If d is a unit vector such that d-b-c-c-a-a-b then - d - a b - c - d -b c-a - d-c a-b - is equal to

The correct option is A | [ a b⃗ c ] | d⃗.a⃗ = λ[ a⃗ amp; b⃗ amp; c⃗ ] d⃗.b⃗ = μ[ c⃗ amp; nbsp;a⃗ amp; b⃗ ] = μ[ a⃗ amp; ...

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Area Method to Find Condition for Co-Linearity

If d is a u...

Question

If $\to d$ is a unit vector such that $\to d = λ \to b \times \to c + μ \to c \times \to a + ν \to a \times \to b$ then $∣ ∣ ∣ (\to d \cdot \to a) (\to b \times \to c) + (\to d \cdot \to b) (\to c \times \to a) + (\to d \cdot \to c) (\to a \times \to b) ∣ ∣ ∣$ is equal to

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

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1

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3 ∣ ∣ [\begin{matrix} \to a & \to b & \to c \end{matrix}] ∣ ∣

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None of these

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Solution

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\to a, \to b, \to c

\to d

\to a \times \to b = \to c \times \to d

\to a \times \to c = \to b \times \to d

(\to a - \to d) \times (\to b - \to c) = \to 0

(\to a \times \to d) \times (\to b \times \to c) = \to 0

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

\to a = \to b + \to c

\to b \times \to d = \to 0

\to c \cdot \to d = 0

\frac{\to d \times (\to a \times \to d)}{| \to d |^{2}}

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

[\to a \times \to b \to a \times \to c \to d]

\to a, \to b, \to c

\to d

\to a + \to b + \to c

\to d = x \to b \times \to c + y \to c \times \to a + z \to a \times \to b

[\to a \times \to b \to b \times \to c \to c \times \to a] = λ [\to a \to b \to c]^{2}

λ

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