The value of -a-b- b-c-c-a- where -a-1-b-5-c-3- isA- 0B- 1C- None of theseD- 6

The correct option is A 0Since,a b+b c+c a=0, nbsp;therefore the vectors a b,b c and c a nbsp;are coplanar. We know that if three vectors are coplanar, the ...

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Standard X

Mathematics

Scalar Triple Product

The value of ...

Question

The value of $[\to a - \to b \to b - \to c \to c - \to a]$ where $| \to a | = 1, | \to b | = 5, | \to c | = 3$ , is

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None of these

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Solution

The correct option is A $0$
Since, $\to a - \to b + \to b - \to c + \to c - \to a = 0$ , therefore the vectors
$\to a - \to b, \to b - \to c and \to c - \to a$ are coplanar. We know that
if three vectors are coplanar, then their box product would be equal to zero. Also try to recollect why
this is so.
Hence for our present case, $[\to a - \to b \to b - \to c \to c - \to a] = 0$ .

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Similar questions

\frac{[(\to a \times \to b) \times (\to b \times \to c) (\to b \times \to c) \times (\to c \times \to a) (\to c \times \to a) \times (\to a \times \to b)]}{[\to a \times \to b \to b \times \to c \to c \times \to a]}

is equal to

Q. If

\to a, \to b, \to c

be any three non - coplanar vectors, then calculate the value of

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

Q. Prove that for any three vectors

\to a, \to b

and

\to c,

[(\to a + \to b) (\to b + \to c) (\to c + \to a)] = 2 [\to a \to b \to c] .

[\to a + \to b \to b + \to c \to c + \to a] = [\to a \to b \to c]

, then

[\begin{matrix} \to a + \to b & \to b + \to c & \to c + \to a \end{matrix}]

is equal to

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The value of [→a−→b →b−→c →c−→a] where |→a|=1,|→b|=5,|→c|=3, is

The value of $[\to a - \to b \to b - \to c \to c - \to a]$ where $| \to a | = 1, | \to b | = 5, | \to c | = 3$ , is