Vectors

Mass, length, distance are some quantities that can be fully described by a magnitude and unit only. These physical quantities are called scalar quantities. However, some quantities cannot be totally described with magnitude alone, its direction must be specified. These quantities which requires magnitude and direction are called vector quantities. Examples of vector quantities are displacement, velocity, and direction.

A vector is represented by an arrow. The length of the arrow represents the magnitude and the arrowhead as the direction of the quantity. A scale is used to indicate the actual magnitude of a quantity. A vector quantity is represented by a letter with arrow above it. Consider the vector below.

The vector represents a quantity 2 m to the East.


Example 1.

What is the velocity of the object as shown by the vector.

The velocity of the object is 3\;m/s to the East.


Example 2.

How much does an object accelerate as described by the vector.

Since \(\vec{a}\) has 4 units, based on the scale \(\vec{a}=4 units \times 5\;m/s^2=20\;m/s^2\). Therefore, the object accelerates at \(20\;m/s^2\; \text{North of East}\).

 


 

Addition: Vectors with the Same Direction and Vectors with Opposite Direction

One and the simplest way to add vectors is to simply use the ordinary arithmetic. However, this process can only be applied when the vectors to be added have the same direction or opposite in direction. In adding vectors, the direction of the quantity is very much considered. It can be represented using the xy-plane.

The x-axis indicates the east and west directions – +x as east and –x as west. The y-axis indicates the north and south directions – +y as north and –y as south.

In vector addition, simply add the magnitude of the quantity indicating the sign of the direction. The sign of the vector sum indicates final direction of the quantity. The sum of vectors is also called resultant vector symbolized by \(\vec{R}\).


Example 1.

Find the sum of the given vectors. \(\vec{a}=\text{18 units East}\) and \(\vec{b}=\text{20 units East}\).

Solution:

\(\vec{R}=18\;\text{units}+20\;\text{units}=38\;\text{units}\)

Hence, \(\vec{R}\) is \(38\;\text{units}\) to the East.


Example 2.

Given: \(\vec{c}=35\;\text{units West}\) and \(\vec {d}=15\;\text{units East}. \) Solve for \(\vec{R}\).

Solution:

\(\vec{R}=-35\;\text{units}+15\;\text{units}=-20\;\text{units}\)

The resultant vector is \(20\;\text{units West}\).


Example 3.

Given: \(\vec{e}=62\;\text{units North}\) and \(\vec {f}=47\;\text{units South}\). Calculate \(\vec{R}\).

Solution:

\(\vec{R}=62\;\text{units}+(-47)\;\text{units}=15\;\text{units}\)

The resultant vector is \(15\;\text{units North}\).

 


 

Vector Addition - Vectors Perpendicular in Direction

Simple arithmetic cannot be applied in adding vectors given that there direction is perpendicular to each other. To solve such problems, the Pythagorean Theorem and Trigonometric Ratios are used. Listed below are the steps to solve the sum of two perpendicular vectors.

  1. Using head-to-tail, represent all the given vectors and the resultant forming a right triangle. Draw first the 1st vector. Draw the second vector, its tail from the head of the 1st one. Lastly, draw the resultant vector, its tail from the tail of the 1st vector and its head to the head of the 2nd vector.
  2. Using Pythagorean theorem, solve for the resultant vector, \(\vec{R}\) as the hypotenuse of the right triangle.
  3. Indicate the direction of the resultant vector using the basic trigonometric ratios. The angle, \(\theta\) with respect to the x-axis.

\(\sin\theta=\frac{\text{opposite side}}{\text{hypotenuse}}\)

\(\cos \theta=\frac{\text{adjacent side}}{\text{hypotentuse}}\)

\(\tan \theta=\frac{\text{opposite side}}{\text{adjacent side}}\)


Example 1.

Given: \(\vec{m}=\text{10 units East}\\ \vec{n}=\text{15 units North}\)

Find: \(\vec{R}\)


Solution 1.

Draw the vectors to form right triangle.


Solution 2.

Solve for \(\vec{R}\) using the Pythagorean theorem, \(c^2=a^2+b^2\)\(c=\vec{R};\; a=\vec{m}\; \text{and}\; b=\vec{n}\).

\((\vec{R})^2=(\vec{m})^2+(\vec{n})^2\\ \quad\quad=(10\;units)^2+(15\;units)^2\\ \quad\quad=100\;units+225\;units\\ \quad\quad=325\;units\)

hence, \(\vec{R}=\sqrt{325\;units}=18.03\;units\)   


Solution 3.

Since the given vectors are the opposite and adjacent sides, we use \(\tan \theta\) to solve for the angle and indicate the direction of \(\vec{R}\).

\(\tan \theta=\frac{\vec{n}}{\vec{m}}=\frac{15\;units}{10\;units}=1.5\)

To solve for \(\theta\), we solve for the value of the inverse function of \(\tan\) given by 

\(\theta=\tan^{-1}1.5=56.31^\circ\).

Therefore, \(\vec{R}\) is \(18.03\;\text{units},\;56.31^\circ\;\text{North of East}\).


Example 2.

Given: \(\vec{r}=\text{63 units West}\\ \vec{s}=\text{50 units South}\)

Solve for \(\vec{R}\).


Solution 1.

Draw the vectors to form right triangle.


Solution 2.

Solve for \(\vec{R}\) using the Pythagorean theorem. Indicate negative sign since the given vectors are in negative direction.

\((\vec{R})^2=(\vec{r})^2+(\vec{s})^2\\ \quad\quad=(-63\;units)^2+(-50\;units)^2\\ \quad\quad=3,969\;units+2,500\;units\\ \quad\quad=6,469\;units\)

thus, \(\vec{R}=\sqrt{6,469\;units}=80.43\;units\).


 

Solution 3.

Solve for the angle and direction of \(\vec{R}\).

\(\tan \theta=\frac{\vec{s}}{\vec{r}}=\frac{-50\;units}{-63\;units}=0.79\)

To solve for \(\theta\), we solve for the value of the inverse function of \(\tan\) given by 

\(\theta=\tan^{-1}0.79=38.31^\circ\).

Therefore, \(\vec{R}\) is \(80.43\;\text{units},\;38.31^\circ\;\text{South of East}\).

 


 

Vector Addition Using Component Method

Another way of adding vectors is resolving the x and y components of the given vectors. This is used when there are more than two vectors to be added with different directions.

The steps in solving resultant vector using this method are as follows:

  1. Given \(\vec{v}\) as shown in the image above and angle \(\theta\), use the appropriate trigonometric ratio to find x-component of \(\vec{v}\) denoted as \(\vec{v_x}\) and y-component as \(\vec{v_y}\).

         Recall:

                      \(\sin \theta=\frac{opposite side}{hypotenuse}=\frac{\vec{v_y}}{\vec{v}}\), thus

                      \(\vec{v_y}=\vec{v}\sin \theta\)

         and

                      \(\cos \theta=\frac{opposite side}{hypotenuse}=\frac{\vec{v_x}}{\vec{v}}\), thus

                      \(\vec{v_x}=\vec{v}\cos \theta\)

  1. Add all x-components denoting the sum as \(\vec{R_x}\) and the sum of y-components as \(\vec{R_y}\). Indicate the sign of each magnitude to represent direction. North and East point to positive direction. South and West are negative directions.
  2. Solve for the magnitude of \(\vec{R}\) using the Pythagorean Theorem

\((\vec{R})^2=(\vec{R_x})^2 + (\vec{R_y})^2\)

  1. Using trigonometric ratio, solve for the angle \(\theta\) to indicate the direction of \(\vec{R}\).

Example 1.

Given:

\(\vec{a}=15\;\text{units North}\\ \vec{b}=30\;\text{units}, \;30^\circ\; \text{North of East}\\ \vec{c}=45\;\text{units}, \;60^\circ\; \text{North of West}\\\)

Soution 1:

Plot the given vectors in the Cartesian plane.


Solution 2:

Solve for the x and y components of \(\vec{a}\)\(\vec{b}\), and \(\vec{c}\).

x-components:

\(\vec{a_x}=0\; \text{units}\)

\(\vec{b_x}= b\cos 30^\circ=(30\;\text{ units})(0.87)=26.1\; \text{units}\)

\(\vec{c_x}= c\cos 60^\circ=(45\; \text{units})(0.5)=22.5\; \text{units} \)

y-components:    

\(\vec{a_y}= 15\;\text{units}\)

\(\vec{b_y}= b\sin 30^\circ=(30\;\text{units})(0.5)=15\;\text{units}\)

\(\vec{c_y}= c\sin 60^\circ=(-45\;\text{units})(0.87)=-39.15\;\text{units}\)


Solution 3.

Solve for the resultant vector for each component.

3.1 Add all x-components to solve \(\vec{R_x}\).

\(\vec{R_x}=\vec{a_x}+\vec{b_x}+\vec{c_x}=0+26.1+22.5=48.6\;\text{units}\)

3.2 Add all y-components to solve \(\vec{R_y}\).

\(\vec{R_y}=\vec{a_y}+\vec{b_y}+\vec{c_y}=15+15+(-39.15)=-9.15\;\text{units}\)


Solution 4.

Solve for the resultant vector using the Pythagorean Theorem \((\vec{R})^2=(\vec{R_x})^2+(\vec{R_y})^2 \)

\(\vec{R}=\sqrt{(\vec{R_x})^2+(\vec{R_y})^2}\\ \;\;\,=\sqrt{(48.6)^2+(-9.15)^2}\\ \;\;\,=\sqrt{2,361.96+83.72}\\ \;\;\,=\sqrt{2,445.68}\\ \;\;\,=49.45\;\text{units}\)

\(\vec{R}\) measures \(49.45\;\text{units}\).


Solution 5.

Solve for the value of \(\theta\) using trigonometric ratio.

\(\cos \theta=\frac{R_x}{R}=\frac{48.6\;units}{49.45\;units}=0.98\)

\(\theta=\cos^{-1}\;0.98=11.48^\circ\)

Therefore, \(\vec{R}\) is \(49.45\;\text{units}\), \(11.48^\circ\) South of East.


Example 2.

 

Given:

\(\vec{a}=3\;\text{m North}\\ \vec{b}=6\;\text{m South}\\ \vec{c}=9\;\text{m East}\\ \vec{d}=12\;\text{m West}\\ \vec{e}=15\;\text{m}, \;45^\circ\; \text{South of West}\\ \vec{f}=18\;\text{m}, \;25^\circ\; \text{South of East} \)

Soution 1:

Plot the given vectors in the Cartesian plane.

 


Solution 2:

Solve for the x and y components of \(\vec{a}\)\(\vec{b}\)\(\vec{c}\)\(\vec{d},\;\vec{e},\;\vec{f}\).

x-components:

\(\vec{a_x}=0\; \text{m}\)

\(\vec{b_x}=0\; \text{m}\)

\(\vec{c_x}=9\; \text{m}\)

\(\vec{d_x}=12\; \text{m}\)

\(\vec{e_x}= e\cos 45^\circ=(15\;\text{m})(0.71)=-10.65\; \text{m}\)

\(\vec{f_x}= f\cos 25^\circ=(18\;\text{m})(0.91)=16.38\; \text{m}\)

 

y-components:    

\(\vec{a_y}= 3\;\text{m}\)

\(\vec{b_y}= 6\;\text{m}\)

\(\vec{c_y}= 0\;\text{m}\)

\(\vec{d_y}= 0\;\text{m}\)

\(\vec{e_y}= e\sin 45^\circ=(15\;\text{m})(0.71)=-10.65\; \text{m}\)

 

\(\vec{f_y}= f\sin 25^\circ=(18\;\text{m})(0.42)=-7.56\; \text{m}\)

 


Solution 3.

Solve for the resultant vector for each component.

3.1 Add all x-components to solve \(\vec{R_x}\).

\(\vec{R_x}=\vec{a_x}+\vec{b_x}+\vec{c_x}+\vec{d_x}+\vec{e_x}+\vec{f_x}\\ \quad\;=0+0+9-12-10.65+16.38\\ \quad\;=2.73\;\text{m}\)

3.2 Add all y-components to solve \(\vec{R_y}\).

 

\(\vec{R_y}=\vec{a_y}+\vec{b_y}+\vec{c_y}+\vec{d_y}+\vec{e_y}+\vec{f_y}\\ \quad\;=3-6+0+0-10.65-7.56\\ \quad\;=-21.21\;\text{m}\)


Solution 4.

Solve for the resultant vector using the Pythagorean Theorem \((\vec{R})^2=(\vec{R_x})^2+(\vec{R_y})^2 \)

\(\vec{R}=\sqrt{(\vec{R_x})^2+(\vec{R_y})^2}\\ \;\;\,=\sqrt{(2.73)^2+(-21.21)^2}\\ \;\;\,=\sqrt{7.45+449.86}\\ \;\;\,=\sqrt{457.31}\\ \;\;\,=21.38\;\text{m}\)

\(\vec{R}\) measures \(21.38\;\text{m}\).


Solution 5.

Solve for the value of \(\theta\) using trigonometric ratio.

\(\cos \theta=\frac{R_x}{R}=\frac{2.73\;m}{21.38\;m}=0.13\)

\(\theta=\cos^{-1}\;0.13=82.53^\circ\)

Therefore, \(\vec{R}\) is \(21.38\;\text{m}\)\(82.53^\circ\) South of East.