $	heta$ is the angle between $a$

Vedclass · Accepted Answer

$a \cdot b = b \cdot c = c \cdot a = 0$

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(D) The scalar triple product is defined as $|(a 	imes b) \cdot c| = |a| |b| |c| |\sin 	heta| |\cos \alpha|$,where $	heta$ is the angle between $a$ and $b$,and $\alpha$ is the angle between $(a 	imes b)$ and $c$.Given $|(a 	imes b) \cdot c| = |a| |b| |c|$,we have $|\sin 	heta| |\cos \alpha| = 1$.Since the maximum value of $|\sin 	heta|$ and $|\cos \alpha|$ is $1$,this equality holds only if $|\sin 	heta| = 1$ and $|\cos \alpha| = 1$.$|\sin 	heta| = 1 \Rightarrow 	heta = \frac{\pi}{2}$,which means $a \perp b$ $(a \cdot b = 0)$.$|\cos \alpha| = 1 \Rightarrow \alpha = 0$ or $\pi$,which means $c$ is parallel to the vector $(a 	imes b)$.Since $(a 	imes b)$ is perpendicular to both $a$ and $b$,$c$ must also be perpendicular to both $a$ and $b$.Therefore,$c \perp a$ $(c \cdot a = 0)$ and $c \perp b$ $(c \cdot b = 0)$.Thus,$a, b, c$ are mutually perpendicular,implying $a \cdot b = b \cdot c = c \cdot a = 0$.

$|(a \times b) \cdot c| = |a| |b| |c|$,if

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