Vectors
A vector is a directed line segment:
A vector has magnitude (how long it is) and direction:
The length of the line shows its magnitude and the arrowhead points in
the direction.
You can add two vectors by simply joining them head-to-tail:
It doesn't matter which order you add them, you get the same result:
Example A plane is flying along, pointing North, but there is a wind
coming from the North-West.
The two vectors (the velocity caused by the propeller, and the velocity of
the wind) result in a slightly slower ground speed heading a little East of
North. If you watched the plane from the ground it would seem to be
slipping sideways a little.
You can also subtract one vector from another:


first you reverse the direction of the vector you want to subtract,
then add them as usual:
⃗ + ⃗𝒃 and 𝒂
⃗ − ⃗𝒃.
Example Construct the vectors 𝒂
Solution
To add the vectors, join the vectors head-to-tail as follows:
To subtract the vectors, you reverse the direction of the vector you want
to subtract, and then add the vectors as usual by joining them head-totail:
A vector can be written as the letters of its head and tail with an arrow
above, like this:
Example Given horizontal vector ⃗⃗⃗⃗⃗
𝐴𝐵 and vertical vector ⃗⃗⃗⃗⃗
𝐴𝐶 , find a
vector equal to the difference ⃗⃗⃗⃗⃗
𝐴𝐵 – ⃗⃗⃗⃗⃗
𝐴𝐶 . Both vectors have a magnitude
of 10.
Solution
Consider – ⃗⃗⃗⃗⃗
𝐴𝐶 as a vector equal to ⃗⃗⃗⃗⃗
𝐴𝐶 in magnitude and opposite in
⃗⃗⃗⃗⃗ . From right
direction. Thus we are finding the resultant of ⃗⃗⃗⃗⃗
𝐴𝐵 and –𝐴𝐶
triangle AC’D, m∠ CAD = 45° and AD = 10√2 ≈ 14. ⃗⃗⃗⃗⃗
𝐴𝐷 has a
magnitude of 14 and bearing S 45° E.
Example A plane is flying north at 240 mph when it encounters a west
wind blowing east at 70 mph. In what direction will the plane be going
and with what speed?
Solution
The scale drawing shows vectors for velocities ⃗⃗⃗⃗⃗
𝑃𝑅 and ⃗⃗⃗⃗⃗
𝑃𝑄 . The vector
⃗⃗⃗⃗ represents the actual path of the plane. It is obtained by completing
𝑃𝑆
the rectangle PQSR. The length of ⃗⃗⃗⃗
𝑃𝑆 epresents the actual speed of
the plane.
⃗⃗⃗⃗ | = √702 + 2402 = 260 𝑚𝑝ℎ
|𝑃𝑆
The bearing angle is
70
7
tan∠ RPS =
= ≈ 0.2917 .
240 24
∠ RPS ≈ 16°
The bearing is N 16° W.
Example A force of 100 lb is acting at 30° to the horizontal. Find the
horizontal and vertical components of the given vector.
Solution
⃗⃗⃗⃗⃗ and ⃗⃗⃗⃗⃗
The sca1e drawing shows the components 𝐷𝐸
𝐷𝐹 of the given
vector ⃗⃗⃗⃗⃗
𝐷𝐺 . From right triangle DEG,
𝐺𝐸 = 100 sin 30° = 100 ∙ 0.5 = 50 lb,
𝐷𝐸 = 100 cos 30° = 100 ∙
√3
2
≈ 87 lb,
Example A man, who rows at the rate of 4 mph in still water, wishes to
go straight across the river at 4 mph, in what direction would he have to
head and with what speed?
Solution The correct answer is 5 mph, 37to heading of boat.
In the figure we are trying to find ⃗⃗⃗⃗⃗
𝑂𝐴 – ⃗⃗⃗⃗⃗
𝑂𝐵 . Draw a vector with
magnitude of 3 and opposite in direction to ⃗⃗⃗⃗⃗
𝑂𝐵 . Now find the resultant
of ⃗⃗⃗⃗⃗
𝑂𝐴 and – ⃗⃗⃗⃗⃗
𝑂𝐵 from the 3–4–5 right triangle AOC. Then OC = 5 and
3
tan ∠AOC = so that ∠AOC ≈ 37.
4
Example Determine the magnitude and direction of the resultant of two
forces, one of 9 N whose bearing is S 60° E and the other of 14 N whose
bearing is S 30° W.
Solution
Example An auto weighing 3,000 lb stands on a hill inclined 15° to the
horizontal. What force tending to pull it downhill must be overcome by
the brakes?
Solution
⃗⃗⃗⃗⃗⃗⃗ | = 3000 lb, ⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗⃗ parallel to
In the figure, |𝐴𝑊
𝐴𝐵 is the component of 𝐴𝑊
the incline and ⃗⃗⃗⃗⃗
𝐴𝐶 is perpendicular to the incline. Since the sides of
∠CAW are perpendicular to the sides of the 15angle, ∠CAW = 15.
Then
Example A ship is sailing northward in a calm sea at 30 mph. Suddenly
a north wind starts blowing at 6 mph, and a current of 10 mph starts
moving it eastward. Find the magnitude and direction of the resultant
velocity.
Solution The correct answer is 26 mph, N 23° E.
The diagram shows the three forces acting on the ship. The north and
south forces have a resultant of 24 heading north.
Bearing is N 23E.
Coordinates
If A(𝑥1 , 𝑦1 ) and B(𝑥2 , 𝑦2 ) are the tail and head of the vector, then we can
write
⃗ = ⃗⃗⃗⃗⃗⃗
𝒂
𝑨𝑩 = (𝑎𝑥 , 𝑎𝑦 ),
where 𝑎𝑥 = 𝑥2 – 𝑥1 , , 𝑎𝑦 = 𝑦2 – 𝑦1 – the coordinates of the vector.
Adding Vectors
To add two vectors, add their corresponding coordinates.
⃗ = (8, 13) and ⃗𝒃 = (26, 7).
Example Add the vectors 𝒂
⃗ =𝒂
⃗ + ⃗𝒃
Solution 𝒄
⃗ = (8, 13) + (26, 7) = (8 + 26, 13 + 7) = (34, 20)
𝒄
The vector (8, 13) and the vector (26, 7) add up to the vector (34, 20)
Subtracting Vectors
To subtract two vectors, subtract their corresponding coordinates.
⃗ = (4, 5) from ⃗𝒃 = (12, 2)
Example Subtract 𝒂
⃗ = ⃗𝒃 − 𝒂
⃗
Solution 𝒄
⃗ = (12, 2) − (4, 5) = (12 − 4, 2 − 5) = (8, −3).
𝒄
⃗ = (–5, 3), what is 3𝒂
⃗ ?
⃗ = (3, 2) and 𝒃
⃗ – 2𝒃
Example If 𝒂
Solution
⃗ = 3(3, 2) – 2(–5, 3)
⃗ – 2𝒃
3𝒂
= (9, 6) – (–10, 6)
= (3 + 10, 6 – 6) = (13, 0).
⃗ = (1, –3, 2) and ⃗𝒃 = (3, 4, –5), what is ⃗𝒃 – 𝒂
⃗ ?
If 𝒂
⃗𝒃 – 𝒂
⃗ = (3, 4, –5) – (1, –3, 2) = (3 – 1, 4 – (–3), –5 – 2) = (2, 7, –7)
Magnitude of a Vector
You can find magnitude of a vector by Pythagorean Theorem:
⃗ | = √𝑎𝑥2 + 𝑎𝑦2
|𝒂
Example What is the magnitude of the vector ⃗𝒃 = (6, 8)?
⃗ | = √62 + 82 = √36 + 64 = √100 = 10
Solution |𝒃
A vector with magnitude 1 is called a unit vector.
⃗ = (–3, 5)?
Example What is the magnitude of the vector 𝒂
⃗ | = √(−3)2 + 52 = √0 + 25 = √34.
Solution |𝒂
Multiplying a Vector by a Scalar
To multiply a vector by a scalar multiply each coordinate by this scalar.
⃗⃗⃗ = (7, 3) by the scalar 3.
Example Multiply the vector 𝒎
⃗ = 3𝒎
⃗⃗⃗ = (3 × 7, 3 × 3) = (21, 9).
Solution 𝒂
3 Dimensions
The vectors we have been looking at have been 2 dimensional, but
vectors work perfectly well in 3 or more dimensions:
⃗ = (3, 7, 4) and ⃗𝒃 = (2, 9, 11)
Example Add the vectors 𝒂
⃗ =𝒂
⃗ + ⃗𝒃
Solution 𝒄
⃗ = (3, 7, 4) + (2, 9, 11) = (3 + 2, 7 + 9, 4 + 11) = (5, 16, 15).
𝒄
⃗ = (1, −2, 3)?
Example What is the magnitude of the vector 𝒂
⃗ | = √12 + (−2)2 + 32 = √1 + 4 + 9 = √14.
|𝒂
⃗ , what is the
⃗ = (4, 5, 6) and ⃗𝒃 = (1, 2, 3), and 𝒄
⃗ =𝒂
⃗ –2𝒃
Example If 𝒂
⃗?
magnitude of 𝒄
Solution
⃗ = (4, 5, 6) – 2 × (1, 2, 3) = (4, 5, 6) – (2, 4, 6) = (2, 1, 0).
⃗ =𝒂
⃗ –2𝒃
𝒄
So,
⃗ | = √22 + 12 + 02 = √4 + 1 + 0 = √5 .
|𝒄
Magnitude and Direction
You may know a vector's magnitude and direction, but what are the
coordinates of the vector 𝑎𝑥 and 𝑎𝑦 (and vice versa):
<=>
⃗ in Polar
Vector 𝒂
Coordinates
⃗ in Cartesian
Vector 𝒂
Coordinates
Here is a quick summary:
From Cartesian Coordinates (𝑎𝑥 , 𝑎𝑦 ) to Polar Coordinates (r, θ)
⃗ | = √𝑎𝑥2 + 𝑎𝑦2
r = |𝒂
θ = tan-1(
𝑎𝑥
𝑎𝑦
) (for the I quadrant)
θ = 180° – tan-1|
𝑎𝑥
θ = 180° + tan-1|
𝑎𝑥
θ = 360° – tan-1|
𝑎𝑥
𝑎𝑦
| (for the II quadrant)
𝑎𝑦
𝑎𝑦
| (for the III quadrant)
| (for the IV quadrant)
From Polar Coordinates (r, θ) to Cartesian Coordinates (𝑎𝑥 , 𝑎𝑦 ):
𝑎𝑥 = r ∙ cos θ
𝑎𝑦 = r ∙ sin θ
⃗ = (5, –12), what are the magnitude and direction of 𝒂
⃗?
Example If 𝒂
We will find the Polar coordinates (r, θ) of the point:
Magnitude = r = √52 + (−12)2 = √25 + 144 = √169 = 13
Solution:
12
Direction = θ = 360° − tan-1( ) = 360° − 67.4° = 292.6° since the
5
point is in the fourth quadrant. Therefore (r, θ) = (13, 292.6°).
Example A vector's magnitude and direction are 8 and 125°. What are
the coordinates of the vector correct to 2 decimal places?
Solution In this case r = 8 and θ = 125°.
Then
𝑎𝑥 = 8 × cos125° = 8 × (−0.573...) = −4.588...
𝑎𝑦 = 8 × sin125° = 8 × 0.819... = 6.553...
So the vector is (-4.59, 6.55).
Example:
Sam and Alex are pulling a box.
Sam pulls with 200 N force at 60°
 Alex pulls with 120 N force at 45° as shown

What is the combined force, and its direction?
Solution: Determine the coordinates of both vectors (forces)
Sam's vector: 𝐹𝑆 = (100, 173.21)
200 × cos 60° = 200 × 0.5 = 100
200 × sin 60° = 200 × 0.8660 = 173.21
Alex's vector: 𝐹𝐴 = (84.85, −84.85)
120 × cos (– 45°) = 120 × 0.7071 = 84.85
120 × sin (– 45°) = 120 × (– 0.7071) = −84.85
Now it is easy to add them:
𝐹𝑛𝑒𝑡 = 𝐹𝑆 +𝐹𝐴 = (100, 173.21) + (84.85, −84.85) = (184.85, 88.36)
We can convert that to polar for a final answer:
r = √184.852 + 88.362 = 204.88
θ = tan-1 (
88.36
184.85
) = 25.5°
And we have this (rounded)
result:
And it looks like this for Sam and
Alex:
They might get a better result if they were shoulder-to-shoulder!
Example A ship is heading due north at 20 km/h but is blown off course
by the wind which is blowing from 30° west of south at 5 km/h:
What is the speed of the ship, and in which direction is it travelling?
Solution
⃗ 𝒔 = (0, 20)
Ship's velocity vector: 𝒗
20 × cos (90°) = 20 × 0 = 0
20 × sin (90°) = 20 × 1 = 20
⃗ 𝒘 = (2.5, 4.330)
Wind's velocity vector: 𝒗
5 × cos(60°) = 5 × 0.5 = 2.5
5 × sin(60°) = 5 × 0.8660 = 4.330
Now it is easy to add them:
⃗ = (0, 20) + (2.5, 4.330) = (2.5, 24.330)
𝒗
We can convert that to polar for a final
answer:
r = √2.52 + 24.3302 = 24.46
24.330
θ = tan-1(
) = 84.1°
2.5
This is counter clockwise angle from the positive x-axis (East), so to find
the bearing we need to subtract this angle from 90°:
90° − 84.1° = 5.9°
So, the velocity of the ship is 24.46 km/h and the direction is 5.9° east of
north.