If a - b and c are non-coplanar vectors and a - c is perpendicular to a - b - c - then the value of - a - b - c - - c is equal to

The correct option is B 0 Given that a , b and c are non coplanar #160; ⇒[ a b c #160;] ≠ 0. #160;Again ( a ×( b × c ...

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Intersection of Sets

If a , b an...

Question

If $\to a, \to b$ and $\to c$ are non-coplanar vectors and $\to a \times \to c$ is perpendicular to $\to a \times (\to b \times \to c)$ , then the value of $[\to a \times (\to b \times \to c)] \times \to c$ is equal to

[\to a \to b \to c] \to c

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[\to a \to b \to c] \to b

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0

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[\to a \to b \to c] \to a

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

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, c

(a + b + c) . (p + q + 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 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

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