2D Vectors - IB Mathematics Applications and Interpretation

1. Two forces, $\vec{F}_1 = \begin{pmatrix} -7 \\ 12 \end{pmatrix}$ N and $\vec{F}_2 = \begin{pmatrix} 11 \\ -5 \end{pmatrix}$ N, act on a particle. What is the magnitude of the resultant force?
   Answer: 8.06 N The resultant force is the vector sum of the individual forces: $\vec{F}_R = \vec{F}_1 + \vec{F}_2$. $$\vec{F}_R = \begin{pmatrix} -7 \\ 12 \end{pmatrix} + \begin{pmatrix} 11 \\ -5 \end{pmatrix} = \begin{pmatrix} 4 \\ 7 \end{pmatrix}$$ The magnitude of the resultant force is $|\vec{F}_R| = \sqrt{4^2 + 7^2} = \sqrt{16 + 49} = \sqrt{65} \approx 8.06$ N.
2. Two tugboats are pulling a large ship. The forces are represented by vectors $\vec{F}_1 = \begin{pmatrix} 60 \\ 90 \end{pmatrix}$ kN and $\vec{F}_2 = \begin{pmatrix} p \\ -50 \end{pmatrix}$ kN. The resultant force has a magnitude of 100 kN. Given that the horizontal component of the resultant force is positive, find the value of $p$, correct to one decimal place.
   Answer: 31.7 First, find the resultant force vector by adding the two force vectors: $$\vec{F}_R = \vec{F}_1 + \vec{F}_2 = \begin{pmatrix} 60 \\ 90 \end{pmatrix} + \begin{pmatrix} p \\ -50 \end{pmatrix} = \begin{pmatrix} 60+p \\ 40 \end{pmatrix}$$ The magnitude of this resultant force is given as 100 kN. $$|\vec{F}_R| = \sqrt{(60+p)^2 + 40^2} = 100$$ Square both sides: $$(60+p)^2 + 1600 = 10000$$ $$(60+p)^2 = 8400$$ $$60+p = \pm\sqrt{8400} \approx \pm 91.65$$ The question states the horizontal component, $60+p$, is positive, so we take the positive root. $$60+p = 91.65$$ $$p = 91.65 - 60 = 31.65 \approx 31.7$$
3. A delivery robot starts at a warehouse located at the origin. It travels to point A with position vector $\vec{OA} = \begin{pmatrix} 12 \\ -5 \end{pmatrix}$ km. It then travels to point B, such that the displacement from A to B is $\vec{AB} = \begin{pmatrix} -8 \\ 20 \end{pmatrix}$ km. What is the magnitude of the robot's final displacement vector from the warehouse, $\vec{OB}$?
   Answer: 15.5 km The final displacement vector from the warehouse (origin O) to point B is the sum of the two displacement vectors: $\vec{OB} = \vec{OA} + \vec{AB}$. $$\vec{OB} = \begin{pmatrix} 12 \\ -5 \end{pmatrix} + \begin{pmatrix} -8 \\ 20 \end{pmatrix} = \begin{pmatrix} 12-8 \\ -5+20 \end{pmatrix} = \begin{pmatrix} 4 \\ 15 \end{pmatrix}$$ The magnitude of this vector is $|\vec{OB}| = \sqrt{4^2 + 15^2} = \sqrt{16 + 225} = \sqrt{241} \approx 15.52$ km.
4. The treasure is buried at the midpoint of the line segment AB. What is the position vector of the treasure relative to the starting point O?
   Answer: $\begin{pmatrix} 35 \\ 50 \end{pmatrix}$ First, we need the position vectors of points A and B. The position vector of A is $\vec{OA} = \vec{a} = \begin{pmatrix} 40 \\ 20 \end{pmatrix}$. The position vector of B is $\vec{OB} = \vec{OA} + \vec{AB} = \vec{a} + \vec{b}$. $$\vec{OB} = \begin{pmatrix} 40 \\ 20 \end{pmatrix} + \begin{pmatrix} -10 \\ 60 \end{pmatrix} = \begin{pmatrix} 30 \\ 80 \end{pmatrix}$$ The treasure T is at the midpoint of AB, so its position vector $\vec{OT}$ is the average of the position vectors of A and B. $$\vec{OT} = \frac{1}{2}(\vec{OA} + \vec{OB}) = \frac{1}{2}\left( \begin{pmatrix} 40 \\ 20 \end{pmatrix} + \begin{pmatrix} 30 \\ 80 \end{pmatrix} \right) = \frac{1}{2}\begin{pmatrix} 70 \\ 100 \end{pmatrix} = \begin{pmatrix} 35 \\ 50 \end{pmatrix}$$
5. What is the ground speed of the airplane, correct to the nearest integer?
   Answer: 780 km/h The airplane's velocity relative to the ground, $\vec{v}_g$, is the sum of its velocity relative to the air and the velocity of the wind: $$\vec{v}_g = \vec{v}_a + \vec{w} = \begin{pmatrix} 800 \\ 0 \end{pmatrix} + \begin{pmatrix} -20 \\ 15 \end{pmatrix} = \begin{pmatrix} 780 \\ 15 \end{pmatrix}$$ The ground speed is the magnitude of this resultant vector: $$|\vec{v}_g| = \sqrt{780^2 + 15^2} = \sqrt{608400 + 225} = \sqrt{608625} \approx 780.14\text{ km/h}$$ To the nearest integer, the speed is 780 km/h.
6. An airplane flies from city P to city Q. The displacement vector for this journey is $\vec{PQ} = \begin{pmatrix} 450 \\ -120 \end{pmatrix}$ km. The flight takes 1.5 hours. What is the velocity vector of the airplane?
   Answer: $\begin{pmatrix} 300 \\ -80 \end{pmatrix}$ Velocity is displacement divided by time. The velocity vector is found by multiplying the displacement vector by the scalar $\frac{1}{\text{time}}$. $$\vec{v} = \frac{1}{1.5} \vec{PQ} = \frac{1}{1.5} \begin{pmatrix} 450 \\ -120 \end{pmatrix} = \begin{pmatrix} 450/1.5 \\ -120/1.5 \end{pmatrix} = \begin{pmatrix} 300 \\ -80 \end{pmatrix}$$
7. A submarine travels from its base on two legs. The first displacement is $\vec{d}_1 = \begin{pmatrix} 80 \\ -120 \end{pmatrix}$ km, and the second is $\vec{d}_2 = \begin{pmatrix} -50 \\ 180 \end{pmatrix}$ km. What displacement vector $\vec{d}_3$ must the submarine follow to return directly to its base?
   Answer: $\begin{pmatrix} -30 \\ -60 \end{pmatrix}$ To return to the base, the sum of all displacement vectors must be the zero vector: $\vec{d}_1 + \vec{d}_2 + \vec{d}_3 = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$. First, find the current displacement from the base: $$\vec{d}_1 + \vec{d}_2 = \begin{pmatrix} 80 \\ -120 \end{pmatrix} + \begin{pmatrix} -50 \\ 180 \end{pmatrix} = \begin{pmatrix} 30 \\ 60 \end{pmatrix}$$ The return vector $\vec{d}_3$ must be the negative of this resultant vector: $$\vec{d}_3 = -(\vec{d}_1 + \vec{d}_2) = -\begin{pmatrix} 30 \\ 60 \end{pmatrix} = \begin{pmatrix} -30 \\ -60 \end{pmatrix}$$
8. Two particles start at the origin at the same time. Particle 1 has velocity $\vec{v}_1 = \begin{pmatrix} 5 \\ 12 \end{pmatrix}$ m/s and Particle 2 has velocity $\vec{v}_2 = \begin{pmatrix} 8 \\ -3 \end{pmatrix}$ m/s. What is the distance between the two particles after 4 seconds?
   Answer: 61.2 m First, find the relative velocity of Particle 2 with respect to Particle 1: $$\vec{v}_{rel} = \vec{v}_2 - \vec{v}_1 = \begin{pmatrix} 8 \\ -3 \end{pmatrix} - \begin{pmatrix} 5 \\ 12 \end{pmatrix} = \begin{pmatrix} 3 \\ -15 \end{pmatrix} \text{ m/s}$$ The displacement vector between them after 4 seconds is $\vec{d} = t \times \vec{v}_{rel}$: $$\vec{d} = 4 \begin{pmatrix} 3 \\ -15 \end{pmatrix} = \begin{pmatrix} 12 \\ -60 \end{pmatrix} \text{ m}$$ The distance between them is the magnitude of this displacement vector: $$\text{Distance} = |\vec{d}| = \sqrt{12^2 + (-60)^2} = \sqrt{144 + 3600} = \sqrt{3744} \approx 61.19\text{ m}$$
9. A drone flies along a displacement vector $\vec{d} = \begin{pmatrix} 6 \\ -8 \end{pmatrix}$ km from its base. It then flies back towards its base, covering exactly half of the return journey. What is the drone's final position vector relative to its base?
   Answer: $\begin{pmatrix} 3 \\ -4 \end{pmatrix}$ The initial displacement is $\vec{d} = \begin{pmatrix} 6 \\ -8 \end{pmatrix}$. The vector for the full return journey would be $-\vec{d} = \begin{pmatrix} -6 \\ 8 \end{pmatrix}$. The drone covers half of this return journey, so its second displacement is $\frac{1}{2}(-\vec{d}) = \begin{pmatrix} -3 \\ 4 \end{pmatrix}$. The final position is the sum of the two displacements: $$\text{Final Position} = \begin{pmatrix} 6 \\ -8 \end{pmatrix} + \begin{pmatrix} -3 \\ 4 \end{pmatrix} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}$$ This is equivalent to finding the midpoint of the initial displacement vector, $\frac{1}{2}\vec{d}$.
10. A triangular park has vertices at coordinates P(2, 6), Q(8, 3), and R(5, 10). A person walks from P to Q, and then from Q to R. Which vector represents the person's total displacement from their starting point P?
    Answer: $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ The total displacement is the sum of the individual displacement vectors, $\vec{PQ} + \vec{QR}$. By the triangle rule of vector addition, this sum is equal to the vector $\vec{PR}$. We can calculate $\vec{PR}$ directly: $$\vec{PR} = \vec{OR} - \vec{OP} = \begin{pmatrix} 5 \\ 10 \end{pmatrix} - \begin{pmatrix} 2 \\ 6 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$$ Alternatively, one could calculate each vector and add them: $\vec{PQ} = \begin{pmatrix} 8-2 \\ 3-6 \end{pmatrix} = \begin{pmatrix} 6 \\ -3 \end{pmatrix}$ and $\vec{QR} = \begin{pmatrix} 5-8 \\ 10-3 \end{pmatrix} = \begin{pmatrix} -3 \\ 7 \end{pmatrix}$. Then, $\vec{PQ} + \vec{QR} = \begin{pmatrix} 6-3 \\ -3+7 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$.
11. A particle starts at point A(2, 5) and moves with a constant velocity vector $\vec{v} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}$ m/s. What is the position vector of the particle after 3 seconds?
    Answer: $\begin{pmatrix} 14 \\ -1 \end{pmatrix}$ The final position vector $\vec{r}$ is the sum of the initial position vector $\vec{r}_0$ and the displacement (velocity $\times$ time). $$\vec{r} = \vec{r}_0 + t\vec{v}$$ $$\vec{r} = \begin{pmatrix} 2 \\ 5 \end{pmatrix} + 3 \begin{pmatrix} 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 2 \\ 5 \end{pmatrix} + \begin{pmatrix} 12 \\ -6 \end{pmatrix} = \begin{pmatrix} 2+12 \\ 5-6 \end{pmatrix} = \begin{pmatrix} 14 \\ -1 \end{pmatrix}$$
12. A particle starts at the origin and is displaced by vector $\vec{a} = \begin{pmatrix} 10 \\ -4 \end{pmatrix}$. It then undergoes a second displacement, $\vec{b}$, which is parallel to $\vec{a}$ and has a magnitude of 2.5 times the magnitude of $\vec{a}$. What is the final position vector of the particle?
    Answer: $\begin{pmatrix} 35 \\ -14 \end{pmatrix}$ The second displacement vector $\vec{b}$ is parallel to $\vec{a}$ and its magnitude is 2.5 times that of $\vec{a}$. This means $\vec{b} = 2.5\vec{a}$. $$\vec{b} = 2.5 \begin{pmatrix} 10 \\ -4 \end{pmatrix} = \begin{pmatrix} 25 \\ -10 \end{pmatrix}$$ The final position vector is the sum of the two displacements: $$\text{Final Position} = \vec{a} + \vec{b} = \begin{pmatrix} 10 \\ -4 \end{pmatrix} + \begin{pmatrix} 25 \\ -10 \end{pmatrix} = \begin{pmatrix} 35 \\ -14 \end{pmatrix}$$
13. A new mix is created using 2 servings of nuts and 1.5 servings of dried fruit. What is the total nutritional vector for this new mix?
    Answer: $\begin{pmatrix} 11.5 \\ 53.5 \end{pmatrix}$ The total nutritional vector is a linear combination of the ingredient vectors: $\vec{T} = 2\vec{n} + 1.5\vec{f}$. $$\vec{T} = 2 \begin{pmatrix} 5 \\ 8 \end{pmatrix} + 1.5 \begin{pmatrix} 1 \\ 25 \end{pmatrix}$$ $$\vec{T} = \begin{pmatrix} 10 \\ 16 \end{pmatrix} + \begin{pmatrix} 1.5 \\ 37.5 \end{pmatrix} = \begin{pmatrix} 10+1.5 \\ 16+37.5 \end{pmatrix} = \begin{pmatrix} 11.5 \\ 53.5 \end{pmatrix}$$
14. Two monitoring stations, A and B, have position vectors $\vec{a} = \begin{pmatrix} 5 \\ 12 \end{pmatrix}$ and $\vec{b} = \begin{pmatrix} 13 \\ 4 \end{pmatrix}$ relative to a central command post. An emergency beacon is activated at a point C, which is the midpoint of the line segment AB. What is the position vector of the beacon, $\vec{c}$?
    Answer: $\begin{pmatrix} 9 \\ 8 \end{pmatrix}$ The position vector of the midpoint of a line segment is the average of the position vectors of its endpoints. $$\vec{c} = \frac{1}{2}(\vec{a} + \vec{b})$$ $$\vec{c} = \frac{1}{2}\left( \begin{pmatrix} 5 \\ 12 \end{pmatrix} + \begin{pmatrix} 13 \\ 4 \end{pmatrix} \right) = \frac{1}{2} \begin{pmatrix} 5+13 \\ 12+4 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 18 \\ 16 \end{pmatrix} = \begin{pmatrix} 9 \\ 8 \end{pmatrix}$$
15. A hiker starts at a point with coordinates P(3, -4) and walks to a point Q(-2, 8). What is the magnitude of the displacement vector $\vec{PQ}$?
    Answer: 13.0 The displacement vector $\vec{PQ}$ is found by subtracting the position vector of P from the position vector of Q: $\vec{PQ} = \vec{OQ} - \vec{OP}$. $$\vec{PQ} = \begin{pmatrix} -2 \\ 8 \end{pmatrix} - \begin{pmatrix} 3 \\ -4 \end{pmatrix} = \begin{pmatrix} -2-3 \\ 8-(-4) \end{pmatrix} = \begin{pmatrix} -5 \\ 12 \end{pmatrix}$$ The magnitude of this vector is $|\vec{PQ}| = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13.0$.
16. The velocity of a ship is represented by vector $\vec{u} = \begin{pmatrix} 9 \\ -12 \end{pmatrix}$ km/h. A speedboat is moving parallel to the ship, with its velocity represented by vector $\vec{v} = \begin{pmatrix} x \\ 8 \end{pmatrix}$ km/h. Find the value of $x$.
    Answer: -6.0 Since the vectors are parallel, one must be a scalar multiple of the other. Let $\vec{v} = k\vec{u}$. $$\begin{pmatrix} x \\ 8 \end{pmatrix} = k \begin{pmatrix} 9 \\ -12 \end{pmatrix}$$ From the y-components, we can find the scalar $k$: $$8 = k(-12) \implies k = -\frac{8}{12} = -\frac{2}{3}$$ Now, use this scalar to find $x$ from the x-components: $$x = k(9) = \left(-\frac{2}{3}\right) \times 9 = -6$$
17. A force vector is given by $\vec{F} = \begin{pmatrix} k \\ -12 \end{pmatrix}$ Newtons. The magnitude of the force is 13 N. Given that $k$ is a positive value, find $k$.
    Answer: 5 The magnitude of a vector $\begin{pmatrix} x \\ y \end{pmatrix}$ is $\sqrt{x^2+y^2}$. We are given that the magnitude is 13 N. $$\sqrt{k^2 + (-12)^2} = 13$$ $$k^2 + 144 = 13^2$$ $$k^2 + 144 = 169$$ $$k^2 = 169 - 144 = 25$$ $$k = \pm\sqrt{25} = \pm 5$$ Since the question states that $k$ is a positive value, $k=5$.
18. The displacement vector from point A to point B is $\vec{AB} = \begin{pmatrix} -7 \\ 4 \end{pmatrix}$. The coordinates of point B are (1, -3). What are the coordinates of point A?
    Answer: (8, -7) Let the coordinates of A be $(x, y)$. The position vectors are $\vec{OA} = \begin{pmatrix} x \\ y \end{pmatrix}$ and $\vec{OB} = \begin{pmatrix} 1 \\ -3 \end{pmatrix}$. We know that $\vec{AB} = \vec{OB} - \vec{OA}$. We need to solve for $\vec{OA}$. $$\vec{OA} = \vec{OB} - \vec{AB}$$ $$\vec{OA} = \begin{pmatrix} 1 \\ -3 \end{pmatrix} - \begin{pmatrix} -7 \\ 4 \end{pmatrix} = \begin{pmatrix} 1 - (-7) \\ -3 - 4 \end{pmatrix} = \begin{pmatrix} 8 \\ -7 \end{pmatrix}$$ So, the coordinates of A are (8, -7).
19. What is the magnitude of the balloon's average velocity in km/h during this time interval?
    Answer: 17.9 km/h First, find the displacement vector from time 1 to time 2. $$\Delta\vec{r} = \vec{r}_2 - \vec{r}_1 = \begin{pmatrix} 10 \\ 4 \end{pmatrix} - \begin{pmatrix} 2 \\ 8 \end{pmatrix} = \begin{pmatrix} 8 \\ -4 \end{pmatrix}\text{ km}$$ The time interval is 30 minutes, which is 0.5 hours. The average velocity vector is the displacement divided by the time interval. $$\vec{v}_{avg} = \frac{\Delta\vec{r}}{\Delta t} = \frac{1}{0.5} \begin{pmatrix} 8 \\ -4 \end{pmatrix} = \begin{pmatrix} 16 \\ -8 \end{pmatrix}\text{ km/h}$$ The magnitude of this average velocity vector is: $$|\vec{v}_{avg}| = \sqrt{16^2 + (-8)^2} = \sqrt{256 + 64} = \sqrt{320} \approx 17.89\text{ km/h}$$

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