Question

If the magnitude of two vector are 2 and 3 the magnitude of their scalar product is 3√2 then the angle between the vectors is ?

Answer 1

Answer:

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Explanation:

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Answer 2

Given,

Two vectors, with magnitudes 2 and 3.

Scalar product of given vectors = 3√2.

To find,

The angle between given vectors.

Solution,

We can solve this problem simply by following the process given below.

To solve such problems, one should understand what the scalar product is. So, the scalar product between the two vectors (say $\vec A \hspace{3}and \hspace{3} \vec B$), having the magnitudes $|A| \hspace{3}and \hspace{3} |B|$, is given by

$A\cdot B=|A|\cdot|B|cos\theta,$

where, $\theta$ is the angle between the two vectors $\vec A \hspace{3}and \hspace{3} \vec B$.

This relation can be used to find the angle $\theta$ between the vectors $\vec A \hspace{3}and \hspace{3} \vec B$ when the scalar product and magnitudes are given.

So, for the given problem,

$|A|=2,|B|=3$ and,

$A\cdot B=3\sqrt{2}$.

Now, substituting these given values in the above relation, we get,

3√2 = 2·3·$cos\theta$

Simplifying and rearranging,

$cos\theta=\frac{3\sqrt{2} }{2*3}$

⇒ $cos\theta=\frac{1}{\sqrt{2}}$

⇒ $\theta = 45 \textdegree.$

Therefore, the angle between the given vectors will be 45°.