If a, b, c are three non-coplanar vectors and | vedclass

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(C) Since $a, b, c$ are non-coplanar,the scalar triple product $[a b c] 
eq 0$. Let $V = [a b c]$. The vectors $b 	imes c, c 	imes a, a 	imes b$ f

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$|[a b c]|$

Vedclass · Answer

(C) Since $a, b, c$ are non-coplanar,the scalar triple product $[a b c] 
eq 0$. Let $V = [a b c]$. The vectors $b 	imes c, c 	imes a, a 	imes b$ form a basis for the space of vectors. Any vector $d$ can be expressed as $d = x(b 	imes c) + y(c 	imes a) + z(a 	imes b)$. Taking the dot product with $a$: $a \cdot d = x(a \cdot (b 	imes c)) = x[a b c] \Rightarrow x = \frac{a \cdot d}{[a b c]}$. Similarly,$y = \frac{b \cdot d}{[a b c]}$ and $z = \frac{c \cdot d}{[a b c]}$. Substituting these into the expression for $d$: $d = \frac{(a \cdot d)(b 	imes c) + (b \cdot d)(c 	imes a) + (c \cdot d)(a 	imes b)}{[a b c]}$. Thus,$(a \cdot d)(b 	imes c) + (b \cdot d)(c 	imes a) + (c \cdot d)(a 	imes b) = d [a b c]$. Taking the magnitude on both sides: $|(a \cdot d)(b 	imes c) + (b \cdot d)(c 	imes a) + (c \cdot d)(a 	imes b)| = |d| |[a b c]|$. Since $d$ is a unit vector,$|d| = 1$. Therefore,the expression equals $|[a b c]|$.

If $a, b, c$ are three non-coplanar vectors and $d$ is any unit vector,then $|(a \cdot d)(b \times c) + (b \cdot d)(c \times a) + (c \cdot d)(a \times b)| = $

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