The correct option is D (#160;a.#160;c)(#160;b.#160;d) We have #160; (a⃗×b⃗).(c⃗×d⃗)+(b⃗.c)(a⃗.d⃗) = a⃗· ((b⃗·d⃗)c⃗ (b⃗· #160;c⃗)d⃗)+(b⃗·c⃗)( ...

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Applications of Cross Product

a⃗×b⃗.c⃗×d⃗+b...

Question

$(\to a \times \to b) . (\to c \times \to d) + (\to b . \to c) (\to a . \to d) =$

0

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\to a . \to c

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(\to a . \to c) (\to b . \to d)

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Solution

The correct option is D $(\to a . \to c) (\to b . \to d)$
We have $(\to a \times \to b) . (\to c \times \to d) + (\to b . ¯ ¯ c) (\to a . \to d)$
$= \to a \cdot ((\to b \cdot \to d) \to c - (\to b \cdot \to c) \to d) + (\to b \cdot \to c) (\to a \cdot \to d)$
$= (\to a \cdot \to c) (\to b \cdot \to d)$

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Similar questions

Q.
If

\to a, \to b, \to c

and

\to d

are unit vectors such that

(\to a \times \to b) . (\to c \times \to d) = 1

and

\to a . \to c = \frac{1}{2}

, then

If $\to a, \to b, \to c$ , and $\to d,$ are the unit vectors such that $(\to a \times \to b) . (\to c \times \to d) = 1$ and $\to a . \to c = \frac{1}{2},$ then

(\to b \times \to c) . (\to a \times \to d) + (\to c \times \to a) . (\to b \times \to d)

+ (\to a \times \to b) . (\to c \times \to d) =

(\to a \times \to b) . (\to c \times \to d)

is equal to

Q. The value of

(\to a \times \to b) . (\to c \times \to d) + (\to b \times \to c) . (\to a \times \to d) + (\to c \times \to a) . (\to b \times \to d)

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(→a×→b).(→c×→d)+(→b.→c)(→a.→d)=

The correct option is D (→a.→c)(→b.→d)We have (→a×→b).(→c×→d)+(→b.¯¯c)(→a.→d)=→a⋅((→b⋅→d)→c−(→b⋅→c)→d)+(→b⋅→c)(→a⋅→d)=(→a⋅→c)(→b⋅→d)

$(\to a \times \to b) . (\to c \times \to d) + (\to b . \to c) (\to a . \to d) =$