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Collinear Vectors

For any four ...

Question

For any four vectors $\to a, \to b, \to c$ and $\to d$ prove that
$(\to a \times \to b) \times (\to c \times \to d) = [\to a \to c \to d] \to b - [\to b \to c \to d] \to a$

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Solution

$\begin{matrix} (\to a \times \to b) \times (\to c \times \to d) = - [(\to c \times \to d) \times (\to a \times \to b)] = - [[\to b \to c \to d] \to a - [\to a \to c \to d] \to b] = [\to a \to c \to d] \to b - [\to b \to c \to d] \to a \end{matrix}$

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Similar questions

[\to b \to c \to d] = 4

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

then the value of

k

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

are non-coplanar vectors then the vector

(\to a \times \to b) \times (\to c \times \to d) + (\to a \times \to c) \times (\to d \times \to b) + (\to a \times \to d) \times (\to b \times \to c)

is parallel to:

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

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

, then the vectors

\to a - \to d

\to b - \to c

\to b, \to c, \to d

are non-coplanar vectors then the vector

(\to a \times \to b) \times (\to c \times \to d) + (\to a \times \to c) \times (\to d \times \to b) + (\to a \times \to d) \times (\to b \times \to c)

\to b, \to c, \to d

are non-coplanar vectors, then the vector

(\to a \times \to b) \times (\to c \times \to d) + (\to a \times \to c) \times (\to d \times \to b) + (\to a \times \to d) \times (\to b \times \to c)

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Collinear Vectors

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