Vector Cross Product is Orthogonal to Factors

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Theorem

Let $\mathbf a$ and $\mathbf b$ be vectors in the real Euclidean space $\R^3$.


Let $\mathbf a \times \mathbf b$ denote the vector cross product.


Then:

$(1): \quad$ $\mathbf a$ and $\mathbf a \times \mathbf b$ are orthogonal.
$(2): \quad$ $\mathbf b$ and $\mathbf a \times \mathbf b$ are orthogonal.


Proof

Let $\mathbf a = \begin {bmatrix} a_1 \\ a_2 \\ a_3 \end {bmatrix}$, and $\mathbf b = \begin {bmatrix} b_1 \\ b_2 \\ b_3 \end {bmatrix}$.

Then the dot product of $\mathbf a$ and $\mathbf a \times \mathbf b$ is:

\(\ds \mathbf a \cdot \paren {\mathbf a \times \mathbf b}\) \(=\) \(\ds a_1 \paren {a_2 b_3 - a_3 b_2} + a_2 \paren {a_3 b_1 - a_1 b_3} + a_3 \paren {a_1 b_2 - a_2 b_1}\) Definition of Dot Product and Definition of Vector Cross Product
\(\ds \) \(=\) \(\ds a_1 a_2 b_3 - a_1 a_3 b_2 + a_2 a_3 b_1 - a_1 a_2 b_3 + a_1 a_3 b_2 - a_2 a_3 b_1\)
\(\ds \) \(=\) \(\ds 0\)

Since the dot product is equal to zero, the vectors $\mathbf a$ and $\mathbf a \times \mathbf b$ are orthogonal by definition.


Similarly, $\mathbf b$ and $\mathbf a \times \mathbf b$ are orthogonal:

\(\ds \mathbf b \cdot \paren {\mathbf a \times \mathbf b}\) \(=\) \(\ds b_1 \paren {a_2 b_3 - a_3 b_2} + b_2 \paren {a_3 b_1 - a_1 b_3} + b_3 \paren {a_1 b_2 - a_2 b_1}\)
\(\ds \) \(=\) \(\ds a_2 b_1 b_3 - a_3 b_1 b_2 + a_3 b_1 b_2 - a_1 b_2 b_3 + a_1 b_2 b_3 - a_2 b_1 b_3\)
\(\ds \) \(=\) \(\ds 0\)

$\blacksquare$


Sources