Q111 / 180 · KCET 2024 · Mathematics
If $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are three non-coplanar vectors and $p, q$ and $r$ are vectors defined by
$\mathbf{p}=\frac{\mathbf{a} \times \mathbf{c}}{[\mathbf{a} \mathbf{b} \mathbf{c}]}, \mathbf{q}=\frac{\mathbf{c} \times \mathbf{a}}{[\mathbf{a} \mathbf{b} \mathbf{c}]}, \mathbf{r}=\frac{\mathbf{a} \times \mathbf{b}}{[\mathbf{a} \mathbf{b} \mathbf{c}]}$, then
$(\mathbf{a}+\mathbf{b}) \cdot \mathbf{p}+(\mathbf{b}+\mathbf{c}) \cdot \mathbf{q}+(\mathbf{c}+\mathbf{a}) \cdot \mathbf{r}$ is
- A$0$
- B$1$
- C$2$
- Correct answerD$3$
Solution
$\mathbf{p}=\frac{\mathbf{b} \times \mathbf{c}}{[\mathbf{a} \mathbf{b} \mathbf{c}]}, \mathbf{q}=\frac{\mathbf{c} \times \mathbf{a}}{[\mathbf{a} \mathbf{b} \mathbf{c}]}, \mathbf{r}=\frac{\mathbf{a} \times \mathbf{b}}{[\mathbf{a} \mathbf{b} \mathbf{c}]}$
Now, $(\mathbf{a}+\mathbf{b}) \cdot \mathbf{p}+(\mathbf{b}+\mathbf{q}) \cdot \mathbf{q}+(\mathbf{c}+\mathbf{a}) \cdot \mathbf{r}$
$\begin{aligned}=\frac{\mathbf{a} \cdot(\mathbf{b} \times \mathbf{a})}{[\mathbf{a} \mathbf{b}]} & +\frac{\mathbf{b} \cdot(\mathbf{b} \times \mathbf{a})}{[\mathbf{a} \mathbf{c}]}+\frac{\mathbf{b} \cdot(\mathbf{c} \times \mathbf{a})}{[\mathbf{a} \mathbf{c}]} \\ & +\frac{\mathbf{c} \cdot(\mathbf{c} \times \mathbf{a})}{[\mathbf{a} \mathbf{b} \mathbf{c}]}+\frac{\mathbf{c} \cdot(\mathbf{a} \times \mathbf{b})}{[\mathbf{a} b \mathbf{b}]}+\frac{\mathbf{a} \cdot(\mathbf{a} \times \mathbf{b})}{[\mathbf{a} \mathbf{b}]}\end{aligned}$
$=\frac{[\mathbf{a} \mathbf{b} \mathbf{c}]}{[\mathbf{a} \mathbf{b} \mathbf{c}]}+0+\frac{[\mathbf{b} \mathbf{c} \mathbf{a}]}{[\mathbf{a} \mathbf{b} \mathbf{c}]}+0+\frac{[\mathrm{c} \mathbf{a} \mathbf{b}]}{[\mathbf{a} \mathbf{b} \mathbf{c}]}+0$
[ $\because$ scalar triple product is zero if two vectors are same]
$=\frac{[a b c]}{[\mathbf{a b c} \mathbf{c}]}+\frac{[a \mathbf{b} \mathbf{c}]}{[\mathbf{a b c}]}+\frac{[a \mathbf{b} \mathbf{c}]}{[\mathbf{a b} \mathbf{b}]}=3$
Hence, the correct answer is option: 'D'.