Question

Find the magnitude of each of the two vectors $\overrightarrow a$ and $\overrightarrow b$, having the same magnitude such that the angle between them is ${60^ \circ }$ and their scalar product is $\dfrac{9}{2}$.

Answer 1

This is a problem related to Vector algebra. In this problem, a scalar product of vectors will be used. There are two relations given to solve equations with two variables i.e The magnitudes of both the vectors are the same, and their scalar product value has been given.In vector algebra ,scalar or dot product of two vectors $\overrightarrow a$ and $\overrightarrow b$ can be represented as
$= \overrightarrow a .\overrightarrow b = \left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|.\cos \theta$, Substituting the values in this equation we get the required answer.

Complete step-by-step answer:
In the problem, it is given that vectors $\overrightarrow a$and$\overrightarrow b$have the same magnitude, it means their absolute values are the same. This can be expressed mathematically as,
$\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right|$ …………… (1)
In the Vector Algebra, scalar or dot product of two vectors $\overrightarrow a$ and $\overrightarrow b$ can be represented as
$= \overrightarrow a .\overrightarrow b = \left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|.\cos \theta$ ………………. (2)
Where θ is the angle between vectors $\overrightarrow a$ and $\overrightarrow b$. Now, it is given that the angle between these two vectors $\overrightarrow a$ and $\overrightarrow b$ is ${60^ \circ }$. So the value of θ is ${60^ \circ }$. Therefore, putting the value of θ in the equation (2), we will get
$\Rightarrow \overrightarrow a .\overrightarrow b = \left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|.\cos {60^ \circ } \\\ \Rightarrow \overrightarrow a .\overrightarrow b = \left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|.\dfrac{1}{2} \\\$
Now, in the problem itself, it is given that their scalar product is $\dfrac{9}{2}$, therefore putting this value in the above expression and also using the relationship (1) we will get
$\Rightarrow \overrightarrow a .\overrightarrow b = \left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|.\dfrac{1}{2} \\\ \Rightarrow \dfrac{9}{2} = \left| {\overrightarrow a } \right|.\left| {\overrightarrow a } \right|.\dfrac{1}{2} \\\ \Rightarrow {\left| {\overrightarrow a } \right|^2} = 9 \\\ \Rightarrow \left| {\overrightarrow a } \right| = \pm 3 \\\$
Since the magnitude of the vectors cannot be negative, hence
$\Rightarrow \left| {\overrightarrow a } \right| = 3$.
And therefore,
$\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right| = 3$
This way, the magnitude of both the vectors $\overrightarrow a$and$\overrightarrow b$ is 3.

Note: It is a well-known fact that the magnitude of any vector cannot be negative. Scalar product or dot product of two vectors can be positive or negative, depending upon the angle between them. If the angle is acute, the dot product will be positive whereas if the angle is obtuse then the dot product will be negative.