If a- b- and c- are three non-coplanar vectors- then a-b-c- -a-b- -a-c- equalsA- 0B- 2 -a-b-c-C- -a-b-c-D- -a-b-c-

The correct option is C [ a⃗b⃗c⃗ ] ( a⃗+b⃗+c⃗). [ (a⃗+b⃗) × (a⃗+c⃗) ] =( a⃗+b⃗+c⃗). [ a⃗×a⃗ + a⃗×c⃗ + b⃗×a⃗ + b⃗×c⃗ ] ={a⃗.(a⃗×c⃗)+ a⃗.(b⃗×a⃗ ) + a⃗. (b⃗ ...

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Byju's Answer

Standard IX

Mathematics

Scalar Triple Product

If a⃗, b⃗ and...

Question

If $\to a, \to b$ and $\to c$ are three non-coplanar vectors, then $(\to a + \to b + \to c) \cdot [(\to a + \to b) \times (\to a + \to c)]$ equals

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

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$2. [\to a \to b \to c]$

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$- [\to a \to b \to c]$

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Solution

The correct option is D
$- [\to a \to b \to c]$

$(\to a + \to b + \to c) . [(\to a + \to b) \times (\to a + \to c)] = (\to a + \to b + \to c) . [\to a \times \to a + \to a \times \to c + \to b \times \to a + \to b \times \to c] = {\to a . (\to a \times \to c) + \to a . (\to b \times \to a) + \to a . (\to b \times \to c)} + {\to b . (\to a \times \to c) + \to b . (\to b \times \to a) + \to b . (\to b \times \to c)} + {\to c . (\to a \times \to c) + \to c . (\to b \times \to a) + \to c . (\to b \times \to c)} \Rightarrow [\to a \to b \to c] + [\to b \to a \to c] + [\to c \to b \to a] \Rightarrow - [\to a \to b \to c]$

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Similar questions

If $\to a, \to b$ and $\to c$ are three non-coplanar vectors, then $(\to a + \to b + \to c) \cdot [(\to a + \to b) \times (\to a + \to c)]$ equals

Q. If

\to a, \to b, \to c

are three non coplanar vectors,

\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]}

, then

(2 \to a + 3 \to b + 4 \to c) . \to p + (2 \to b + 3 \to c + 4 \to a) . \to q +

(2 \to c + 3 \to a + 4 \to b) . \to r =

Q. If

\to a, \to b, \to c

are vectors such that

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

then

[\to a \times \to b

\to b \times \to c

\to c \times \to a] = ?

Q. If

\to a, \to b, \to c

are three vectors such that

\to a, \to b, \to c

, then prove that

\to a \times \to b = \to b \times \to c = \to c \times \to a

, and hence show that

[\to a \to b \to c] = 0.

Q. If

\to a, \to b, \to c

are three vectors such that

\to a + \to b + \to c = \to 0,

then prove that

\to a \times \to b = \to b \times \to c = \to c \times \to a,

and hence show that

[\to a \to b \to c] = 0.

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If →a,→b and →c are three non-coplanar vectors, then (→a+→b+→c)⋅[(→a+→b)×(→a+→c)] equals

If $\to a, \to b$ and $\to c$ are three non-coplanar vectors, then $(\to a + \to b + \to c) \cdot [(\to a + \to b) \times (\to a + \to c)]$ equals