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If $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are three non-coplanar vectors and
$\mathbf{p}, \mathbf{q}$ and $\mathbf{r}$ are vectors defined by $\mathbf{p}=\frac{\mathbf{b} \times \mathbf{c}}{[\mathbf{a} \mathbf{b} \mathbf{c}]}, \mathbf{q}=\frac{\mathbf{c} \times \mathbf{a}}{[\mathbf{a} \mathbf{b} \mathbf{c}]}$ and $\mathbf{r}=\frac{\mathbf{a} \times \mathbf{b}}{[\mathbf{a} \mathbf{b} \mathbf{c}]}$, then the value of $(\mathbf{a}+\mathbf{b}) \cdot \mathbf{p}+(\mathbf{b}+\mathbf{c}) \cdot \mathbf{q}+(\mathbf{c}+\mathbf{a}) \cdot \mathbf{r}$ is equal to
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$\mathbf{p}, \mathbf{q}$ and $\mathbf{r}$ are vectors defined by $\mathbf{p}=\frac{\mathbf{b} \times \mathbf{c}}{[\mathbf{a} \mathbf{b} \mathbf{c}]}, \mathbf{q}=\frac{\mathbf{c} \times \mathbf{a}}{[\mathbf{a} \mathbf{b} \mathbf{c}]}$ and $\mathbf{r}=\frac{\mathbf{a} \times \mathbf{b}}{[\mathbf{a} \mathbf{b} \mathbf{c}]}$, then the value of $(\mathbf{a}+\mathbf{b}) \cdot \mathbf{p}+(\mathbf{b}+\mathbf{c}) \cdot \mathbf{q}+(\mathbf{c}+\mathbf{a}) \cdot \mathbf{r}$ is equal to
Solution:
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Verified Answer
The correct answer is:
3
Let $\mathrm{T}_{1}=(\mathbf{a}+\mathbf{b}) \cdot \mathbf{p}$
$=\mathbf{a} \cdot \mathbf{p}+\mathbf{b} \cdot \mathbf{p}$
$=\mathbf{a} \cdot \frac{\mathbf{b} \times \mathbf{c}}{[\mathbf{a} \mathbf{b} \mathbf{c}]}+\frac{\mathbf{b} \cdot(\mathbf{b} \times \mathbf{c})}{[\mathbf{a} \mathbf{b} \mathbf{c}]}$
$=\frac{[\mathbf{a} \mathbf{b} \mathbf{c}]}{[\mathbf{a} \mathbf{b} \mathbf{c}]}+\frac{[\mathbf{b} \mathbf{b} \mathbf{c}]}{[\mathbf{a} \mathbf{b} \mathbf{c}]}=1+0=1$
Similarly, $\mathrm{T}_{2}=(\mathbf{b}+\mathbf{c}) \cdot \mathbf{q}=1$
and $\quad \mathrm{T}_{3}=(\mathbf{c}+\mathbf{a}) \cdot \mathbf{r}=1$
$\therefore \quad \mathrm{T}_{1}+\mathrm{T}_{2}+\mathrm{T}_{3}=3$
$=\mathbf{a} \cdot \mathbf{p}+\mathbf{b} \cdot \mathbf{p}$
$=\mathbf{a} \cdot \frac{\mathbf{b} \times \mathbf{c}}{[\mathbf{a} \mathbf{b} \mathbf{c}]}+\frac{\mathbf{b} \cdot(\mathbf{b} \times \mathbf{c})}{[\mathbf{a} \mathbf{b} \mathbf{c}]}$
$=\frac{[\mathbf{a} \mathbf{b} \mathbf{c}]}{[\mathbf{a} \mathbf{b} \mathbf{c}]}+\frac{[\mathbf{b} \mathbf{b} \mathbf{c}]}{[\mathbf{a} \mathbf{b} \mathbf{c}]}=1+0=1$
Similarly, $\mathrm{T}_{2}=(\mathbf{b}+\mathbf{c}) \cdot \mathbf{q}=1$
and $\quad \mathrm{T}_{3}=(\mathbf{c}+\mathbf{a}) \cdot \mathbf{r}=1$
$\therefore \quad \mathrm{T}_{1}+\mathrm{T}_{2}+\mathrm{T}_{3}=3$
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