Vectors in the Plane: Precalculus Math Worksheet Answer Key – PDF Download

Precalculus Vectors Worksheet Answer Key: problems 1–5

Problem 1: Component form from two points

1. Subtract coordinates: \(\langle 7-2,\ 13-1\rangle=\langle 5, 12\rangle\). Then \(\sqrt{25+144}=\sqrt{169}=13\). \(\langle 5, 12\rangle\), magnitude 13.
2. \(\langle 1-(-3),\ -2-4\rangle=\langle 4, -6\rangle\). Then \(\sqrt{16+36}=\sqrt{52}=2\sqrt{13}\approx 7.21\). \(\langle 4, -6\rangle\), magnitude \(2\sqrt{13}\approx 7.21\).
3. \(\langle 0-5,\ 0-5\rangle=\langle -5, -5\rangle\). Then \(\sqrt{25+25}=5\sqrt{2}\approx 7.07\). \(\langle -5, -5\rangle\), magnitude \(5\sqrt{2}\approx 7.07\).

Problem 2: Magnitude of a vector

1. \(\sqrt{36+64}=\sqrt{100}=\)\(10\)
2. \(\sqrt{25+144}=\sqrt{169}=\)\(13\)
3. \(\sqrt{81+1600}=\sqrt{1681}=\)\(41\)
4. \(\sqrt{4+9}=\)\(\sqrt{13}\approx 3.61\)

Problem 3: Direction angles

1. \(\tan\theta=\frac{3}{3}=1\), quadrant I: \(45^{\circ}\).
2. Reference angle \(\tan^{-1}(1)=45^{\circ}\), quadrant II: \(180^{\circ}-45^{\circ}=\)\(135^{\circ}\).
3. Reference angle \(45^{\circ}\), quadrant III: \(180^{\circ}+45^{\circ}=\)\(225^{\circ}\).
4. \(\tan\theta=-\sqrt{3}\), reference angle \(60^{\circ}\), quadrant IV: \(360^{\circ}-60^{\circ}=\)\(300^{\circ}\).

Problem 4: Components from magnitude and angle

1. \(\langle 10\cos 30^{\circ},\ 10\sin 30^{\circ}\rangle=\)\(\langle 5\sqrt{3},\ 5\rangle\) \(\approx\langle 8.66, 5\rangle\).
2. \(\langle 8\cos 120^{\circ},\ 8\sin 120^{\circ}\rangle=\)\(\langle -4,\ 4\sqrt{3}\rangle\) \(\approx\langle -4, 6.93\rangle\).
3. \(\langle 12\cos 270^{\circ},\ 12\sin 270^{\circ}\rangle=\)\(\langle 0, -12\rangle\).
4. \(\langle 6\sqrt{2}\cdot\frac{\sqrt{2}}{2},\ 6\sqrt{2}\cdot\frac{\sqrt{2}}{2}\rangle=\)\(\langle 6, 6\rangle\).

Problem 5: Add and subtract vectors

1. \(\langle 3+(-1),\ -2+5\rangle=\)\(\langle 2, 3\rangle\)
2. \(\langle 3-(-1),\ -2-5\rangle=\)\(\langle 4, -7\rangle\)
3. \(\langle -1-3,\ 5-(-2)\rangle=\)\(\langle -4, 7\rangle\). Notice \(\vec{v}-\vec{u}\) is the opposite of \(\vec{u}-\vec{v}\).

Precalculus Vectors Worksheet Answer Key: problems 6–10

Problem 6: Scalar multiplication

1. \(\langle 12, -8\rangle\)
2. \(\langle 3, -15\rangle\)
3. \(\langle 3, -5\rangle\)
4. \(\langle -6, 4\rangle\)

Problem 7: Linear combinations

1. \(\langle 6, -4\rangle-\langle -3, 15\rangle=\)\(\langle 9, -19\rangle\)
2. \(\langle 9, -6\rangle+\langle -2, 10\rangle=\)\(\langle 7, 4\rangle\)
3. \(\sqrt{81+361}=\sqrt{442}\approx\) \(21.02\)

Problem 8: Unit vectors

1. \(|\vec{v}|=10\), so \(\langle \frac{4}{5}, -\frac{3}{5}\rangle\).
2. \(|\vec{v}|=13\), so \(\langle \frac{5}{13}, \frac{12}{13}\rangle\).
3. \(|\vec{v}|=\sqrt{49+576}=25\), so \(\langle -\frac{7}{25}, \frac{24}{25}\rangle\).
4. \(|\vec{v}|=9\), so \(\langle 0, -1\rangle\).

Problem 9: Unit vector notation i and j

1. \(4\vec{i}-3\vec{j}\)
2. \(\langle 7, 2\rangle\), magnitude \(\sqrt{49+4}=\sqrt{53}\approx 7.28\).
3. \(\langle -1, 6\rangle\)
4. \(6\vec{i}-3\vec{j}+\vec{i}+4\vec{j}=\)\(7\vec{i}+\vec{j}\)

Problem 10: Vectors with a given length

1. \(|\langle 3, 4\rangle|=5\), so multiply the unit vector by 20: \(20\langle \frac{3}{5}, \frac{4}{5}\rangle=\)\(\langle 12, 16\rangle\).
2. Reverse the sign: \(\langle -12, -16\rangle\).
3. \(|\langle -5, 12\rangle|=13\), so scale by \(\frac{26}{13}=2\): \(\langle -10, 24\rangle\). Check: \(\sqrt{100+576}=26\).

Precalculus Vectors Worksheet Answer Key: problems 11–15

Problem 11: Parallelogram rule

Slide each arrow to the tip of the other. The fourth corner is \((5, 4)\), so \(\vec{u}+\vec{v}=\langle 4+1,\ 1+3\rangle=\)\(\langle 5, 4\rangle\).

\(\vec{u}-\vec{v}=\langle 4-1,\ 1-3\rangle=\)\(\langle 3, -2\rangle\).

Length of the diagonal: \(\sqrt{25+16}=\sqrt{41}\approx\) \(6.40\).

Problem 12: Tip-to-tail path

\(\vec{a}+\vec{b}+\vec{c}=\langle 2+4-5,\ 3-1+2\rangle=\)\(\langle 1, 4\rangle\), which matches the end point \((1, 4)\) in the figure.

Magnitude \(\sqrt{1+16}=\sqrt{17}\approx 4.12\) blocks. Direction \(\tan^{-1}(4)\approx\) \(76.0^{\circ}\).

Problem 13: Dot products

1. \(8-3=\)\(5\)
2. \(12-12=\)\(0\)
3. \(-2+12=\)\(10\)
4. \(-6+20=\)\(14\)

Only pair (b) has a dot product of 0, so pair (b) is perpendicular.

Problem 14: Angle between two vectors

1. \(\cos\theta=\frac{1}{1\cdot\sqrt{2}}\), so \(\theta=45.0^{\circ}\).
2. \(\cos\theta=\frac{6}{\sqrt{10}\sqrt{10}}=0.6\), so \(\theta\approx 53.1^{\circ}\).
3. \(\cos\theta=\frac{-10}{\sqrt{5}\sqrt{20}}=\frac{-10}{10}=-1\), so \(\theta=180.0^{\circ}\) (opposite directions).
4. \(\vec{u}\cdot\vec{v}=-3+2=-1\), \(\cos\theta=\frac{-1}{\sqrt{5}\sqrt{10}}\approx -0.1414\), so \(\theta\approx 98.1^{\circ}\).

Problem 15: Perpendicular and parallel vectors

1. \(4k-18=0\), so \(k=\frac{9}{2}\).
2. Parallel means a scalar multiple. Since \(6=2\cdot 3\), also \(k=2\cdot 2=\)\(4\).
3. Swap the components and change one sign: \(\langle 2, 5\rangle\). Check: \(10-10=0\) and \(\sqrt{4+25}=\sqrt{29}\). \(\langle 2, 5\rangle\) (or \(\langle -2, -5\rangle\)).

Precalculus Vectors Worksheet Answer Key: problems 16–20

Problem 16: Angles of a triangle

At A: \(\vec{AB}=\langle 4, 0\rangle\), \(\vec{AC}=\langle 1, 3\rangle\), \(\cos A=\frac{4}{4\sqrt{10}}\approx 0.3162\), so \(A\approx 71.57^{\circ}\).

At B: \(\vec{BA}=\langle -4, 0\rangle\), \(\vec{BC}=\langle -3, 3\rangle\), \(\cos B=\frac{12}{4\cdot 3\sqrt{2}}\approx 0.7071\), so \(B=45.00^{\circ}\).

At C: \(\vec{CA}=\langle -1, -3\rangle\), \(\vec{CB}=\langle 3, -3\rangle\), \(\cos C=\frac{6}{\sqrt{10}\cdot 3\sqrt{2}}\approx 0.4472\), so \(C\approx 63.43^{\circ}\).

Sum: \(71.57+45.00+63.43=180.00\). The angles check out.

Problem 17: Hiking displacement

\(\langle 3, 0\rangle+\langle 0, 4\rangle+\langle -6, 0\rangle=\)\(\langle -3, 4\rangle\).

Distance: \(\sqrt{9+16}=\)5 miles.

Direction: reference angle \(\tan^{-1}(\frac{4}{3})\approx 53.13^{\circ}\) in quadrant II, so \(180^{\circ}-53.13^{\circ}=\)\(126.87^{\circ}\) (about \(36.9^{\circ}\) west of north).

Problem 18: Boat crossing a river

Ground velocity \(\langle 5, 12\rangle\), speed \(\sqrt{25+144}=\)13 mph.

Time to cross: only the across component counts, \(\frac{0.6}{12}=0.05\) hour, which is 3 minutes.

Drift: \(5\times 0.05=0.25\) mile, which is \(0.25\times 5280=\)1,320 feet downstream.

Problem 19: Airplane and crosswind

Ground velocity (east, north): \(\langle 50, 300\rangle\). Speed \(\sqrt{2500+90000}=\sqrt{92500}\approx\) 304.1 mph.

Angle east of north: \(\tan^{-1}(\frac{50}{300})\approx\) \(9.5^{\circ}\).

Distance in 2 hours: \(2\times 304.14\approx\) 608.3 miles.

Problem 20: Forces in equilibrium

Resultant: \(\langle 30, 0\rangle+\langle 0, 40\rangle=\langle 30, 40\rangle\), magnitude \(\sqrt{900+1600}=\)50 pounds, direction \(\tan^{-1}(\frac{4}{3})\approx\) \(53.1^{\circ}\).

The balancing force is the opposite vector: \(\langle -30, -40\rangle\), which is 50 pounds at \(180^{\circ}+53.1^{\circ}=233.1^{\circ}\).

Precalculus Vectors Worksheet Answer Key: problems 21–22

Problem 21: Work done by a force

Force \(\vec{F}=\langle 40\cos 60^{\circ},\ 40\sin 60^{\circ}\rangle=\langle 20,\ 20\sqrt{3}\rangle\) pounds; displacement \(\vec{d}=\langle 25, 0\rangle\) feet.

Work \(=\vec{F}\cdot\vec{d}=20\cdot 25+20\sqrt{3}\cdot 0=\)500 foot-pounds.

Problem 22: Crate on a ramp

\(\vec{W}\cdot\vec{u}=0\cdot\cos 30^{\circ}+(-100)(-\sin 30^{\circ})=100\cdot\frac{1}{2}=\)50 pounds down the ramp.

To hold the crate still, push up the ramp with a force of 50 pounds. The remaining \(100\cos 30^{\circ}\approx 86.6\) pounds presses into the ramp.

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