Let a- b and c be three non-coplanar vectors and let p- q and r be vectors defined by the relations p-b- cabc- q-c- aabc and r - a- babc Then the value of the expression a-b -p - b -c -q- c - a-r is equal to

The correct option is C 3 Given that, #160; p=b× c/abc, q=c× a/abc and r = a× b/abc ∴ a.p=a.(b× c)/abc=a.(b× c)/abc=1 And a.q =a.(c× ...

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Let a, b an...

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Let $a, b$ and $c$ be three non-coplanar vectors and let $p, q$ and $r$ be vectors defined by the relations
$p = \frac{b \times c}{a b c}, q = \frac{c \times a}{a b c}$ and $r = \frac{a \times b}{a b c}$
Then the value of the expression
$(a + b) . p + (b + c) . q + (c + a) . r$ is equal to

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\to a, \to b, \to c

\to p, \to q, \to r

\to p = \frac{\to b \times \to c}{[\to a \to b \to c]}, \to q = \frac{\to c \times \to a}{[\to a \to b \to c]}, \to r = \frac{\to a \times \to b}{[\to a \to b \to c]}

(\to a + \to b) . \to p + (\to b + \to c) . \to q + (\to c + \to a) . \to r

\to a, \to b, \to c

\to p = \frac{\to b \times \to a}{[\to a \to b \to c]}, \to q = \frac{\to c \times \to a}{[\to a \to b \to c]}, \to r = \frac{\to a \times \to b}{[\to a \to b \to c]}

(\to a + \to b) . \to p + (\to b + \to c) . \to q + (\to c + \to a) . \to r =

a, b

c

p, q

r

p = \frac{b \times c}{[a b c]}, q = \frac{c \times a}{[a b c]},

r = \frac{a \times b}{[a b c]},

(a + b) \cdot p + (b + c) \cdot q + (c + a) \cdot r

\to a, \to b, \to c

\to p, \to p, \to r

\to p = \frac{\to b \times \to c}{[\to b \to c \to a]}, \to q = \frac{\to c \times \to a}{[\to c \to a \to b]}, \to r = \frac{\to a \times \to b}{[\to a \to b \to c]}, (\to a + \to b) . \to p + (\to b + \to c) . \to q + (\to c + \to a) . \to r

\to a, \to b, \to c

\to p, \to q, \to r

\to p = \frac{\to b \times \to c}{[\to a \to b \to c]}, \to q = \frac{\to c \times \to a}{[\to a \to b \to c]}, \to r = \frac{\to a \times \to b}{[\to a \to b \to c]},

(\to a + \to b + \to c), (\to p + \to q + \to r)

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