Q24DQ - University Physics with Modern Physics

Question

Show that, no matter what $\overrightarrow{A}$ and $\overrightarrow{B}$are, $\overrightarrow{A}.\left(\overrightarrow{A}\times \overrightarrow{B}\right)$ =0

Step-by-Step Solution

Answer

Given-

$\overrightarrow{A}$ and $\overrightarrow{B}$ vectors.

Step1- To Prove

$\overrightarrow{A}.\left(\overrightarrow{A}\times \overrightarrow{B}\right)=0$

Magnitude is equal to zero

Step 2 Explaination ( A → × B → ) =0

We know that  $\overrightarrow{A}\times \overrightarrow{B}$ is vector Perpendicular to both  $\overrightarrow{A}$and $\overrightarrow{B}$ Let’s say that $\overrightarrow{A}$and $\overrightarrow{B}$ in xy plane. Then  $\overrightarrow{A}$and $\overrightarrow{B}$ will be along   z-axis.

Hence,  $\left(\overrightarrow{A}\times \overrightarrow{B}=0\right)$ 

 

Step3 (Explaination for A → . A → × B → = 0 )

Since, $\overrightarrow{A}$ which lies in xy plane and $\overrightarrow{A}$and $\overrightarrow{B}$  lies along zaxis are perpendicular to each other.

$$\begin{gathered} \stackrel{\overrightarrow{A}.\left(\overrightarrow{A}\times \overrightarrow{B}=0\right)}{ \\ } \end{gathered}$$

Step 4- (Conclusion)

It is proved 

$\overrightarrow{A}.\left(\overrightarrow{A}\times \overrightarrow{B}=0\right)$