If a-b-c- are three non coplanar vectors- p-b-c- a-b-c- q-c-a- a-b-c- r-a-b- a-b-c- then 2a-3b-4c-p- 2b-3c-4a-q- 2c-3a-4b-r-

The correct option is A 6 P = b × c #160; [ a × b × c #160; ] #160; = b × c #160; a ( b × c ) =a q = b ^ 1 r = c ^ 1 ...

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Linear Combination of Vectors

If a⃗,b⃗,c⃗...

Question

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

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\to a, \to b, \to c

[\to a \to b \to c] = \frac{4}{7}

[2 \to a - \to b, 2 \to b - \to c, 2 \to c - \to a]

¯ ¯ ¯ 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, \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

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

2 \to a + 3 \to b - \to c, \to a - 2 \to b + 3 \to c, 3 \to a + 4 \to b - 2 \to c a n d \to a - 6 \to b + 6 \to c

\to a, \to b a n d \to c

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