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Work Done as Dot Product

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Question

If $\to a \times (\to a \times \to b) = \to b \times (\to b \times \to c)$ and $\to a . \to b \neq 0$ , then $[\begin{matrix} \to a & \to b & \to c \end{matrix}]$ is equal to

A

0

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B

1

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C

2

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D

None of these

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Solution

The correct option is A $0$
$\to a \times \to b$ is a vector perpendicular to the plane containing $\to a$ and $\to b$ .
Hence,
$\to a \times (\to a \times \to b)$ is a vector in the same plane as $\to a$ and $\to b$ and perpendicular to $\to a$
Similarly,
$\to b \times (\to b \times \to c)$ is a vector in the same plane as $\to b$ and $\to c$ and perpendicular to $\to b$
For the above two vectors to be equal, $\to a \times \to b$ and $\to b \times \to c$ should be in the same direction.
$\Rightarrow \to c$ has to be in the same plane as $\to a$ and $\to b$ .
$\Rightarrow$ $\to a, \to b, \to c$ are coplanar. Hence their scalar triple product is equal to $0$ .
$\Rightarrow [\begin{matrix} \to a & \to b & \to c \end{matrix}] = 0$
Hence, option $A$ .

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Similar questions

\to a \times (\to a \times \to b) = \to b \times (\to b \times \to c)

(\to a . \to b) \neq 0

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

[\begin{matrix} \to a & \to b & \to c \end{matrix}] = 1

\frac{\to a . \to b \times \to c}{\to c \times \to a . \to b} + \frac{\to b . \to c \times \to a}{\to a \times \to b . \to c} + \frac{\to c . \to a \times \to b}{\to b \times \to c . \to a}

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

\frac{\to a . (\to b \times \to c)}{(\to c \times \to a) . \to b} + \frac{\to b . (\to c \times \to a)}{(\to a \times \to b) . \to c} + \frac{\to c . (\to a \times \to b)}{(\to b \times \to c) . \to a} =

| \to a | = 1, | \to b | = 2, | \to c | = 3

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

, then the value of

\to a . \to b + \to b . \to c + \to c . \to a

\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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