Our universe has three dimensions of space and one dimension of time. The three different spacial dimensions make communicating information about space ambiguous.
Information about the three dimensions of space needs a direction in addition to the magnitude. The pairing of direction and magnitude is called a vector.
If I say "bike 10 km to get to the store", you still can't find the store because you don't know the direction. I should say "bike 10 km west".
Scalars
A scalar is a variable that has a magnitude, but not a direction in space. Temperature is a scalar. You couldn't say it is 50°C to the left.
Variable | Magnitude |
---|---|
air pressure | 101.3 kPa |
temperature | 21° C |
price | $50 |
speed | 10 m/s |
distance | 3000 m |
Speed and distance seem like they could have a direction, but they are defined as only the magnitude of velocity and displacement vectors.
Vectors →
A vector is a variable that has a magnitude and direction in space. Velocity is a vector. You could say a velocity is 10 m/s in the west direction.
Variable | Magnitude | Direction |
---|---|---|
displacement | 10 m | west |
displacement | 5.5 m | up |
velocity | 20 m/s | 20° above the x-axis |
velocity | 3 m/s | 10 left |
acceleration | 9.8 m/s² | ↓ |
Think of a vector like the hypotenuse of a right triangle. This makes the other two sides of the triangle the components of the vector. Often the components are horizontal and vertical, but they don't have to be.
We can use the pythagorean theorem to solve for the magnitude of the sides.
$$a^{2} + b^{2} = c^{2}$$ Example: You walk 3 miles north and then 4 miles west. How far away are you from your starting location?solution
$$a^{2}+b^{2} = c^{2}$$ $$3^{2}+4^{2} = c^{2}$$ $$25 = c^{2}$$ $$5 \, \mathrm{miles} = c$$Vectors and Angles
We can use trig functions (SOH-CAH-TOA) to find how the vector components relate to the angle of the vector.
For a velocity vector the hypotenuse is v. The adjacent and opposite sides of the triangle are the x and y part of v.
For displacement the equations are the same
solution
$$ v_{y} = (v) \sin \theta $$ $$ v_{y} = (250 \, \mathrm{\tfrac{m}{s}}) \sin(14 \degree) $$ $$ v_{y} = 61 \mathrm{\tfrac{m}{s}} $$solution
Northeast is an angle of 45 degrees to the north.
$$ d_\mathrm{north} = d \cos \theta $$ $$ d_\mathrm{north} = (75 \, \mathrm{m}) \cos(45 \degree) $$ $$ d_\mathrm{north} = 53 \, \mathrm{m} $$