Question

The value of $\frac{{\left(a\times b\right)}^2+{\left(a·b\right)}^2}{2{\left|a\right|}^2{\left|b\right|}^2}$is

• A. $a·b$
• B. $1$
• C. $0$
• D. $\frac{1}{2}$

Answer 1

The correct option is D

$\frac{1}{2}$

Explanation for the correct option:

Find the value of the given expression:

An expression $\frac{{\left(a\times b\right)}^2+{\left(a·b\right)}^2}{2{\left|a\right|}^2{\left|b\right|}^2}$ is given.

We know that, $\left|a\times b\right|\mathbf{=}\left|a\right|\left|b\right|\mathbf{sin\theta }$ and $\mathbf{a}\mathbf{·}\mathbf{b}\mathbf{=}\left|a\right|\left|b\right|\mathbf{cos\theta }$

From the given equation.

$$\begin{gathered} \frac{{\left(a\times b\right)}^2+{\left(a·b\right)}^2}{2{\left|a\right|}^2{\left|b\right|}^2}=\frac{{\left(\left|a\right|\left|b\right|\sin \theta \right)}^2+{\left(\left|a\right|\left|b\right|\cos \theta \right)}^2}{2{\left|a\right|}^2{\left|b\right|}^2} \\ \Rightarrow \frac{{\left(a\times b\right)}^2+{\left(a·b\right)}^2}{2{\left|a\right|}^2{\left|b\right|}^2}=\frac{{\left|a\right|}^2{\left|b\right|}^2\left[{\sin }^2\theta +{\cos }^2\theta \right]}{2{\left|a\right|}^2{\left|b\right|}^2}\left[\because {\sin }^2\theta +{\cos }^2\theta =1\right] \\ \Rightarrow \frac{{\left(a\times b\right)}^2+{\left(a·b\right)}^2}{2{\left|a\right|}^2{\left|b\right|}^2}=\frac{{\left|a\right|}^2{\left|b\right|}^2}{2{\left|a\right|}^2{\left|b\right|}^2} \\ \Rightarrow \frac{{\left(a\times b\right)}^2+{\left(a·b\right)}^2}{2{\left|a\right|}^2{\left|b\right|}^2}=\frac{1}{2} \end{gathered}$$

Therefore, the value of $\frac{{\left(a\times b\right)}^2+{\left(a·b\right)}^2}{2{\left|a\right|}^2{\left|b\right|}^2}$is $\frac{1}{2}$.

Hence, option (D) is the correct answer.