The Physics
Opus in profectus

Vector Resolution & Components

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practice problem 1

A laser beam is aimed 15.95° above the horizontal at a mirror 11,648 m away. It glances off the mirror and continues for an additional 8570. m at 11.44° above the horizon until it hits its target. What is the resultant displacement of the beam to the target?


Vectors that are not at nice angles need to be dealt with. Break them up into their components.

x1 =  r1 cos θ1
x1 =  (11,648 m)cos(15.95°)
x1 =  11,200 m
y1 =  r1 sin θ1
y1 =  (11,648 m)sin(15.95°)
y1 =  3,200 m
x2 =  r2 cos θ2
x2 =  (8,570 m)cos(11.44°)
x2 =  8,400 m
y2 =  r2 sin θ2
y2 =  (8,570 m)sin(11.44°)
y2 =  1,700 m

Add vectors in the same direction with "ordinary" addition.

x =  11,200 m + 8,400 m
x =  19,600 m
y =  3,200 m + 1,700 m
y =  4,900 m

Add vectors at right angles with a combination of pythagorean theorem for magnitude…

r =  √(x2 + y2)
r =  √[(19,600 m)2 + (4,900 m)2]
r =  20,200 m

and tangent for direction.

tan θ =  y  =  4,900 m  
x 19,600 m  
θ =  14.04°  

Don't forget to answer the question.

The target of the laser beam is 20,200 m away at an angle of elevation of 14.04°.

Oh yeah, and don't forget to make a drawing. I probably should have told you to do that earlier. I feel bad that I've done that twice in the sections on vectors. I should set a better example. Let me make it up to you by giving you an animated drawing.

If you have a feeling of déjà vu, do not be alarmed. The Matrix is fine. I recycled the solution to this problem from an earlier one. The idea was to show a common problem solving method used in physics. Whenever possible, take a difficult problem that you haven't solved and reduce it one that you have solved.

How does one add parallel vectors? Very simple — add them. How does one add antiparallel vectors? Also simple — subtract them. How does one add vectors at right angles. Reasonably simple — use pythagorean theorem and tangent. How does one add vectors that aren't at 0°, 180°, or 90°? Astoundingly simple — Make them! Make them point in a direction that's convenient for you. Break them up into components.

And then the students learned that there really was no such thing as a "bad" vector and everyone lived happily ever after. The End.

practice problem 2

Three forces act on a point: 3 N at 0°, 4 N at 90°, and 5 N at 217°. What is the net force?


Resolve the vectors into their components along the x and y axes. (Watch the signs.) Then add the components along each axis to get the components of the resultant. Use these to get the magnitude and direction of the resultant. Problems with a lot of components are easier to work on when the values are written in table form like this…

  magnitude direction x-component y-component
first force 3 N +3 N 0 N
second force 4 N 90° 0 N +4 N
third force 5 N 217° −4 N −3 N
resultant 1.4 N 135° −1 N +1 N

A drawing or animation may be helpful.

practice problem 3

A cyclist heads due west on a straight road. The wind is blowing from 248° at 10 m/s.
  1. Is this wind more like a headwind or a tailwind?
  2. What is the headwind/tailwind speed?
  3. What is the crosswind speed?



Start with a diagram. You can draw a top view of a cyclist like I did if you'd like, but it isn't necessary. Do draw an arrow pointed to the right, however, to represent the direction of the cyclist. Wind directions are measured clockwise from due north. North is 0°, east is 90°, south is 180°, and west is 270°. The wind is coming from 248°, which lies somewhere between south and west. Draw an arrow from the lower left corner to the upper right corner to represent the wind. The angle between the two arrows is…

270° −  248° = 22°

Add this info to the diagram.

  1. This is why you need a diagram. It makes it easy to see the answer. This wind is more like a headwind than a tailwind.
  2. The headwind is given by the "x" component.
  3. vx = v cos θ = (10 m/s)cos(22°) = 9.2 m/s

  4. The crosswind is given by the "y" component.
  5. vy = v sin θ = (10 m/s)sin(22°) = 3.7 m/s

practice problem 4

One unfortunate winter day I happened to slip on an icy ramp inclined 37° to the horizontal. Find my acceleration down the ramp given that the acceleration due to gravity points straight down and has a value of 9.8 m/s2. (Assume the ice is perfectly frictionless.)


This is an example of an inclined plane problem — something very common in introductory physics classes. Solution…

Start with a diagram. Draw a diagonal line to represent the ramp. Draw a tilted box to represent poor unfortunate me. Draw an arrow pointing down and label it g for acceleration due to gravity.

I can't accelerate down in this problem since the solid surface of the ramp is in the way, but I can accelerate down the ramp; that is, parallel to the ramp. This sets a natural direction for a rotated coordinate system.

x'  parallel to the surface
y'  perpendicular to the surface

Add the rotated coordinate axes to the drawing, then project the acceleration vector onto them. (I've drawn this with dashed lines.) With a little bit of geometric reasoning, it can be shown that the angle between a horizontal line and the parallel axis (also known as the angle of inclination) is equal to the angle between a vertical line and the perpendicular axis. That gives us a right triangle with the following sides…

g  hypotenuse
gx'  opposite side
gy'  adjacent side

which means…

gx'  = g sin θ
gy'  = g cos θ

Adding these details to the diagram puts everything in perspective.

We only care about the component parallel to the ramp, so we'll only do one calculation.

gx' = 9.8 m/s2 sin 37° = 5.9 m/s2