Question

The angle between the vectors $\overrightarrow{A},\overrightarrow{B}$ is $\theta$. The value of triple product $\overrightarrow{A.}\left(\overrightarrow{A}\times \overrightarrow{B}\right)$ is

• A. $A^{2}B$
• B. $0$
• C. $A^{2}B\sin \theta$
• D. $A^{2}B\cos \theta$

Answer 1

The correct option is B

$0$

Explanation for the correct option:

Let

$\overrightarrow{A.}\left(\overrightarrow{A}\times \overrightarrow{B}\right)=\overrightarrow{A}.\overrightarrow{C}$ where $\overrightarrow{C}=\overrightarrow{A}\times \overrightarrow{B}$

So,

$$\begin{gathered} \stackrel{\rightharpoonup }{A.}\left(\stackrel{\rightharpoonup }{A}\times \stackrel{\rightharpoonup }{B}\right) \\ \Rightarrow \stackrel{\rightharpoonup }{A}.\stackrel{\rightharpoonup }{C} \end{gathered}$$

$\Rightarrow \stackrel{\rightharpoonup }{C}$is perpendicular to $\stackrel{\rightharpoonup }{A}and\stackrel{\rightharpoonup }{B}$

$\Rightarrow$Angle between $\stackrel{\rightharpoonup }{A}and\stackrel{\rightharpoonup }{C}$ is${90}^{°}$

Therefore,

$\begin{array}{rcl}\stackrel{\rightharpoonup }{A}.\stackrel{\rightharpoonup }{C}& =& AC\cos \theta \\ & =& 0\end{array}$

$\therefore \stackrel{\rightharpoonup }{A}.\left(\stackrel{\rightharpoonup }{B}\times \stackrel{\rightharpoonup }{A}\right)=0$

Hence, the correct option is B.