If a- b- c- be three non-coplanar vectors and r- be any arbitrary vector- then a-b- r-c-b-c- r-a-c-a- r-b- is equal to-

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

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Test for Coplanarity

If a⃗, b⃗, ...

Question

If $\to a, \to b, \to c$ be three non-coplanar vectors and $\to r$ be any arbitrary vector, then $(\to a \times \to b) \times (\to r \times \to c) + (\to b \times \to c) \times (\to r \times \to a) + (\to c \times \to a) \times (\to r \times \to b)$ is equal to?

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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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3 [\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 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)

¯ ¯ ¯ a, ¯ ¯ b, ¯ ¯ c

¯ ¯ ¯ p, ¯ ¯ ¯ q, ¯ ¯ ¯ r

\to p = \frac{\to b \times \to c}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]}, \to q = \frac{\to c \times \to a}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]}, \to r = \frac{\to a \times \to b}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]}

(\to a + \to b + \to c) . (\to p + \to q + \to r)

\to a, \to b, \to c

\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]}

(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 =

\to a, \to b, \to c

[\to a \to b \to c] \neq

{\to a}^{'}, {\to b}^{'}, {\to c}^{'}

\to r

\to a, \to b, \to c

\to p, \to q, \to r

\to p = \frac{\to b \times \to c}{[\begin{matrix} \to b & \to c & \to a \end{matrix}]}

\to q = \frac{\to c \times \to a}{[\begin{matrix} \to c & \to a & \to b \end{matrix}]}

\to r = \frac{\to a \times \to b}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]}

(\to a + \to b) . \to p + (\to b + \to c) . \to q + (\to c + \to a) . \to r

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