Vectors in the Plane Worksheet Answer Key – Precalculus

Problem 1 Component Form from Two Points

Subtract the initial point from the terminal point: ⟨4 − (−3), 1 − 5⟩ = ⟨7, −4⟩.

Magnitude = √(7² + (−4)²) = √65 ≈ 8.06.

Problem 2 Magnitude and Direction Angle

(a) |v| = √(49 + 49) = 7√2 ≈ 9.90. The point is in Quadrant II with reference angle 45°, so θ = 135°.

(b) |v| = √97 ≈ 9.85. In Quadrant IV, θ = 360° − arctan(4/9) ≈ 336.04°.

Problem 3 Components from Magnitude and Angle

(a) ⟨18 cos 40°, 18 sin 40°⟩ ≈ ⟨13.79, 11.57⟩.

(b) ⟨25 cos 215°, 25 sin 215°⟩ ≈ ⟨−20.48, −14.34⟩. Both components are negative in Quadrant III.

Problem 4 Add and Subtract from a Figure

u = ⟨4, 1⟩ and v = ⟨1, 3⟩.

u + v = ⟨4 + 1, 1 + 3⟩ = ⟨5, 4⟩. u − v = ⟨4 − 1, 1 − 3⟩ = ⟨3, −2⟩.

Problem 5 Scalar Multiples

5v = ⟨−15, 40⟩. −¼v = ⟨3/4, −2⟩.

|v| = √73, so |5v| = 5√73 ≈ 42.72.

Problem 6 Find a Unit Vector

|v| = √(4 + 25) = √29 ≈ 5.385.

Unit vector = ⟨2/√29, −5/√29⟩ ≈ ⟨0.371, −0.928⟩. Its length is 1.

Problem 7 Linear Combination of Vectors

2u = ⟨−2, 8⟩ and 3v = ⟨9, −6⟩.

2u − 3v = ⟨−2 − 9, 8 + 6⟩ = ⟨−11, 14⟩. Magnitude = √(121 + 196) = √317 ≈ 17.80.

Problem 8 Angle Between Two Vectors

u · v = 6(−2) + 1(5) = −7.

|u| = √37 ≈ 6.083 and |v| = √29 ≈ 5.385. cos θ = −7/(6.083 × 5.385) ≈ −0.2137, so θ ≈ 102.3° (obtuse because the dot product is negative).

Problem 9 Perpendicular Vectors

(a) Set the dot product to 0: 9k − 24 = 0, so k = 8/3.

(b) 2(−7) + 7(2) = −14 + 14 = 0, so yes, they are perpendicular.

Problem 10 Vectors Between Three Points

PQ = ⟨5, 5⟩, PR = ⟨−5, 3⟩, QR = ⟨−10, −2⟩.

PR − PQ = ⟨−5 − 5, 3 − 5⟩ = ⟨−10, −2⟩ = QR ✓.

Side lengths: PQ = 5√2 ≈ 7.07, PR = √34 ≈ 5.83, QR = √104 ≈ 10.20.

Problem 11 Two Forces on a Crate

F₁ = ⟨80 cos 20°, 80 sin 20°⟩ ≈ ⟨75.18, 27.36⟩ and F₂ = ⟨120 cos 110°, 120 sin 110°⟩ ≈ ⟨−41.04, 112.76⟩.

Resultant ≈ ⟨34.13, 140.12⟩. Magnitude ≈ 144.2 lb (check: the forces are 90° apart, so √(80² + 120²) = 144.2). Direction ≈ 76.3°.

Problem 12 Plane and Crosswind

With east as +x and north as +y, the plane is ⟨0, 310⟩ and the wind blows toward the east, ⟨45, 0⟩. Sum: ⟨45, 310⟩.

Ground speed = √(45² + 310²) ≈ 313.2 mph. Angle east of north = arctan(45/310) ≈ 8.3°.

Problem 13 Boat Crossing a River

Velocity = ⟨3, 7⟩. Speed = √(7² + 3²) = √58 ≈ 7.62 mph. Angle = arctan(3/7) ≈ 23.2° downstream.

Crossing time = 0.8 ÷ 7 h = 6.9 minutes (the current does not change it). Drift = 3 × (0.8/7) ≈ 0.343 mile.

Problem 14 Force for Equilibrium

The sum of the first two is ⟨12, 28⟩. Equilibrium needs a total of ⟨0, 0⟩, so the third force is ⟨−12, −28⟩.

Magnitude = √(144 + 784) = √928 ≈ 30.5 lb.

Problem 15 Diagonals of a Parallelogram

u + v = ⟨3, 5⟩ with magnitude √34 ≈ 5.83. u − v = ⟨7, −3⟩ with magnitude √58 ≈ 7.62.

Check: 34 + 58 = 92, and 2(26 + 20) = 92 ✓.

Problem 16 Work Done Pulling a Wagon

Force ⟨45 cos 30°, 45 sin 30°⟩ ≈ ⟨38.97, 22.5⟩. Displacement ⟨60, 0⟩.

Work = 38.97(60) + 22.5(0) ≈ 2,338 ft·lb.

Problem 17 Glider in a Wind

A bearing of 060° is a standard angle of 30°, so the glider is ⟨85 cos 30°, 85 sin 30°⟩ ≈ ⟨73.61, 42.50⟩. A wind from the north blows toward the south: ⟨0, −20⟩.

Ground velocity ≈ ⟨73.61, 22.5⟩, ground speed ≈ 77.0 mph.

Standard angle ≈ 17.0°, so the bearing is 90° − 17.0° ≈ 73.0°.

Problem 18 Tension in Two Ropes

Horizontal balance: T₁ cos 40° = T₂ cos 55°. Vertical: T₁ sin 40° + T₂ sin 55° = 150.

Substituting T₂ = T₁ cos 40° ÷ cos 55°: T₁(sin 40° + cos 40° tan 55°) = 150, so T₁ ≈ 86.4 lb and T₂ ≈ 115.3 lb.

Problem 19 Angle Between Two Trails

Dot product: 8(2) + 3(7) = 37. Magnitudes: √73 ≈ 8.544 and √53 ≈ 7.280.

cos θ = 37 ÷ (8.544 × 7.280) ≈ 0.5948, so θ ≈ 53.5°.

Problem 20 Drone Flight Path

AB = ⟨9, 7⟩, BC = ⟨6, −10⟩, and AC = AB + BC = ⟨15, −3⟩.

BA = ⟨−9, −7⟩. BA · BC = −54 + 70 = 16, with |BA| = √130 and |BC| = √136. cos θ = 16 ÷ 132.96 ≈ 0.1203, so the angle is ≈ 83.1°.

Total distance = √130 + √136 ≈ 23.06 units.