Question

If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ form a left handed orthogonal system and $\overrightarrow{a}\cdot \overrightarrow{a}=4,\overrightarrow{b}\cdot \overrightarrow{b}=9,\overrightarrow{c}\cdot \overrightarrow{c}=16$ then $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$ is equal to

• A. $-24$
• B. $24$
• C. $12$
• D. $-12$

Answer 1

The correct option is A $-24$

Given: $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ form a left handed orthogonal system. $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=\overrightarrow{a}\cdot (\overrightarrow{b}\times \overrightarrow{c})$
$\Rightarrow |\overrightarrow{a}||\overrightarrow{b}\times \overrightarrow{c}|\cos {180}^0\hat{n}(\because \theta ={180}^0\text{Left hand Orthogonal})$
$\Rightarrow -|\overrightarrow{a}||\overrightarrow{b}\times \overrightarrow{c}|$
$\Rightarrow -|\overrightarrow{a}||\overrightarrow{b}||\overrightarrow{c}|\sin {90}^0$
$\Rightarrow -2\times 3\times 4\times 1=-24$