If a, b, c are three non-coplanar vectors, p=b×ca b c, q=c×aa b c, r=a×ba b c then(a + b) · p + (b + c) · q + (c + a) · r =

a·p=[a b c][a b c]=1 a·q=0=a·rb·p=0 b·q=1 b·r=0c·p=0 c·q=0 c·r=1 =(a+b)·p+(b+c)·q+(c+a)·r =1+0+1+0+1+0 =3

If a, b, c are three non-coplanar vectors, p=b×ca b c, q=c×aa b c, r=a×ba b c then(a + b) · p + (b + c) · q + (c + a) · r =

a·p=[a b c][a b c]=1 a·q=0=a·rb·p=0 b·q=1 b·r=0c·p=0 c·q=0 c·r=1 =(a+b)·p+(b+c)·q+(c+a)·r =1+0+1+0+1+0 =3

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If $a, b, c$ are three non-coplanar vectors,
$p = \frac{b \times c}{[a b c]}, q = \frac{c \times a}{[a b c]}, r = \frac{a \times b}{[a b c]}$ then
$(a + b) \cdot p + (b + c) \cdot q + (c + a) \cdot r =$

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answer is C.

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Detailed Solution

$\begin{array}{rcl} a \cdot p & = & \frac{[a b c]}{[a b c]} = 1 a \cdot q = 0 = a \cdot r \\ b \cdot p & = & 0 b \cdot q = 1 b \cdot r = 0 \\ c \cdot p & = & 0 c \cdot q = 0 c \cdot r = 1 \\ = & (a + b) \cdot p + (b + c) \cdot q + (c + a) \cdot r \\ = & 1 + 0 + 1 + 0 + 1 + 0 \\ = & 3 \end{array}$