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:
Example Construct the vectors a+ b and a- b.
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-to-tail:
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 direction. Thus we are finding the resultant of AB and –AC. From right triangle AC’D, m∠ CAD = 45° and AD = 102 ≈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 PS represents the actual path of the plane. It is obtained by completing the rectangle PQSR. The length of PS epresents the actual speed of
the plane.
PS= 702+ 2402=260 mph
The bearing angle is
tan∠ RPS = 70240 = 724≈0.2917 .
∠ 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
The sca1e drawing shows the components DE and DF of the given vector DG. From right triangle DEG,
GE = 100 sin 30°= 100 ∙ 0.5 = 50 lb,
DE = 100 cos 30°= 100 ∙ 32 ≈ 87 lb,
Example A man who rows at the rate of 4 mph relative to the shore, goes straight across the river which current speed is 3 mph. In what direction does he head and what is his speed relative to the water?
Solution The correct answer is 37° to heading of the boat, the water speed is 5 mph.
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 tan ∠AOC = 34 so that ∠AOC ≈ 37°.
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
In the figure, |AW| = 3000 lb, AB is the component of AW parallel to 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.
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.
Example Add the vectors a = (8, 13) and b = (26, 7).
Solution c = a + b
c = (8, 13) + (26, 7) = (8 + 26, 13 + 7) = (34, 20)
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.
Example Subtract a = (4, 5) from b = (12, 2)
Solution c = b − a
c = (12, 2) − (4, 5) = (12 − 4, 2 − 5) = (8, −3).
Example If a = (3, 2) and b = (–5, 3), what is 3a – 2b ?
Solution
3a – 2b = 3(3, 2) – 2(–5, 3)
= (9, 6) – (–10, 6)
= (3 + 10, 6 – 6) = (13, 0).
If a = (1, –3, 2) and b = (3, 4, –5), what is b – 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:
|a| = ax2+ ay2
Example What is the magnitude of the vector b = (6, 8)?
Solution |b| = 62+ 82 = 36 + 64 = 100 = 10
A vector with magnitude 1 is called a unit vector.
Example What is the magnitude of the vector a = (–3, 5)?
Solution |a| = -32+ 52 = 0+25 = 34.
Multiplying a Vector by a Scalar
To multiply a vector by a scalar multiply each coordinate by this scalar.
Example Multiply the vector m = (7, 3) by the scalar 3.
Solution a = 3m = (3 × 7, 3 × 3) = (21, 9).
3 Dimensions
The vectors we have been looking at have been 2 dimensional, but vectors work perfectly well in 3 or more dimensions:
Example Add the vectors a = (3, 7, 4) and b = (2, 9, 11)
Solution c = a + b
c = (3, 7, 4) + (2, 9, 11) = (3 + 2, 7 + 9, 4 + 11) = (5, 16, 15).
Example What is the magnitude of the vector a = (1, −2, 3)?
|a| = 12+ -22+ 32 = 1 + 4 + 9 = 14.
Example If a = (4, 5, 6) and b = (1, 2, 3), and c = a –2b, what is the magnitude of c?
Solution
c = a –2b = (4, 5, 6) – 2 × (1, 2, 3) = (4, 5, 6) – (2, 4, 6) = (2, 1, 0).
So,
|c| = 22+ 12+ 02 = 4 + 1 + 0 = 5 .
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):
/ <=> /Vector a in Polar
Coordinates / Vector a in Cartesian
Coordinates
Here is a quick summary:
From Cartesian Coordinates (ax, ay) to Polar Coordinates (r, θ)
r = |a| = ax2+ ay2
θ = tan-1ax ay (for the I quadrant)
θ = 180° – tan-1ax ay (for the II quadrant)
θ = 180° + tan-1ax ay (for the III quadrant)
θ = 360° – tan-1ax ay (for the IV quadrant)
From Polar Coordinates (r, θ) to Cartesian Coordinates (ax, ay):
ax = r ∙ cos θ ay= r ∙ sin θ
Example If a = (5, –12), what are the magnitude and direction of a?
We will find the Polar coordinates (r, θ) of the point:
Magnitude = r = 52+ -122 = 25+144 = 169 = 13
Solution:
Direction = θ = 360° - tan-1125 = 360° - 67.4° = 292.6° since the 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:
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
Ship's velocity vector: vs = (0, 20)
20 × cos (90°) = 20 × 0 = 0
20 × sin (90°) = 20 × 1 = 20
Wind's velocity vector: vw = (2.5, 4.330)
5 × cos(60°) = 5 × 0.5 = 2.5
5 × sin(60°) = 5 × 0.8660 = 4.330
Now it is easy to add them:
v= (0, 20) + (2.5, 4.330) = (2.5, 24.330)
We can convert that to polar for a final
answer:
r = 2.52+ 24.3302 = 24.46
θ = tan-124.3302.5 = 84.1°
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.