Question

How do you find magnitude and direction angle of a vector v = 3(cos 60i + sin 60j)?

Answer 1

Hint: Vectors are normally a scalar quantity given with direction, or we can say that vectors are normally magnitudes of any quantity that are given with directions. Vectors are shown with an arrow symbol above them but this arrow does not define their direction. Do not confuse that the arrow points towards
the direction of the vector.

Complete answer:
The term magnitude includes only the value of the term or variable and it does not define any other thing, like for example if we say that a car is moving with $10m{{s}^{-1}}$ towards the east then the term “10” is known as the magnitude. While east is the direction and $m{{s}^{-1}}$ is the unit for this velocity.
To find the magnitude of a vector quantity we can use the simple formula
$\left| c \right|=\sqrt{{{(x\hat{i})}^{2}}+{{(y\hat{j})}^{2}}}$
Now, we have been given a vector v = 3(cos 60i + sin 60j)
Or, v = 3cos 60i + 3sin 60j
$3\cos (60)=3(\dfrac{1}{2})=\dfrac{3}{2}$
$3\sin (60)=3(\dfrac{\sqrt{3}}{2})=\dfrac{3\sqrt{3}}{2}$
Magnitude of “v” can be given as
$\left| v \right|=\sqrt{{{(\dfrac{3}{2})}^{2}}+{{(\dfrac{3\sqrt{3}}{2})}^{2}}}$
$\left| v \right|=+3,-3$
Now, the direction angle can be given by
Let’s assume that we have taken the positive value if magnitude then,
$\tan \theta =\dfrac{3\sin (60)}{3\cos (60)}$
$\tan \theta =\sqrt{3}$
$\tan \theta $gives this value at ${{60}^{\circ }}$, hence we can say that the direction angle is ${{60}^{\circ }}$

Note:
Even though a vector is a quantity with direction, this does not define its position. Vector consists of only direction with magnitude and not the location coordinates. Do not confuse the direction of the vector with its location. Both are different things. Location can only be defined with time coordinates for a particular instant of time (In case of moving objects) while directions can be defined without time coordinates.