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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
$(\to a \times \to b) \times (\to c \times \to d) = [\to a \to b \to d] \to c - [\to a \to b \to c] \to d$

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Solution

$\begin{matrix} W e h a v e (\to a \times \to b) \times (\to c \times \to d) L e t (\to a \times \to b) = p N o w (\to a \times \to b) \times (\to c \times \to d) = \to p \times (\to c \times \to d) = \to c (\to p . \to d) - \to d (\to p . \to c) = \to c [(\to a \times \to b) . \to d] - \to d [(\to a \times \to b) . \to c] = \to c [\to a \to b \to d] - \to d [\to a \to b \to c] = [\to a \to b \to d] \to c - [\to a \to b \to c] \to d H e n c e, (\to a \times \to b) \times (\to c \times \to d) = [\to a \to b \to d] \to c - [\to a \to b \to c] \to d p r o v e d \end{matrix}$

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Similar questions

\to a = \to i + \to j + \to k, \to a = 2 \to i + \to k, \to c = 2 \to i + \to j + \to k; \to d = \to i + \to j + 2 \to k

, then verify that

(\to a \times \to b) \times (\to c \times \to d) = [\to a \to b \to d] \to c - [\to a \to b \to c] \to d

\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

Standard XII Mathematics

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