If a- b- c- are non-coplanar vector and d-a-b-c- - then - is equal to -

The correct option is B [ b⃗ c⃗ d⃗ ] [ b⃗c⃗ a⃗ ] Given #160; d⃗=λa⃗+μb⃗+νc⃗ Forming dot product with b⃗×c⃗ ( b⃗×c⃗ ).d⃗=λ [ b⃗ c⃗ a⃗ ] ...

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Test for Coplanarity

If a⃗, b⃗, ...

Question

If $\to a, \to b, \to c$ are non-coplanar vector and $\to d = λ \to a + μ \to b + ν \to c,$ then $λ$ is equal to -

\frac{[\to a \to b \to c]}{[\to b \to a \to c]}

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\frac{[\to b \to c \to d]}{[\to b \to c \to a]}

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\frac{[\to b \to d \to c]}{[\to a \to b \to c]}

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\frac{[\to c \to b \to d]}{[\to a \to b \to c]}

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Solution

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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

\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

¯ ¯ ¯ 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 b \to c \to d] \to a - [\to a \to c \to d] \to b + [\to a \to d \to b] \to c - [\to a \to b \to c] \to a =

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