First slide

Scalar triple product of vectors

Question

If $\vec{r} = x_{1} (\vec{a} \times \vec{b}) + x_{2} (\vec{b} \times \vec{a}) + x_{3} (\vec{c} \times \vec{d}) and 4 [\vec{a} \vec{b} \vec{c}] = 1, then x_{1} + x_{2} + x_{3}$ is equal to

Moderate

A

$\frac{1}{2} \vec{r} \cdot (\vec{a} + \vec{b} + \vec{c})$
B

$- \vec{r} \cdot (\vec{a} + \vec{b} + \vec{c})$
C

$2 \vec{r} \cdot (\vec{a} + \vec{b} + \vec{c})$
D

$4 \vec{r} \cdot (\vec{a} + \vec{b} + \vec{c})$

Solution

$\begin{array}{l} \vec{r} = x_{1} (\vec{a} \times \vec{b}) + x_{2} (\vec{b} \times \vec{c}) + x_{3} (\vec{c} \times \vec{a}) \\ \Rightarrow \vec{r} \cdot \vec{a} = x_{2} [\vec{a} \vec{b} \vec{c}], \vec{r} \cdot \vec{b} = x_{3} [\vec{b} \vec{c} \vec{a}] \\ and \vec{r} \cdot \vec{c} = x_{1} [\vec{c} \vec{a} \vec{b}] = x_{1} [\vec{a} \vec{b} \vec{c}] \\ \Rightarrow x_{1} + x_{2} + x_{3} = 4 \vec{r} \cdot (\vec{a} + \vec{b} + \vec{c}) \end{array}$

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