Question

If $\left[\overline{a}\overline{b}\overline{c}\right]\ne 0$ and $\overline{p}=\frac{\overline{b}\times \overline{c}}{\left[\overline{a}\overline{b}\overline{c}\right]},\overline{q}=\frac{\overline{c}\times \overline{a}}{\left[\overline{a}\overline{b}\overline{c}\right]},\overline{r}=\frac{\overline{a}\times \overline{b}}{\left[\overline{a}\overline{b}\overline{c}\right]}$ then $\overline{a}.\overline{p}+\overline{b}.\overline{q}+\overline{c}.\overline{r}$ is equal to _____

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is D $3$

Given

$\overrightarrow{p}=\frac{\overrightarrow{b}\times \overrightarrow{c}}{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}$

$\overrightarrow{q}=\frac{\overrightarrow{c}\times \overrightarrow{a}}{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}$

$\overrightarrow{r}=\frac{\overrightarrow{a}\times \overrightarrow{b}}{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}$

$\overrightarrow{a}\cdot \overrightarrow{p}+\overrightarrow{b}\cdot \overrightarrow{q}+\overrightarrow{c}\cdot \overrightarrow{r}$

$\overrightarrow{a}\cdot \frac{\overrightarrow{b}\times \overrightarrow{c}}{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}+\overrightarrow{b}\cdot \frac{\overrightarrow{c}\times \overrightarrow{a}}{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}+\overrightarrow{c}\cdot \frac{\overrightarrow{a}\times \overrightarrow{b}}{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}$

$\frac{\overrightarrow{a}\cdot (\overrightarrow{b}\times \overrightarrow{c})}{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}+\frac{\overrightarrow{b}\cdot (\overrightarrow{c}\times \overrightarrow{a})}{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}+\frac{\overrightarrow{c}\cdot (\overrightarrow{a}\times \overrightarrow{b})}{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}$

$\frac{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}+\frac{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}+\frac{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}$

$=1+1+1$

$=3$

hence option D is correct