Question

${\left(\overrightarrow{a}\times \overrightarrow{b}\right)}^2+{\left(\overrightarrow{a}·\overrightarrow{b}\right)}^2$ is equal to?

• A. $a^2{b}^2$
• B. $a^2+b^2$
• C. $1$
• D. $2a·b$

Answer 1

The correct option is A

$a^2{b}^2$

Explanation for correct option:

Given, ${\left(\overrightarrow{a}\times \overrightarrow{b}\right)}^2+{\left(\overrightarrow{a}·\overrightarrow{b}\right)}^2$

We know that $\overrightarrow{\mathbf{a}}\mathbf{.}\overrightarrow{\mathbf{b}}\mathbf{=}\mathbf{abcos}\left(\theta \right)\mathbf{}\mathbf{\&}\mathbf{}\left|\overrightarrow{a}\times \overrightarrow{b}\right|\mathbf{=}\mathbf{absin}\left(\theta \right)$

$\begin{array}{rcl}& \Rightarrow & {\left(\overrightarrow{a}\times \overrightarrow{b}\right)}^2+{\left(\overrightarrow{a}·\overrightarrow{b}\right)}^2={\left(ab\cos \left(\theta \right)\right)}^2+{\left(ab\sin \left(\theta \right)\right)}^2\\ & =& a^2{b}^2\left({\cos }^2\left(\theta \right)+{\sin }^2\left(\theta \right)\right)\\ & =& a^2{b}^2\left(\because co{s}^2\left(\theta \right)+si{n}^2\left(\theta \right)=1\right)\end{array}$

Therefore, correct answer is option $\left(A\right)$