If a-b- c- are three non-coplanar non-zero vectors and r- is any vector- then a-b-r-c-b-c-r-a-c-a-r-b-

The correct option is B 2 [ a⃗b⃗c⃗] r⃗ We have, (a×b)×(r×c)+(b×c)×(r×a)+(c×a)×(r×b) =[ a #160;b #160;c] r [ a #160;b #160;r] #160;c + ...

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Scalar Triple Product

If a⃗,b⃗, c...

Question

If $\to a, \to b, \to c$ are three non-coplanar non-zero vectors and $\to r$ is any vector, then
$(\to a \times \to b) \times (\to r \times \to c) + (\to b \times \to c) \times (\to r \times \to a) + (\to c \times \to a) \times (\to r \times \to b) =$

3 [\to a \to b \to c] \to r

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

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2 [\to a \to b \to c] \to r

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4 [\to a \to b \to c] \to r

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Solution

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Similar questions

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

\to a = \to i + \to j - \to k, \to b = 5 \to i - 3 \to j - 3 \to k, \to c = 3 \to i - \to j + 2 \to k

\to r

\to c

| \to r | = ∣ ∣ \to a + \to b ∣ ∣

\to r

\to a, \to b, \to c

\to r

(\to a \times \to b) \times (\to r \times \to c) + (\to b \times \to c) \times (\to r \times \to a) + (\to c \times \to a) \times (\to r \times \to b)

\to r \times \to a = \to b, \to r . \to a = 1,

\to r = \to a + \to a \times \to b .

\to r \times \to a = \to b

\to r = \to a \times \to b

\to A = 2 \to i + \to k, \to B = \to i + \to j + \to k

\to C = 4 \to i - 3 \to j + 7 \to k

\to R

\to R \times \to B = \to C \times \to B

\to R \cdot \to A = 0

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