Question

If the magnitude of two vectors are 4 and 6 and the magnitude of their scalar product is 12√2 , then what is the angle between these vectors?

Answer 1

Step 1. Given data:

Let the vectors be $\overrightarrow{m}$and $\overrightarrow{n}$.

So $\left|\overrightarrow{m}\right|=4$, $\left|\overrightarrow{n}\right|=6$.

Scaler product, $\overrightarrow{m}.\overrightarrow{n}$=$12\sqrt{2}$.

Step 2: Calculation:

We know the scaler product between these two vectors is given by:

$\overrightarrow{m}.\overrightarrow{n}=\left|\overrightarrow{m}\right|\left|\overrightarrow{n}\right|\cos \left(\theta \right)$, where $\theta$ is the angle between these two vectors.

Hence putting the given values in the equation of scaler product, we get:

$$\begin{gathered} 12\sqrt{2}=4.6\cos \left(\theta \right) \\ \cos \left(\theta \right)=\frac{1}{\sqrt{2}} \\ \theta ={\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right) \\ \theta =45° \end{gathered}$$

Hence, the angle between two vectors is $\theta =45°$