Figure 1, below, shows two vectors on a plane. To add the two vectors, translate one of the vectors so that the terminal point of one vector coincides with the starting point of the second vector and the sum is a vector whose starting point is the starting point of the first vector and the terminal point is the terminal point of the second vector as shown in figure 2.

When the components of the two vectors are known, the sum of two vectors is found by adding corresponding components.

Example 1

Given vectors \( A = \lt 2 , -4 > \) and \( B = \lt 4 , 8 > \), find the components of \( \vec A + \vec B \).

Solution

\( \vec A + \vec B = \; \lt 2 , -4 > + \lt 4 , 8> = \; \lt 2 + 4 ,-4 + 8 > = \; \lt 6 , 4 > \)

The subtraction of two vectors is shown in figure 3. The idea is to change the subtraction into an addition as follows: \[ \vec A - \vec B = \vec A + (- \vec B) \]

Example 2

Given vectors \( A = \; \lt 2 , 6 > \) and \( B = \; \lt -2 , 8 > \), find the components of \( \vec A - \vec B \).

Solution

\( \vec A - \vec B = \; \lt 2 , 6 > - \lt -2 , 8 > = \; \lt 2 - (-2),-2 - 8 > = \; \lt 0 , -6 > \)

The relationships between the magnitude \( |\vec A| \), the direction \( \theta \), which is the angle made by the vector and the positive direction of the x-axis, and the components \( A_x \) and \( A_y \) of vector \( \vec A \) are given by

\( A_x = |\vec A| \cos \theta \)

\( A_y = |\vec A| \sin \theta \)

\( |\vec A| = \sqrt {A_x^2 + A_y^2} \)

\( \tan \theta = \dfrac{A_y}{A_x} \)

NOTE that in detrmining \( \theta \), we need to take into account the signs of the components \( A_x \) and \( A_y \) as will be shown in example 3.

Example 3

The magnitudes of two vectors \( \vec U \) and \( \vec V \) are equal to 5 and 8 respectively. \( \vec U \) makes an angle of 20° with the positive direction of the x-axis and \( \vec V \) makes an angle of 80° with the positive direction of the x-axis. Both angles are measured counterclockwise. Find the magnitudes and directions of vectors \( \vec U + \vec V\) and \( \vec U - \vec V\).

Solution

Let us first use the magnitudes and directions to find the components of vectors \( \vec U \) and \( \vec V \).

\( \vec U = \lt 5 \cos(20^{\circ}) , 5 \sin(20^{\circ}) > \)

\( \vec V = \lt 10 \cos(80^{\circ}) , 10 \sin(80^{\circ}) > \)

__Components of vector \( \vec U + \vec V \) __

\( \vec U + \vec V = \lt 5 \cos(20^{\circ}) , 5 \sin(20^{\circ})> + \lt 10 \cos(80^{\circ}) , 10 \sin(80^{\circ})> \)

\( = \; \lt 5 \cos(20^{\circ}) + 10 \cos(80^{\circ}) , 5 \sin(20^{\circ}) + 10 \sin(80^{\circ}) > \)

__Magnitude of vector \( \vec U + \vec V \) __

Now that we have the components of vector the \( \vec U + \vec V \), we can calculate the magnitude as follows:

\( |\vec U + \vec V| = \sqrt{ (5 \cos(20^{\circ}) + 10 \cos(80^{\circ}))^2 + (5 \sin(20^{\circ})+10 \sin(80^{\circ}))^2} \approx 13.22\)

__\( \theta \) : Direction of vector \( \vec U + \vec V \) __

We now approximate the components of vector \( \vec U + \vec V \) so that we can determine the quadrant of this vector

\( \vec U + \vec V = \; \lt 5 \cos(20^{\circ}) + 10 \cos(80^{\circ}) , 5 \sin(20^{\circ}) + 10 \sin(80^{\circ}) > \; \approx \; \lt 6.43 , 11.6 > \)

The approximated components of vector \( \vec U + \vec V \) are both positive and therefore the direction given by angle \( \theta \) in standard position is calculated as follows:

\( \tan \theta = \dfrac{y-component \; of \; \vec U + \vec V}{x-component \; of \; \vec U + \vec V} \)

Substitute

\( \tan \theta = \dfrac{5 \sin(20^{\circ})+10 \sin(80^{\circ})}{5 \cos(20^{\circ}) + 10 \cos(80^{\circ})} \)

The angle \( \theta \) which is the direction is given by

\( \theta = \arctan (\dfrac{5 \sin(20^{\circ})+10 \sin(80^{\circ})}{5 \cos(20^{\circ}) + 10 \cos(80^{\circ})}) \approx 60.9^{\circ} \)

__Components of vector \( \vec U - \vec V \) __

\( \vec U - \vec V = \lt 5 \cos(20^{\circ}) , 5 \sin(20^{\circ})> - \lt 10 \cos(80^{\circ}) , 10 \sin(80^{\circ})> \)

\( = (5 \cos(20^{\circ}) - 10 \cos(80^{\circ}) , 5 \sin(20^{\circ}) - 10 \sin(80^{\circ})) \)

__Magnitude of vector \( \vec U - \vec V \) __

Now that we have the components of vector the \( \vec U - \vec V \), we can calculate the magnitude as follows:

\( |\vec U + \vec V| = \sqrt{ (5 \cos(20^{\circ}) - 10 \cos(80^{\circ}))^2 + (5 \sin(20^{\circ}) - 10 \sin(80^{\circ}))^2} \approx 8.66\)

__\( \beta \) : Direction of vector \( \vec U - \vec V \) __

We now approximate the components of vector \( \vec U - \vec V \) so that we can determine the quadrant of this vector

\( \vec U - \vec V = \; \lt 5 \cos(20^{\circ}) - 10 \cos(80^{\circ}) , 5 \sin(20^{\circ}) - 10 \sin(80^{\circ}) > \; \approx \; \lt 2.96 , -8.14 > \)

The signs of the components of vector \( \vec U - \vec V \) indicate that terminal side of angle \( \beta \) is in quadrant IV. So we first determine the reference angle \( \alpha \) given by

\( \tan \alpha = \dfrac{|5 \sin(20^{\circ})-10 \sin(80^{\circ})|}{|5 \cos(20^{\circ}) - 10 \cos(80^{\circ})|} \)

\( \alpha = \arctan (\dfrac{|5 \sin(20^{\circ})-10 \sin(80^{\circ})|}{|5 \cos(20^{\circ}) - 10 \cos(80^{\circ})|}) \approx 70^{\circ} \)

The signs of the components of vector \( \vec U - \vec V \) indicate that terminal side of angle \( \beta \) is in quadrant IV and therefore the direction of the angle given by the angle in standard position \( \theta \) is given by

\( \beta = 360^{\circ} - \alpha = 290^{\circ} \)

Example 4

The components of three vectors \( \vec A \), \( \vec B \) and \( \vec C \) are given as follows:

\( \vec A = \; \lt 2 , -1 > \), \( \vec B = \; \lt -3 , 2 > \)
and \( \vec C = \; \lt 13 , - 8> \)

Find real numbers a and b such that : \( \vec C = a \vec A + b \vec B \)

Solution

We first rewrite the equation \( \vec C = a \vec A + b \vec B \) substituting the components of the vectors

\( \lt 13, - 8 > = a \lt 2 , -1> + b \lt -3 , 2> \)

Multiply components on left side

\( \lt 13, - 8 > = \; \lt 2 a , - a> + \lt -3 b , 2b > \)

Add components on the left side of the above and rewrite as

\( \lt 13, - 8 > = \; \lt 2a - 3b , -a + 2b >\)

Vectors are equal when their components area equal; hence the equations

\( 13 = 2 a - 3 b \) and \( - 8 = - a + 2 b \)

Solve the above equations in a and b to obtain

\( a = 2 \) and \( b = -3 \).

- Introduction to Vectors
- Formulas for Vectors
- Scalar Product of Vectors with questions (some of which may be challenging) that explains the application of the scalar product.
- Parallel and Perpendicular Vectors with questions (some of which may be challenging) and detailed solutions.
- Vector Direction and Bearing With examples, applications and questions with solutions.
- Free SAT II Physics Practice Questions Vectors with detailed solutions and explanations