Vectors, Polar Coordinates, and Complex Numbers: practice solutions, Grade 12 – download the PDF

1 Magnitude of a vector ★★★

Use \(\|\langle a,b\rangle\|=\sqrt{a^2+b^2}\).

1. \(\sqrt{25+144}=\sqrt{169}=13\).
2. \(\sqrt{64+36}=\sqrt{100}=10\).
3. \(\sqrt{49+576}=\sqrt{625}=25\).

2 Direction angles ★★★

1. Magnitude \(\sqrt{1+3}=2\); \(\tan\theta=\sqrt3\) in quadrant I, so \(\theta=60^\circ\).
2. Magnitude \(\sqrt{8}=2\sqrt2\); \(\tan\theta=1\) but the vector is in quadrant III, so \(\theta=225^\circ\).
3. Magnitude \(\sqrt{9+27}=6\); \(\tan\theta=-\sqrt3\) in quadrant IV, so \(\theta=300^\circ\).

3 Combining vectors ★★★

\(\mathbf{u}+\mathbf{v}=\langle 2,5\rangle\). \(\mathbf{u}-\mathbf{v}=\langle 6,-7\rangle\). \(3\mathbf{u}-2\mathbf{v}=\langle 12,-3\rangle-\langle -4,12\rangle=\langle 16,-15\rangle\).

4 Computing dot products ★★★

1. \(-6+20=14\); not perpendicular.
2. \(6-6=0\); the vectors are perpendicular.
3. \(-2-2=-4\); not perpendicular (the vectors point in opposite directions).

5 Polar to rectangular ★★★

1. \(x=8\cos60^\circ=4\), \(y=8\sin60^\circ=4\sqrt3\): \((4,\,4\sqrt3)\).
2. \(x=5\cos\pi=-5\), \(y=5\sin\pi=0\): \((-5,\,0)\).
3. \(x=6\cos270^\circ=0\), \(y=6\sin270^\circ=-6\): \((0,\,-6)\).

6 Rectangular to polar ★★★

1. On the positive \(y\)-axis: \(\left(9,\dfrac{\pi}{2}\right)\).
2. On the negative \(x\)-axis: \((4,\pi)\).
3. \(r=\sqrt{18}=3\sqrt2\), \(\tan\theta=1\) in quadrant I: \(\left(3\sqrt2,\dfrac{\pi}{4}\right)\).

7 Polar form of complex numbers ★★★

1. \(r=\sqrt8=2\sqrt2\), the point is on the line \(y=x\) in quadrant I: \(2\sqrt2\operatorname{cis}45^\circ\).
2. On the negative real axis: \(6\operatorname{cis}180^\circ\).
3. On the negative imaginary axis: \(3\operatorname{cis}270^\circ\).

8 True or false? ★★★

1. False. With \(\mathbf{u}=\langle1,0\rangle\) and \(\mathbf{v}=\langle0,1\rangle\): \(\|\mathbf{u}+\mathbf{v}\|=\sqrt2\) but \(1+1=2\).
2. False. \(\langle1,0\rangle\cdot\langle0,1\rangle=0\) and neither vector is zero; they are perpendicular.
3. False. \((-3,\theta)=(3,\theta+\pi)\) is the point on the opposite side of the pole.

9 Kayak in a current ★★★

1. Paddling \(\langle0,5\rangle\), current \(\langle3,0\rangle\), resultant \(\langle3,5\rangle\).
2. Speed \(=\sqrt{9+25}=\sqrt{34}\approx5.83\) mph. In metric, \(5.83\times1.609\approx9.38\) km/h.
3. The angle from north satisfies \(\tan\alpha=\dfrac{3}{5}\), so \(\alpha\approx31.0^\circ\) east of north (equivalently, a direction angle of \(59.0^\circ\) from the east axis).

The kayaker travels at about 5.83 mph, \(31^\circ\) east of north.

10 Pulling a sled ★★★

Force: \(\mathbf{F}=\langle 50\cos30^\circ,\ 50\sin30^\circ\rangle=\langle 25\sqrt3,\ 25\rangle\). Displacement: \(\mathbf{d}=\langle 20,0\rangle\).

\(W=\mathbf{F}\cdot\mathbf{d}=25\sqrt3\cdot20=500\sqrt3\approx866.0\) ft·lb (about 1,174 joules).

The rope does about 866 foot-pounds of work.

11 Angle between two vectors ★★★

\(\mathbf{u}\cdot\mathbf{v}=2+3=5\), \(\|\mathbf{u}\|=\sqrt5\), \(\|\mathbf{v}\|=\sqrt{10}\).

\(\cos\theta=\dfrac{5}{\sqrt{50}}=\dfrac{5}{5\sqrt2}=\dfrac{\sqrt2}{2}\), so \(\theta=45^\circ\).

12 Polar equation to rectangular ★★★

1. Multiply by \(r\): \(r^2=6r\cos\theta\), so \(x^2+y^2=6x\). Completing the square: \((x-3)^2+y^2=9\), a circle with center \((3,0)\) and radius 3.
2. Since \(y=r\sin\theta\), the equation is \(y=4\): a horizontal line.

13 Rectangular equation to polar ★★★

1. \(r^2=25\), so \(r=5\).
2. \(\tan\theta=\dfrac{y}{x}=\sqrt3\), so \(\theta=\dfrac{\pi}{3}\) (a line through the pole).
3. \(r\sin\theta=-3\), so \(r=-\dfrac{3}{\sin\theta}\).

14 Products and quotients ★★★

\(z_1z_2=4\cdot2\operatorname{cis}(50^\circ+20^\circ)=8\operatorname{cis}70^\circ\).

\(\dfrac{z_1}{z_2}=\dfrac42\operatorname{cis}(50^\circ-20^\circ)=2\operatorname{cis}30^\circ=2\cos30^\circ+2i\sin30^\circ=\sqrt3+i\).

15 Powers with De Moivre ★★★

1. \(\cos180^\circ+i\sin180^\circ=-1\).
2. \(2^4\operatorname{cis}60^\circ=16\left(\dfrac12+i\dfrac{\sqrt3}{2}\right)=8+8\sqrt3\,i\).

16 Cardioid and roses ★★★

(a) \(r=4,\ 3,\ 2,\ 1,\ 0\). Replacing \(\theta\) by \(-\theta\) does not change \(\cos\theta\), so the graph is symmetric about the polar axis; its farthest point is \(r=4\) at \(\theta=0\).

(b) \(n=3\) is odd, so there are 3 petals, each of length 5 (tips at \(\theta=0,\ 2\pi/3,\ 4\pi/3\)). The curve passes through the pole when \(\cos3\theta=0\), for example at \(\theta=\pi/6,\ \pi/2,\ 5\pi/6\).

(c) \(n=2\) is even, so there are \(2n=4\) petals, each of length 4.

17 Where two curves meet ★★★

Set \(6\cos\theta=3\): \(\cos\theta=\dfrac12\), so \(\theta=\pm\dfrac\pi3\). Both points have \(r=3\).

Polar: \(\left(3,\dfrac\pi3\right)\) and \(\left(3,-\dfrac\pi3\right)\). Rectangular: \(\left(\dfrac32,\dfrac{3\sqrt3}{2}\right)\) and \(\left(\dfrac32,-\dfrac{3\sqrt3}{2}\right)\).

18 Projection of a vector ★★★

\(\mathbf{u}\cdot\mathbf{v}=24-6=18\) and \(\|\mathbf{v}\|=5\). The scalar component is \(\dfrac{18}{5}=3.6\).

\(\operatorname{proj}_{\mathbf{v}}\mathbf{u}=\dfrac{18}{25}\langle4,-3\rangle=\langle2.88,\,-2.16\rangle\).

\(\mathbf{u}-\operatorname{proj}=\langle3.12,\,4.16\rangle\), and \(3.12\cdot4+4.16\cdot(-3)=12.48-12.48=0\), so it is perpendicular to \(\mathbf{v}\).

19 A triangle from vectors ★★★

\(\overrightarrow{BA}=\langle-3,-4\rangle\) and \(\overrightarrow{BC}=\langle4,-3\rangle\). Their dot product is \(-12+12=0\), so the angle at \(B\) is a right angle.

\(\|\overrightarrow{BA}\|=\sqrt{9+16}=5\) and \(\|\overrightarrow{BC}\|=\sqrt{16+9}=5\), so \(BA=BC\): the triangle is isosceles.

Area \(=\dfrac12\cdot5\cdot5=12.5\) square units.

20 A tenth power ★★★

\(-1+i=\sqrt2\operatorname{cis}135^\circ\). Then \((-1+i)^{10}=(\sqrt2)^{10}\operatorname{cis}1350^\circ=32\operatorname{cis}270^\circ=-32i\) (since \(1350^\circ-1080^\circ=270^\circ\)).

Check: \((-1+i)^2=1-2i-1=-2i\), and \((-2i)^5=-32\,i^5=-32i\). Both methods agree.

21 Sixth roots of 64 ★★★

\(64=64\operatorname{cis}0^\circ\), so the modulus is \(\sqrt[6]{64}=2\) and the arguments are \(0^\circ,60^\circ,120^\circ,180^\circ,240^\circ,300^\circ\).

\(z_0=2\), \(z_1=1+i\sqrt3\), \(z_2=-1+i\sqrt3\), \(z_3=-2\), \(z_4=-1-i\sqrt3\), \(z_5=1-i\sqrt3\): the vertices of a regular hexagon of radius 2.

The sum is \(0\), because the imaginary parts cancel in pairs and the real parts add up to \(2+1-1-2-1+1=0\).

22 Fourth roots ★★★

\(r=\sqrt{64+192}=16\) and the point is in quadrant II with reference angle \(60^\circ\), so \(\theta=120^\circ\). The roots have modulus \(\sqrt[4]{16}=2\) and arguments \(\dfrac{120^\circ+360^\circ k}{4}=30^\circ+90^\circ k\).

\(z_0=\sqrt3+i\), \(z_1=-1+i\sqrt3\), \(z_2=-\sqrt3-i\), \(z_3=1-i\sqrt3\).

Check: \((\sqrt3+i)^2=2+2\sqrt3\,i\) and \((2+2\sqrt3\,i)^2=4-12+8\sqrt3\,i=-8+8\sqrt3\,i\).

23 Triple-angle formulas ★★★

By the binomial theorem, with \(c=\cos\theta\) and \(s=\sin\theta\): \((c+is)^3=c^3+3c^2(is)+3c(is)^2+(is)^3=(c^3-3cs^2)+i(3c^2s-s^3)\).

De Moivre gives \(\cos3\theta+i\sin3\theta\). Matching real parts: \(\cos3\theta=c^3-3c(1-c^2)=4c^3-3c\).

Matching imaginary parts: \(\sin3\theta=3(1-s^2)s-s^3=3\sin\theta-4\sin^3\theta\).