DWG NO. 6.5 — Lesson 5 of 6

Vector Operations and Applications

Unit 6: Additional Topics in Trigonometry · ~25 min

Objective Add, subtract, and scale vectors in component form, and use vector addition to solve applied problems involving forces and velocities.

Lesson 6.4 defined a vector's component form, magnitude, and direction. This lesson builds the arithmetic on top of that: combining vectors is how you find the net effect of two forces, or a plane's actual path once wind is added to its heading.

Addition, subtraction, and scalar multiplication

All three operations work component by component:

\(\langle a_{1}, a_{2} \rangle\) + \(\langle b_{1}, b_{2} \rangle\) = \(\langle a_{1}+b_{1}, a_{2}+b_{2} \rangle\)
\(\langle a_{1}, a_{2} \rangle\) − \(\langle b_{1}, b_{2} \rangle\) = \(\langle a_{1}-b_{1}, a_{2}-b_{2} \rangle\)
k\(\langle v_{1}, v_{2} \rangle\) = \(\langle kv_{1}, kv_{2} \rangle\)

Geometrically, adding two vectors means placing them tip-to-tail: draw the first, then start the second where the first ends. The sum — the resultant — runs from the very start to the very end. A negative scalar reverses a vector's direction; scaling by |k| stretches or shrinks its magnitude without changing the line it lies on.

Unit vectors i and j

The vectors i = \(\langle 1, 0 \rangle\) and j = \(\langle 0, 1 \rangle\) are the standard basis vectors, so any vector can be written v = \(\langle v_{1}, v_{2} \rangle\) = \(v_{1}\)i + \(v_{2}\)j — a form you'll see often in physics.

Tip-to-tail addition: place v's tail at u's tip, and the resultant u + v runs start to end

Worked Example 1 · Vector arithmetic

Problemu = \(\langle 4, 1 \rangle\), v = \(\langle -2, 5 \rangle\). Find u+v, u−v, and 3u.

1u+v = \(\langle 4+(-2), 1+5 \rangle\) = \(\langle 2, 6 \rangle\).

2u−v = \(\langle 4-(-2), 1-5 \rangle\) = \(\langle 6, -4 \rangle\).

33u = \(\langle 3(4), 3(1) \rangle\) = \(\langle 12, 3 \rangle\).

u+v = \(\langle 2, 6 \rangle\), u−v = \(\langle 6, -4 \rangle\), 3u = \(\langle 12, 3 \rangle\)

Worked Example 2 · Airplane ground velocity

ProblemA plane flies with airspeed 300 mph due east: velocity \(\langle 300, 0 \rangle\). Wind blows 40 mph at 120° from east. Find the actual ground velocity.

1Wind vector: \(\langle 40cos120^\circ, 40sin120^\circ \rangle\) = \(\langle -20, 34.64 \rangle\).

2Resultant = \(\langle 300-20, 0+34.64 \rangle\) = \(\langle 280, 34.64 \rangle\).

3Ground speed = \(\sqrt{280^{2} + 34.64^{2}}\) = \(\sqrt{79600}\) ≈ 282.1 mph.

4Direction above due east: arctan(\(\frac{34.64}{280}\)) ≈ 7.1°.

Ground velocity ≈ 282.1 mph at about 7.1° north of east

Guided practice

1.u = \(\langle 3, -2 \rangle\), v = \(\langle 1, 5 \rangle\). Find u+v.
Show answer

\(\langle 4, 3 \rangle\).
2.u = \(\langle 6, 4 \rangle\), v = \(\langle 2, -1 \rangle\). Find u − 2v.
Show answer

2v = \(\langle 4, -2 \rangle\). u − 2v = \(\langle 6-4, 4-(-2) \rangle\) = \(\langle 2, 6 \rangle\).
3.Two perpendicular forces \(\langle 8, 0 \rangle\) and \(\langle 0, 6 \rangle\) act on an object. Find the magnitude of the resultant force.
Show answer

Resultant = \(\langle 8, 6 \rangle\). Magnitude = \(\sqrt{64+36}\) = \(\sqrt{100}\) = 10.
4.A boat's engine gives velocity \(\langle 12, 0 \rangle\) (12 mph east) while a current pushes \(\langle 0, -5 \rangle\) (5 mph south). Find the resultant velocity and its magnitude.
Show answer

Resultant = \(\langle 12, -5 \rangle\). Magnitude = \(\sqrt{144+25}\) = \(\sqrt{169}\) = 13 mph.
5.Explain the effect of multiplying a vector by a negative scalar −k (k > 0) on its direction and magnitude.
Show answer

Multiplying by −k reverses the vector's direction (points it exactly opposite, i.e. adds 180° to its direction angle) while scaling its magnitude by a factor of k — the negative sign flips direction, the |k| controls length.

← 6.4 Introduction to Vectors Unit Overview 6.6 Polar Coordinates and Polar Graphs →