If - -160-a-160- -160-b-160- c- are non-coplanar- non-zero vectors and - -160- -160-r- is any vector in space- then- -a-b-r-c-b-c-r-a-c-a-r-b- 3- -a-b-c-r- -a-b-c- -10- 0- -10-

The correct option is B [a⃗b⃗c⃗]r⃗ We have #160; r⃗=λa⃗+μb⃗+γc⃗ [ a⃗ b̅ r̅ ]=γ [ a̅ b̅ c̅ ] [ b̅ c̅ r̅ ]=λ #160; [ a̅ b̅ c̅ ] [ c̅ ...

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Dot Product

If #160;...

Question

If $\to a, \to b, \to c$ are non-coplanar, non-zero vectors and $\to r$ is any vector in space, then
$[\to a \to b \to r] \to c + [\to b \to c \to r] \to a + [\to c \to a \to r] \to b =$

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[\to a \to b \to c] \to r

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[\to a \to b \to c]

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\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 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

\to a, \to b, \to c

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

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

\to r

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