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Answer the following.
Prove that a⃗ · [(b⃗ + c⃗) × (a⃗ + 3b⃗ + 4c⃗)] = [a⃗ b⃗ c⃗].
Prove that $\vec{a} \cdot [(\vec{b} + \vec{c}) \times (\vec{a} + 3\vec{b} + 4\vec{c})] = [\vec{a} \quad \vec{b} \quad \vec{c}]$.
Given: $\vec{a} \cdot [(\vec{b} + \vec{c}) \times (\vec{a} + 3\vec{b} + 4\vec{c})]$
To show: $[\vec{a} \quad \vec{b} \quad \vec{c}]$
Answer
Answer
Hence $\vec{a}\cdot[(\vec{b} + \vec{c}) \times (\vec{a} + 3\vec{b} + 4\vec{c})] = [\vec{a}\ \vec{b}\ \vec{c}]$.
- Expand the cross product by the distributive law: $(\vec{b} + \vec{c}) \times (\vec{a} + 3\vec{b} + 4\vec{c}) = \vec{b}\times\vec{a} + 3\vec{b}\times\vec{b} + 4\vec{b}\times\vec{c} + \vec{c}\times\vec{a} + 3\vec{c}\times\vec{b} + 4\vec{c}\times\vec{c}$.
- Since $\vec{u}\times\vec{u} = \vec{0}$, the terms $\vec{b}\times\vec{b}$ and $\vec{c}\times\vec{c}$ vanish.
- Using $\vec{c}\times\vec{b} = -\vec{b}\times\vec{c}$: $4\vec{b}\times\vec{c} + 3\vec{c}\times\vec{b} = 4\vec{b}\times\vec{c} - 3\vec{b}\times\vec{c} = \vec{b}\times\vec{c}$.
- So the cross product $= \vec{b}\times\vec{a} + \vec{c}\times\vec{a} + \vec{b}\times\vec{c}$.
- Take the dot product with $\vec{a}$: $\vec{a}\cdot(\vec{b}\times\vec{a}) + \vec{a}\cdot(\vec{c}\times\vec{a}) + \vec{a}\cdot(\vec{b}\times\vec{c})$.
- A scalar triple product with two equal vectors is zero, so $\vec{a}\cdot(\vec{b}\times\vec{a}) = [\vec{a}\ \vec{b}\ \vec{a}] = 0$ and $\vec{a}\cdot(\vec{c}\times\vec{a}) = [\vec{a}\ \vec{c}\ \vec{a}] = 0$.
- Hence the expression $= \vec{a}\cdot(\vec{b}\times\vec{c}) = [\vec{a}\ \vec{b}\ \vec{c}]$.