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Multiplication of a Vector by a Scalar

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Question

Show that $^i \times (\to a \times^i) +^j \times (\to a \times^j) +^k \times (\to a \times^k) = 2 \to a$ for any vector $\to a$

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Solution

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$\begin{matrix} \to a = a_{1}^j + a_{2}^j + a_{3}^k [(^i \cdot^i) \to a - (^i \cdot \to a)^i] + [(^j \cdot^j) \to a - (^j . \to a)^j] + [(^k \cdot^k) \to a - (^k \cdot \to a) \cdot^k] = 2 \to a = [\to a - a_{1}^i] + [\to a - a_{2}^j] + [\to a - a_{3}^k] = \to a - a_{1}^i + \to a - a_{2}^j + \to a - a_{3}^k = 3 \to a - (a_{1}^i + a_{2}^j + a_{3}^k) = 3 \to a - \to a = 2 \to a \end{matrix}$

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Similar questions

\to a =^i +^j +^k

\to b =^i -^j +^k

\to c =^i +^j -^k

\to a =^i -^j -^k

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

is a vector orthogonal to both

\to a = 3^i + 4^j - 5^k

|^i \times (\to a \times^i) +^j \times (\to a \times^j) +^k \times (\to a \times^k) | =

\to a =^i +^j +^k

\to b =^i +^j -^k

\to c =^i -^j +^k

\to d =^i -^j -^k

| (\to a \times \to b) . (\to c \times \to d) | =

\to a =^i +^j +^k, \to b =^i +^j -^k

\to c =^i -^j +^k

\to d =^i -^j -^k

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

\to a =^i +^j +^k, \to b =^i -^j +^k, \to c =^2 i +^3 j -^k

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

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