Vectors and Components
A free Trigonometry lesson from the “Applications of Trigonometry” unit, with a worked example and practice problems including step-by-step solutions.
A vector can be written as components (x, y), where x = r*cos(theta) and y = r*sin(theta), with r the magnitude and theta the direction angle measured from the positive x-axis. Going the other way, the magnitude is sqrt(x^2 + y^2) and the direction angle comes from tan(theta) = y/x. To add two vectors, add their x-components and their y-components separately to get the resultant.
What you'll learn
- Find a vector's x and y components from its magnitude and direction
- Compute magnitude and direction angle from components
- Add two vectors and describe the resultant
Worked example
Problem. A vector has magnitude 10 and direction angle 30 degrees. Find its x and y components (exact values).
- Use x = r*cos(theta) = 10*cos(30) = 10*(sqrt(3)/2).
- Simplify: x = 5*sqrt(3).
- Use y = r*sin(theta) = 10*sin(30) = 10*(1/2) = 5.
Answer: x = 5*sqrt(3), y = 5
Practice problems
1. Find the magnitude of the vector (3, 4).
Show solution
- Warm-up: First identify exactly what the question is asking: Find the magnitude of the vector (3, 4).
- Choose the operation or relationship that matches the wording, then carry it out one clear step at a time.
- Use magnitude = sqrt(x^2 + y^2).
- Compute sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25).
- The magnitude is 5.
- Check the result by substituting or estimating: the response should match 5 and make sense in the original problem.
Answer: 5
2. A vector has magnitude 8 and direction angle 0 degrees. Find its x-component.
Show solution
- Warm-up: First identify exactly what the question is asking: A vector has magnitude 8 and direction angle 0 degrees. Find its x-component.
- Use the relevant geometric relationship first, then set up an equation from the angle measures or side relationships.
- Use x = r*cos(theta) = 8*cos(0).
- cos(0) = 1.
- So x = 8.
- Check the result by substituting or estimating: the response should match 8 and make sense in the original problem.
Answer: 8
3. A vector has magnitude 6 and direction angle 90 degrees. Find its y-component.
Show solution
- Warm-up: First identify exactly what the question is asking: A vector has magnitude 6 and direction angle 90 degrees. Find its y-component.
- Use the relevant geometric relationship first, then set up an equation from the angle measures or side relationships.
- Use y = r*sin(theta) = 6*sin(90).
- sin(90) = 1.
- So y = 6.
- Check the result by substituting or estimating: the response should match 6 and make sense in the original problem.
Answer: 6
4. A vector has magnitude 12 and direction angle 60 degrees. Find its x and y components (exact values).
Show solution
- Core Practice: First identify exactly what the question is asking: A vector has magnitude 12 and direction angle 60 degrees. Find its x and y components (exact values).
- Use the relevant geometric relationship first, then set up an equation from the angle measures or side relationships.
- x = 12*cos(60) = 12*(1/2) = 6.
- y = 12*sin(60) = 12*(sqrt(3)/2).
- So y = 6*sqrt(3).
- Check the result by substituting or estimating: the response should match x = 6, y = 6*sqrt(3) and make sense in the original problem.
Answer: x = 6, y = 6*sqrt(3)
5. Find the magnitude of the vector (5, 12).
Show solution
- Core Practice: First identify exactly what the question is asking: Find the magnitude of the vector (5, 12).
- Choose the operation or relationship that matches the wording, then carry it out one clear step at a time.
- Use magnitude = sqrt(x^2 + y^2).
- Compute sqrt(5^2 + 12^2) = sqrt(25 + 144) = sqrt(169).
- The magnitude is 13.
- Check the result by substituting or estimating: the response should match 13 and make sense in the original problem.
Answer: 13
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