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

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(D) Given that $p=\frac{b 	imes c}{[a b c]}, q=\frac{c 	imes a}{[a b c]}, r=\frac{a 	imes b}{[a b c]}$.We know that $[a b c] = a \cdot (b 	imes c)

Vedclass · Accepted Answer

$3$

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(D) Given that $p=\frac{b 	imes c}{[a b c]}, q=\frac{c 	imes a}{[a b c]}, r=\frac{a 	imes b}{[a b c]}$.We know that $[a b c] = a \cdot (b 	imes c) = b \cdot (c 	imes a) = c \cdot (a 	imes b)$.Let $T_{1} = (a+b) \cdot p = a \cdot p + b \cdot p$.$T_{1} = a \cdot \frac{b 	imes c}{[a b c]} + b \cdot \frac{b 	imes c}{[a b c]} = \frac{[a b c]}{[a b c]} + \frac{[b b c]}{[a b c]} = 1 + 0 = 1$.Similarly,$T_{2} = (b+c) \cdot q = b \cdot q + c \cdot q = b \cdot \frac{c 	imes a}{[a b c]} + c \cdot \frac{c 	imes a}{[a b c]} = \frac{[b c a]}{[a b c]} + 0 = 1$.Similarly,$T_{3} = (c+a) \cdot r = c \cdot r + a \cdot r = c \cdot \frac{a 	imes b}{[a b c]} + a \cdot \frac{a 	imes b}{[a b c]} = \frac{[c a b]}{[a b c]} + 0 = 1$.Therefore,$(a+b) \cdot p + (b+c) \cdot q + (c+a) \cdot r = T_{1} + T_{2} + T_{3} = 1 + 1 + 1 = 3$.

If $a, b$ and $c$ are three non-coplanar vectors and $p, q$ and $r$ are vectors defined by $p=\frac{b \times c}{[a b c]}, q=\frac{c \times a}{[a b c]}$ and $r=\frac{a \times b}{[a b c]}$,then the value of $(a+b) \cdot p+(b+c) \cdot q+(c+a) \cdot r$ is equal to

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