If a- b- c are three non-coplanar vectors- then a - b - c - a - b - a - c is equal to

The correct option is D [a b c] The given expression is equal to #160; (a+b+c)·((a+b)×(a+c)) =( a+b+c ) .( a× a+a× c+b× a+b× c ) =b.( a× ...

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Unit Vectors

If a, b, c ...

Question

If $a, b, c$ are three non-coplanar vectors, then $(a + b + c) \cdot ((a + b) \times (a + c))$ is equal to

0

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[a b c]

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2 [a b c]

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- [a b c]

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\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 =

\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 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 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)

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