Prove that - a- b- c- - d- - - a- b- c- - - - a- b- d- -

[a⃗b⃗c⃗+d⃗]=[a⃗b⃗c⃗]+[a⃗b⃗d⃗] If a⃗=x_1î+y_1ĵ+z_1k̂ b⃗=x_2î+y_2ĵ+z_2k̂ c⃗=x_3î+y_3ĵ+z_3k̂ d⃗=x_4î+y_4ĵ+z_4k̂ [a⃗b⃗.c⃗+d⃗]= x _ 1 ...

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Addition of Vectors

Prove that ...

Question

Prove that $[\to a \to b \to c + \to d] = [\to a \to b \to c] + [\to a \to b \to d]$

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Solution

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\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

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

\to a \to b \to c

\to d

(\to a \times \to b)

(\to c \times \to d) = 1

\to a . \to c = \frac{1}{2}

\to a \cdot (\to b + \to c) \times (\to a + 2 \to b + 3 \to c) = [\to a \to b \to c]

\to d

\to d . \to a = \to d . \to b = \to d . \to c = 0

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

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