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Let $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ be non-coplanar vectors and $\overrightarrow{\mathrm{p}}=\frac{\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}}{[\overrightarrow{\mathrm{abc}}]}$,
$\overrightarrow{\mathrm{q}}=\frac{\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}}{[\overrightarrow{\mathrm{abc}}]}, \overrightarrow{\mathrm{r}}=\frac{\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}}{[\overrightarrow{\mathrm{abc}}]}$
What is the value of
$(\vec{a}-\vec{b}-\vec{c}) \cdot \vec{p}+(\vec{b}-\vec{c}-\vec{a}) \cdot \vec{q}+(\vec{c}-\vec{a}-\vec{b}) \cdot \vec{r} ?$
Options:
$\overrightarrow{\mathrm{q}}=\frac{\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}}{[\overrightarrow{\mathrm{abc}}]}, \overrightarrow{\mathrm{r}}=\frac{\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}}{[\overrightarrow{\mathrm{abc}}]}$
What is the value of
$(\vec{a}-\vec{b}-\vec{c}) \cdot \vec{p}+(\vec{b}-\vec{c}-\vec{a}) \cdot \vec{q}+(\vec{c}-\vec{a}-\vec{b}) \cdot \vec{r} ?$
Solution:
2953 Upvotes
Verified Answer
The correct answer is:
3
As given $\overrightarrow{\mathrm{p}}=\frac{\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}}{[\overrightarrow{\mathrm{abc}}]}, \overrightarrow{\mathrm{q}}=\frac{\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}}{[\overrightarrow{\mathrm{abc}}]}$, and $\overrightarrow{\mathrm{r}}=\frac{\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}}{[\overrightarrow{\mathrm{abc}}]}$
$\begin{aligned} \therefore &(\vec{a}-\vec{b}-\vec{c}) \cdot \vec{p}+(\vec{b}-\vec{c}-\vec{a}) \cdot \vec{q}+(\vec{c}-\vec{a}-\vec{b}) \cdot \vec{r} \\ &=\frac{a \cdot(\vec{b} \times \vec{c})}{[\overrightarrow{a b c}]}+\frac{\vec{b} \cdot(\vec{c} \times \vec{a})}{[a \vec{b} c]}+\frac{\vec{c} \cdot(\vec{a} \times \vec{b})}{[\overrightarrow{a b c}]} \end{aligned}$
$[$ Since $\overrightarrow{\mathrm{b}} \cdot(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=0, \overrightarrow{\mathrm{c}} \cdot(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=0, \overrightarrow{\mathrm{c}} \cdot(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}})=0$
$\overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}})=0, \overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})=0$ and $\overrightarrow{\mathrm{b}} \cdot(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})=0]$
$=\frac{[\vec{a} \vec{b} \vec{c}]}{[\vec{a} \vec{b} c]}+\frac{[\vec{a} \vec{b} \vec{c}]}{[\vec{a} \vec{b} c]}+\frac{[\vec{a} \vec{b} \vec{c}]}{[\vec{a} \vec{b} c]}=3$
$\begin{aligned} \therefore &(\vec{a}-\vec{b}-\vec{c}) \cdot \vec{p}+(\vec{b}-\vec{c}-\vec{a}) \cdot \vec{q}+(\vec{c}-\vec{a}-\vec{b}) \cdot \vec{r} \\ &=\frac{a \cdot(\vec{b} \times \vec{c})}{[\overrightarrow{a b c}]}+\frac{\vec{b} \cdot(\vec{c} \times \vec{a})}{[a \vec{b} c]}+\frac{\vec{c} \cdot(\vec{a} \times \vec{b})}{[\overrightarrow{a b c}]} \end{aligned}$
$[$ Since $\overrightarrow{\mathrm{b}} \cdot(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=0, \overrightarrow{\mathrm{c}} \cdot(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=0, \overrightarrow{\mathrm{c}} \cdot(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}})=0$
$\overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}})=0, \overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})=0$ and $\overrightarrow{\mathrm{b}} \cdot(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})=0]$
$=\frac{[\vec{a} \vec{b} \vec{c}]}{[\vec{a} \vec{b} c]}+\frac{[\vec{a} \vec{b} \vec{c}]}{[\vec{a} \vec{b} c]}+\frac{[\vec{a} \vec{b} \vec{c}]}{[\vec{a} \vec{b} c]}=3$
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