Unit 6: Additional Topics in Trigonometry · ~20 min
Objective
Represent vectors in component form, and compute their magnitude and direction angle.
Every number you've graphed so far — a function value, a coordinate — is a scalar: a single quantity with size but no direction. Some real quantities need more: wind blowing 20 mph isn't fully described until you also say which way it's blowing. A vector carries both a magnitude and a direction, and this lesson develops the algebra for working with them.
Component form
Placed in standard position with its tail at the origin, a vector's head lands at some point (\(v_{1}\), \(v_{2}\)), and that pair is its component form, written v = \(\langle v_{1}, v_{2} \rangle\). For a vector drawn from an arbitrary point P(\(x_{1}\), \(y_{1}\)) to Q(\(x_{2}\), \(y_{2}\)), the component form is found by subtracting tail from head:
The magnitude of v = \(\langle v_{1}, v_{2} \rangle\) is its length, found the same way as any distance:
|v| = \(\sqrt{v_{1}^{2} + v_{2}^{2}}\)
The direction angle θ is the angle the vector makes with the positive x-axis — the same setup as Unit 4's unit circle. Find a reference angle from tanθ = \(\frac{v_{2}}{v_{1}}\), then adjust for the quadrant of (\(v_{1}\), \(v_{2}\)), exactly as in Lesson 4.3.
Going the other direction — from magnitude and angle to components — reuses the unit-circle definitions directly:
\(v_{1}\) = |v|cosθ and \(v_{2}\) = |v|sinθ
A vector in standard position: the head sits at (v₁, v₂), with magnitude \(\sqrt{v_{1}^{2}+v_{2}^{2}}\) and direction angle θ
Worked Example 1 · Vector from two points
ProblemFind the component form and magnitude of the vector from P(2, 3) to Q(7, −1).