Question

Two vectors have magnitudes as $$3$$ unit and $$4$$ unit respectively. What should be the angle between them of the magnitude of the resultant is $$5$$ unit.

Answer 1

Hint: Here we are asked to find the angle between two vectors whose magnitudes are given. We are also provided with the magnitude of the resultant vector of those two vectors. So, we will use the formula of the magnitude of the resultant vector since it involves the angle between them. For us, the required value is the angle between the vectors so we use this formula to find it.
Formula: Formula that we need to know before solving this problem:
let $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$ be two vectors then the magnitude of the resultant of these two vectors is $$\overrightarrow{\left|R\right|}=\sqrt{a^2+b^2+2ab\cos \theta }$$ where $$\theta$$ the angle between the vectors $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$.

Angle in degrees	$$0^{\circ }$$	$${30}^{\circ }$$	$${45}^{\circ }$$	$${60}^{\circ }$$	$${90}^{\circ }$$
$$\cos$$	$$1$$	$${\displaystyle \frac{\sqrt{3}}{2}}$$	$${\displaystyle \frac{1}{\sqrt{2}}}$$	$${\displaystyle \frac{1}{2}}$$	$$0$$

Complete step-by-step solution:
It is given that the two vectors have magnitude $$3$$ unit and $$4$$ unit respectively and the magnitude of its resultant is $$5$$ units. We aim to find the angle between the vectors.
Since we are given the magnitude of the resultant vector, we will use the formula to find the angle between the two vectors.
We know that the magnitude of the resultant vector of two vectors $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$ $$\overrightarrow{\left|R\right|}=\sqrt{a^2+b^2+2ab\cos \theta }$$ where $$\theta$$ the angle between the vectors $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$.
We have that $$\left|\overrightarrow{a}\right|=a=3$$, $$\left|\overrightarrow{b}\right|=b=4$$ and $$\left|\overrightarrow{R}\right|=5$$ substituting these in the formula we get
$$5=\sqrt{3^2+4^2+2\left(3\right)\left(4\right)\cos \theta }$$ where $$\theta$$ the angle between the vectors $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$.
On simplifying the above, we get
$$5=\sqrt{9+16+24\cos \theta }$$
On simplifying further, we get
$$5=\sqrt{25+24\cos \theta }$$
Now let us square the above expression on both sides
$$5^2=25+24\cos \theta$$
$$25=25+24\cos \theta$$
$$0=24\cos \theta$$
$$\cos \theta =0$$
We know that the value $$\cos \theta$$equals zero when the angle is $${90}^{\circ }$$
Thus, $$\theta ={90}^{\circ }$$
Therefore, the angle between the two vectors $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$ is $${90}^{\circ }$$.

Note: We have found that the angle between the two angles is $${90}^{\circ }$$. We can also picture the angle, since the angle is $${90}^{\circ }$$ the two angles are perpendicular to each other. Either the vector $$\overrightarrow{a}$$ will be horizontal and the vector $$\overrightarrow{b}$$ will be vertical or the vector $$\overrightarrow{b}$$ will be horizontal and the vector $$\overrightarrow{a}$$ will be vertical.