IB exam questions: topic 5

a)

1. $|\mathbf{a}|=\sqrt{12^2+(-5{)}^2}=13$.
2. By the triangle inequality, $||\mathbf{b}|-|\mathbf{a}||\le |\mathbf{a}+\mathbf{b}|\le |\mathbf{a}|+|\mathbf{b}|$, that is, $15-13\le |\mathbf{a}+\mathbf{b}|\le 13+15$.
3. Both extremes are attained (vectors parallel and in opposite directions, and vectors parallel and in the same direction).

b)

1. The minimum ($2$) occurs when $\mathbf{b}$ has the opposite direction to $\mathbf{a}$: $\mathbf{b}=-\frac{15}{13}\mathbf{a}$.
2. $\mathbf{p}=\mathbf{a}+\mathbf{b}=\left(1-\frac{15}{13}\right)\mathbf{a}=-\frac{2}{13}\left(\begin{array}{c}12\\ -5\end{array}\right)=\left(\begin{array}{c}-24/13\\ 10/13\end{array}\right)$. Check: $|\mathbf{p}|=\frac{2}{13}\cdot 13=2$.

c)

1. Perpendicular to $\mathbf{a}$: $12x-5y=0$, so $y=\frac{12}{5}x$.
2. Magnitude $15$: $x^2+\frac{144}{25}{x}^2=225$, that is, $\frac{169}{25}{x}^2=225$, so $x^2=\frac{5625}{169}$.
3. As $x>0$: $x=\frac{75}{13}$ and $y=\frac{12}{5}\cdot \frac{75}{13}=\frac{180}{13}$ (positive, as the question requires).

Answer: a) $2\le |\mathbf{a}+\mathbf{b}|\le 28$; b) $\mathbf{p}=\left(\begin{array}{c}-24/13\\ 10/13\end{array}\right)$; c) $\mathbf{q}=\left(\begin{array}{c}75/13\\ 180/13\end{array}\right)$

a) (i) Since $\overrightarrow{OF}=\frac{1}{6}\overrightarrow{FB}$, we have $\overrightarrow{FB}=6\overrightarrow{OF}$, and $\overrightarrow{OB}=\overrightarrow{OF}+\overrightarrow{FB}=7\overrightarrow{OF}$. So $$\overrightarrow{OF}=\frac{1}{7}\mathbf{b}.$$

(ii) $\overrightarrow{AF}=\overrightarrow{AO}+\overrightarrow{OF}=-\mathbf{a}+\frac{1}{7}\mathbf{b}$.

b) (i) $\overrightarrow{OD}=\overrightarrow{OA}+\lambda \overrightarrow{AF}=\mathbf{a}+\lambda \left(-\mathbf{a}+\frac{1}{7}\mathbf{b}\right)$, that is, $$\overrightarrow{OD}=(1-\lambda )\mathbf{a}+\frac{\lambda }{7}\mathbf{b}.$$

(ii) With $\overrightarrow{OC}=\frac{1}{2}\mathbf{a}$ and $\overrightarrow{CB}=\mathbf{b}-\frac{1}{2}\mathbf{a}$: $$\overrightarrow{OD}=\overrightarrow{OC}+\mu \overrightarrow{CB}=\frac{1}{2}\mathbf{a}+\mu \left(\mathbf{b}-\frac{1}{2}\mathbf{a}\right)=\frac{1-\mu }{2}\mathbf{a}+\mu \mathbf{b}.$$

c)

1. As $\mathbf{a}$ and $\mathbf{b}$ are not parallel, we equate coefficients: $$\frac{\lambda }{7}=\mu ,1-\lambda =\frac{1-\mu }{2}.$$
2. From the first, $\lambda =7\mu$. Substituting: $1-7\mu =\frac{1-\mu }{2}$, so $2-14\mu =1-\mu$, that is, $1=13\mu$.
3. Therefore $\mu =\frac{1}{13}$ and $\lambda =7\mu =\frac{7}{13}$.

d) $\overrightarrow{CD}=\mu \overrightarrow{CB}=\frac{1}{13}\left(\mathbf{b}-\frac{1}{2}\mathbf{a}\right)$, that is, $$\overrightarrow{CD}=-\frac{1}{26}\mathbf{a}+\frac{1}{13}\mathbf{b}.$$

e)

1. Area of $△OAB$: $\frac{1}{2}|\mathbf{a}\times \mathbf{b}|$ (using the modulus of the vector (cross) product in the plane, that is, $\frac{1}{2}|\mathbf{a}||\mathbf{b}|\sin \widehat{AOB}$).
2. In $△CAD$: $\overrightarrow{CA}=\frac{1}{2}\mathbf{a}$ and $\overrightarrow{CD}=\frac{1}{13}\left(\mathbf{b}-\frac{1}{2}\mathbf{a}\right)$. The $\mathbf{a}$ term of $\overrightarrow{CD}$ is parallel to $\overrightarrow{CA}$ and contributes no area, so the area is $\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{13}|\mathbf{a}\times \mathbf{b}|=\frac{1}{52}|\mathbf{a}\times \mathbf{b}|$.
3. So $k=\frac{\frac{1}{2}|\mathbf{a}\times \mathbf{b}|}{\frac{1}{52}|\mathbf{a}\times \mathbf{b}|}=26$.

Answer: $k=26$

1. Square the condition: $|\mathbf{a}+\mathbf{b}{|}^2=|\mathbf{a}-\mathbf{b}{|}^2$.
2. Using $|\mathbf{u}{|}^2=\mathbf{u}\cdot \mathbf{u}$: $$|\mathbf{a}{|}^2+2\mathbf{a}\cdot \mathbf{b}+|\mathbf{b}{|}^2=|\mathbf{a}{|}^2-2\mathbf{a}\cdot \mathbf{b}+|\mathbf{b}{|}^2.$$
3. Simplify: $4\mathbf{a}\cdot \mathbf{b}=0$.

Answer: $\mathbf{a}\cdot \mathbf{b}=0$

1. The position vectors are $\overrightarrow{OP}=\left(\begin{array}{c}3\\ -1\end{array}\right)$ and $\overrightarrow{OQ}=\left(\begin{array}{c}\lambda \\ -4-\lambda \end{array}\right)$.
2. Two vectors are perpendicular if their scalar product is zero: $$3\lambda +(-1)(-4-\lambda )=0.$$
3. Expanding: $3\lambda +4+\lambda =0$, that is, $4\lambda =-4$.
4. Solving, $\lambda =-1$ (this is the only value, because the equation is linear).

Answer: $\lambda =-1$

1. Scalar product: $\mathbf{u}\cdot \mathbf{v}=\cos \frac{1}{n}+\sin \frac{1}{n}$.
2. Magnitudes: $|\mathbf{u}|=\sqrt{1+1}=\sqrt{2}$ and $|\mathbf{v}|=\sqrt{{\cos }^2\frac{1}{n}+{\sin }^2\frac{1}{n}}=1$.
3. Angle formula: $\cos \theta =\frac{\mathbf{u}\cdot \mathbf{v}}{|\mathbf{u}||\mathbf{v}|}=\frac{\cos \frac{1}{n}+\sin \frac{1}{n}}{\sqrt{2}}$.

Answer: $\cos \theta =\frac{\cos \frac{1}{n}+\sin \frac{1}{n}}{\sqrt{2}}$

a) 1. To go from $C$ to $B$ we pass through $O$: $\overrightarrow{CB}=\overrightarrow{CO}+\overrightarrow{OB}=-\mathbf{c}+\mathbf{b}$. 2. Since $[AB]$ is a diameter, $O$ is the midpoint and $\overrightarrow{AO}=\overrightarrow{OB}=\mathbf{b}$. Therefore $\overrightarrow{AC}=\overrightarrow{AO}+\overrightarrow{OC}=\mathbf{b}+\mathbf{c}$.

b) 1. Two non-zero vectors are perpendicular if their scalar (dot) product is $0$. We calculate $$\overrightarrow{AC}\cdot \overrightarrow{CB}=(\mathbf{b}+\mathbf{c})\cdot (\mathbf{b}-\mathbf{c})=|\mathbf{b}{|}^2-|\mathbf{c}{|}^2.$$ 2. $B$ and $C$ lie on the circumference, so $|\mathbf{b}|=|\mathbf{c}|=r$ (the radius). Hence $\overrightarrow{AC}\cdot \overrightarrow{CB}=r^2-r^2=0$. 3. Therefore $\overrightarrow{AC}\perp \overrightarrow{CB}$, that is, the angle at $C$ is a right angle.

Answer: $\overrightarrow{CB}=\mathbf{b}-\mathbf{c}$, $\overrightarrow{AC}=\mathbf{b}+\mathbf{c}$ and $\widehat{ACB}=90^{\circ }$. As required.

a) 1. If $\mathbf{a}\perp \mathbf{b}$, then $\mathbf{a}\cdot \mathbf{b}=0$: $$(3\cos \theta +6)(\cos \theta -2)+7(1+\sin \theta )=0.$$ 2. Expanding: $3{\cos }^2\theta -6\cos \theta +6\cos \theta -12+7+7\sin \theta =0$, that is, $3{\cos }^2\theta +7\sin \theta -5=0$. 3. We use ${\cos }^2\theta =1-{\sin }^2\theta$: $3-3{\sin }^2\theta +7\sin \theta -5=0$, so $-3{\sin }^2\theta +7\sin \theta -2=0$. 4. Multiplying by $-1$: $3{\sin }^2\theta -7\sin \theta +2=0$. As required.

b) 1. Factorise: $3{\sin }^2\theta -7\sin \theta +2=(3\sin \theta -1)(\sin \theta -2)=0$. 2. $\sin \theta =2$ is impossible, so $\sin \theta =\frac{1}{3}$. 3. The smallest positive value is $\theta =\arcsin \frac{1}{3}\approx 0.340$ rad ($19.5^{\circ }$).

Answer: $\theta =\arcsin \frac{1}{3}\approx 19.5^{\circ }$

a) 1. In rhombus $PQRS$, the side $\overrightarrow{SR}$ equals $\overrightarrow{PQ}=\mathbf{a}$, so $\overrightarrow{RS}=-\mathbf{a}$. 2. $\overrightarrow{PR}=\overrightarrow{PQ}+\overrightarrow{QR}=\mathbf{a}+\mathbf{b}$. 3. $\overrightarrow{QS}=\overrightarrow{QR}+\overrightarrow{RS}=\mathbf{b}-\mathbf{a}$.

b) 1. We calculate the scalar (dot) product of the diagonals: $$\overrightarrow{PR}\cdot \overrightarrow{QS}=(\mathbf{a}+\mathbf{b})\cdot (\mathbf{b}-\mathbf{a})=|\mathbf{b}{|}^2-|\mathbf{a}{|}^2.$$ 2. All sides of a rhombus have the same length: $|\mathbf{a}|=|\mathbf{b}|$. Therefore the scalar product equals $0$. 3. Hence the diagonals are perpendicular.

Answer: $\overrightarrow{PR}=\mathbf{a}+\mathbf{b}$, $\overrightarrow{QS}=\mathbf{b}-\mathbf{a}$ and the diagonals intersect at right angles.

1. The line $x+y=1$ has normal vector $(1,1)$ and direction vector $(1,-1)$.
2. $\mathbf{a}$ is perpendicular to the line if it is perpendicular to its direction vector: $\mathbf{a}\cdot (1,-1)=0$.
3. $2\sin \theta -(1-\sin \theta )=0\Rightarrow 3\sin \theta -1=0\Rightarrow \sin \theta =\frac{1}{3}$.
4. This holds for all $\theta$ with $\sin \theta =\frac{1}{3}$; the principal value is $\theta =\arcsin \frac{1}{3}\approx 0.340$ rad.

Answer: $\theta =\arcsin \frac{1}{3}\approx 19.5^{\circ }$

1. The scalar product is $\mathbf{a}\cdot \mathbf{b}=\cos \theta \sin \theta +\sin \theta \cos \theta =2\sin \theta \cos \theta =\sin 2\theta$.
2. The magnitudes: $|\mathbf{a}|=\sqrt{{\cos }^2\theta +{\sin }^2\theta }=1$ and $|\mathbf{b}|=1$.
3. Therefore $\cos \alpha =\frac{\mathbf{a}\cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}=\sin 2\theta$.
4. We use $\sin 2\theta =\cos \left(\frac{\pi }{2}-2\theta \right)$, so that $\cos \alpha =\cos \left(\frac{\pi }{2}-2\theta \right)$.
5. If $0<\theta <\frac{\pi }{4}$, then $0<\frac{\pi }{2}-2\theta <\frac{\pi }{2}$, which lies in the interval $[0,\pi ]$ where the cosine is injective. Hence $\alpha =\frac{\pi }{2}-2\theta$.

Answer: $\alpha =\frac{\pi }{2}-2\theta$

a)

1. Perpendicular vectors: $\mathbf{a}\cdot \mathbf{b}=0$, that is, $(3\cos \theta +6)(\cos \theta -2)+7(1+\sin \theta )=0$.
2. Expanding: $3{\cos }^2\theta -6\cos \theta +6\cos \theta -12+7+7\sin \theta =0$, so $3{\cos }^2\theta -5+7\sin \theta =0$.
3. Using ${\cos }^2\theta =1-{\sin }^2\theta$: $3-3{\sin }^2\theta -5+7\sin \theta =0$, that is, $-3{\sin }^2\theta +7\sin \theta -2=0$.
4. Multiplying by $-1$: $3{\sin }^2\theta -7\sin \theta +2=0$.

As required.

b)

1. Factorise: $(3\sin \theta -1)(\sin \theta -2)=0$.
2. $\sin \theta =2$ is impossible, so $\sin \theta =\frac{1}{3}$.
3. The smallest positive value is $\theta =\arcsin \frac{1}{3}\approx 0.340$ rad.

Answer: $\theta =\arcsin \frac{1}{3}\approx 19.5^{\circ }$ ($\approx 0.340$ rad)

1. $Q$ lies on $x-2y-4=0$: we write $Q=(2q+4,q)$.
2. $P(1,1)$ is the midpoint of $[QR]$, so $R=2P-Q=(2-(2q+4),2-q)=(-2q-2,2-q)$.
3. $R$ lies on $x+y=5$: $(-2q-2)+(2-q)=5\Rightarrow -3q=5\Rightarrow q=-\frac{5}{3}$.
4. $Q=\left(2\left(-\frac{5}{3}\right)+4,-\frac{5}{3}\right)=\left(\frac{2}{3},-\frac{5}{3}\right)$.
5. $R=\left(\frac{10}{3}-2,2+\frac{5}{3}\right)=\left(\frac{4}{3},\frac{11}{3}\right)$.
6. Check: $\frac{4}{3}+\frac{11}{3}=5$ and the midpoint of $Q$ and $R$ is $(1,1)$.

Answer: $Q\left(\frac{2}{3},-\frac{5}{3}\right)$ and $R\left(\frac{4}{3},\frac{11}{3}\right)$

1. The line is $kx-y+1=0$. It is tangent if the distance from the centre $(5,1)$ to the line equals the radius $3$.
2. Distance: $\frac{|5k-1+1|}{\sqrt{k^2+1}}=\frac{|5k|}{\sqrt{k^2+1}}=3$.
3. Square both sides: $25k^2=9(k^2+1)\Rightarrow 16k^2=9\Rightarrow k^2=\frac{9}{16}$.

Answer: $k=\pm \frac{3}{4}$

a)

1. $\overrightarrow{BC}=(-14-4,12-6)=(-18,6)$, with gradient $\frac{6}{-18}=-\frac{1}{3}$.
2. A perpendicular line has gradient $m$ with $m\cdot \left(-\frac{1}{3}\right)=-1$, so $m=3$.
3. Through $A(-2,20)$: $y-20=3(x+2)$, that is, $y=3x+26$.

b) Substituting $(k,2)$: $2=3k+26$, so $3k=-24$ and $k=-8$.

Answer: a) $y=3x+26$; b) $k=-8$

a)

1. At $t=0$ Ed is at $P$, so $\mathbf{a}=\overrightarrow{OP}=\left(\begin{array}{c}-1\\ 4\end{array}\right)$.
2. At $t=3$ he is at $Q$: $\mathbf{a}+3\mathbf{b}=\left(\begin{array}{c}4\\ 16\end{array}\right)$, so $3\mathbf{b}=\left(\begin{array}{c}5\\ 12\end{array}\right)$ and $\mathbf{b}=\left(\begin{array}{c}5/3\\ 4\end{array}\right)$.

b)

1. The point closest to $C$ is the one where $\overrightarrow{CR}\perp \mathbf{b}$, where $R$ is Ed's position. With $\mathbf{r}-\overrightarrow{OC}=\left(\begin{array}{c}-12+\frac{5}{3}t\\ -5+4t\end{array}\right)$.
2. Scalar product with $\mathbf{b}$ equal to zero: $\frac{5}{3}\left(-12+\frac{5}{3}t\right)+4(-5+4t)=0$.
3. Expanding: $-20+\frac{25}{9}t-20+16t=0$, that is, $\frac{169}{9}t=40$.
4. $t=\frac{360}{169}\approx 2.13$ h, which lies between $0$ and $3$, so it is valid.

Answer: $t=\frac{360}{169}\approx 2.13$ hours

1. The line travelled by $S_1$ has direction $\mathbf{d}=(1,2)$ and passes through $A(0,0)$. Its speed is $\sqrt{10^2+20^2}=10\sqrt{5}$ km/h, so $S_2$ travels at $30\sqrt{5}$ km/h.
2. The shortest distance from $B$ to that line is the perpendicular. The foot $P=s(1,2)$ satisfies $\overrightarrow{BP}\cdot \mathbf{d}=0$: $(s-70)+2(2s-30)=0$, so $5s=130$ and $s=26$. Thus $P=(26,52)$.
3. Distance: $\overrightarrow{BP}=(-44,22)$, $|\overrightarrow{BP}|=\sqrt{1936+484}=22\sqrt{5}$ km.
4. $S_1$ reaches $P$ when $10t=26$, that is, $t=2.6$ h.
5. $S_2$ takes $\frac{22\sqrt{5}}{30\sqrt{5}}=\frac{11}{15}$ h to reach $P$.
6. It must leave at the latest at $t=2.6-\frac{11}{15}=\frac{39}{15}-\frac{11}{15}=\frac{28}{15}$ h $=1$ h $52$ min after 10:00.

Answer: at 11:52 (1 h 52 min after 10:00)