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Prove the following.
Show that: a⃗ · (b⃗ + c⃗) × (a⃗ + 2b⃗ + 3c⃗) = [a⃗ b⃗ c⃗]
Show that:
$\vec{a} \cdot (\vec{b} + \vec{c}) \times (\vec{a} + 2\vec{b} + 3\vec{c}) = [\vec{a} \; \vec{b} \; \vec{c}]$
Given: $\vec{a} \cdot (\vec{b} + \vec{c}) \times (\vec{a} + 2\vec{b} + 3\vec{c})$
To show: $[\vec{a} \; \vec{b} \; \vec{c}]$
Answer
Answer
$(\vec{b}+\vec{c})\times(\vec{a}+2\vec{b}+3\vec{c}) = \vec{b}\times\vec{a} + 2(\vec{b}\times\vec{b}) + 3(\vec{b}\times\vec{c}) + \vec{c}\times\vec{a} + 2(\vec{c}\times\vec{b}) + 3(\vec{c}\times\vec{c})$
$= \vec{b}\times\vec{a} + 3(\vec{b}\times\vec{c}) + \vec{c}\times\vec{a} - 2(\vec{b}\times\vec{c})$ (as $\vec{x}\times\vec{x} = \vec{0}$)
$= \vec{b}\times\vec{a} + \vec{b}\times\vec{c} + \vec{c}\times\vec{a}$.
Dot with $\vec{a}$: $\vec{a}\cdot(\vec{b}\times\vec{a}) = 0$, $\vec{a}\cdot(\vec{c}\times\vec{a}) = 0$ (two equal vectors in a scalar triple product), leaving $\vec{a}\cdot(\vec{b}\times\vec{c})$.
Hence the expression $= [\vec{a}\ \vec{b}\ \vec{c}]$.
- Expand the cross product using the distributive law: $\vec{b}\times\vec{a} + 2\vec{b}\times\vec{b} + 3\vec{b}\times\vec{c} + \vec{c}\times\vec{a} + 2\vec{c}\times\vec{b} + 3\vec{c}\times\vec{c}$.
- Use $\vec{x}\times\vec{x} = \vec{0}$ and $\vec{c}\times\vec{b} = -\vec{b}\times\vec{c}$ to get $\vec{b}\times\vec{a} + \vec{b}\times\vec{c} + \vec{c}\times\vec{a}$.
- Take the dot product with $\vec{a}$: $\vec{a}\cdot(\vec{b}\times\vec{a}) = [\vec{a}\ \vec{b}\ \vec{a}] = 0$, two vectors being equal.
- Similarly $\vec{a}\cdot(\vec{c}\times\vec{a}) = 0$.
- What remains is $\vec{a}\cdot(\vec{b}\times\vec{c}) = [\vec{a}\ \vec{b}\ \vec{c}]$.