IB Mathematics AHL 3.7 area of sector, length of arc AI HL Paper 3- Exam Style Questions- New Syllabus

Question

This problem evaluates the variations in distance and orientation when comparing a Euclidean flat-plane model to a spherical model for planetary geography.

First, consider a simplified model where the cities of Bogotá, Moscow, and Nairobi are situated on a perfectly flat surface. In this representation, Nairobi is located $6000\text{ km}$ due south of Moscow, while Bogotá is positioned $12\,500\text{ km}$ due west of Nairobi.

(a) (i) Calculate the direct distance between Bogotá and Moscow in this flat model.

(ii) Determine the bearing of Moscow from Bogotá, expressing your result in degrees.

To improve accuracy, we must account for the Earth’s curvature. We now employ a $3\text{D}$ coordinate system $(x,y,z)$ with its origin $O$ at the Earth’s centre. Units are defined in thousands of kilometres, and the Earth is approximated as a sphere with radius $R = 6000\text{ km}$. The North Pole, $P$, is located on the positive $z$-axis, and Nairobi, $N$, is positioned on the equator along the $y$-axis.

The position vectors are defined as $\vec{OP} = \mathbf{p} = \begin{pmatrix} 0 \\ 0 \\ 6 \end{pmatrix}$ and $\vec{ON} = \mathbf{n} = \begin{pmatrix} 0 \\ 6 \\ 0 \end{pmatrix}$.

(b) (i) Utilise the scalar product to determine the angle between vectors $\mathbf{p}$ and $\mathbf{n}$.

(ii) Demonstrate that the geodesic distance (arc length) between $P$ and $N$ is $3000\pi\text{ km}$.

A third point $A$, also on the equator, has the position vector $\mathbf{a} = \begin{pmatrix} 6 \\ 0 \\ 0 \end{pmatrix}$. The points $P$, $N$, and $A$ form the vertices of a spherical triangle.

The interior angle at vertex $A$ is defined as the angle between the normal vectors $\mathbf{a} \times \mathbf{p}$ and $\mathbf{a} \times \mathbf{n}$.

(ii) Show that the interior angle at vertex $A$ in this spherical triangle is exactly $90^\circ$.

Moscow, $M$, is assigned the position vector $\vec{OM} = \mathbf{m} = \begin{pmatrix} 0 \\ 6\cos\theta \\ 6\sin\theta \end{pmatrix}$.

On a sphere, the shortest path between two points follows a great circle arc. In this model, the shortest distance from Moscow to Nairobi is exactly $6000\text{ km}$.

(d) Show that $\theta \approx 57.3^\circ$, rounded to three significant figures.

Bogotá, $B$, is located west of Nairobi with the position vector $\vec{OB} = \mathbf{b} = \begin{pmatrix} 6\sin120^\circ \\ 6\cos120^\circ \\ 0 \end{pmatrix}$.

(e) Calculate the shortest spherical distance from Bogotá to Moscow.

The bearing from $B$ to $M$ is defined as the angle at vertex $B$ within the spherical triangle $BMP$. Given that $\mathbf{b} \times \mathbf{p} = \begin{pmatrix} 36\cos120^\circ \\ -36\sin120^\circ \\ 0 \end{pmatrix}$:

(f) Using the methodology established in part (c), find the bearing from Bogotá to Moscow.

Most-appropriate topic codes (IB Mathematics Applications and Interpretation HL):

• AHL 3.13: Vector product and scalar product — parts (b)(i), (c), (f)
• AHL 3.7: Radian measure — parts (b)(ii), (d)
• AHL 3.10: Vectors — parts (b), (c), (e), (f)
• SL 3.3: Applications of trigonometry — part (a)
• SL 3.2: Sine and cosine rules — part (a)

▶️ Answer/Explanation

(a)(i)
Method: Using Pythagoras’ theorem in the right-angled triangle with sides 6000 km and 12500 km.
Distance = $\sqrt{6000^2 + 12500^2} = \sqrt{36000000 + 156250000} = \sqrt{192250000} = 13865.24…$
$\boxed{13900 \text{ km}}$ (to 3 significant figures)

(a)(ii)
Method 1: Using trigonometry $\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}} = \frac{12500}{6000}$
Angle = $\arctan\left(\frac{12500}{6000}\right) = \arctan(2.0833…) = 64.4^\circ$
Bearing is measured clockwise from north, so bearing = $064^\circ$
$\boxed{064^\circ}$ or $\boxed{64.4^\circ}$

(b)(i)
Method: Using scalar product formula $p \cdot n = |p||n|\cos\theta$
$p \cdot n = (0)(0) + (0)(6) + (6)(0) = 0$
$|p| = \sqrt{0^2 + 0^2 + 6^2} = 6$, $|n| = \sqrt{0^2 + 6^2 + 0^2} = 6$
$\cos\theta = \frac{p \cdot n}{|p||n|} = \frac{0}{6 \times 6} = 0$
∴ $\theta = 90^\circ$ or $\frac{\pi}{2}$ radians
$\boxed{90^\circ}$

(b)(ii)
Method: Using arc length formula $s = r\theta$ with $\theta$ in radians
From part (b)(i), angle = $90^\circ = \frac{\pi}{2}$ radians
Arc length = $6000 \times \frac{\pi}{2} = 3000\pi$ km
$\boxed{3000\pi \text{ km}}$

(c)(i)
Method: Using cross product formula $a \times p = \begin{vmatrix} i & j & k \\ 6 & 0 & 0 \\ 0 & 0 & 6 \end{vmatrix}$
$a \times p = (0 \times 6 – 0 \times 0)i – (6 \times 6 – 0 \times 0)j + (6 \times 0 – 0 \times 0)k$
$= 0i – 36j + 0k = \begin{pmatrix} 0 \\ -36 \\ 0 \end{pmatrix}$
$\boxed{\begin{pmatrix} 0 \\ -36 \\ 0 \end{pmatrix}}$

(c)(ii)
Method 1 (Scalar product):
First find $a \times n = \begin{vmatrix} i & j & k \\ 6 & 0 & 0 \\ 0 & 6 & 0 \end{vmatrix} = (0 \times 0 – 0 \times 6)i – (6 \times 0 – 0 \times 0)j + (6 \times 6 – 0 \times 0)k = \begin{pmatrix} 0 \\ 0 \\ 36 \end{pmatrix}$
$(a \times p) \cdot (a \times n) = (0)(0) + (-36)(0) + (0)(36) = 0$
Since scalar product = 0, the vectors are perpendicular, so angle = $90^\circ$

Method 2 (Cross product magnitude):
$|a \times p| = \sqrt{0^2 + (-36)^2 + 0^2} = 36$
$|a \times n| = \sqrt{0^2 + 0^2 + 36^2} = 36$
$(a \times p) \times (a \times n) = \begin{pmatrix} (-36)(36) – (0)(0) \\ (0)(0) – (0)(36) \\ (0)(0) – (-36)(0) \end{pmatrix} = \begin{pmatrix} -1296 \\ 0 \\ 0 \end{pmatrix}$
$|(a \times p) \times (a \times n)| = 1296$
$\sin\theta = \frac{|(a \times p) \times (a \times n)|}{|a \times p||a \times n|} = \frac{1296}{36 \times 36} = 1$
∴ $\theta = 90^\circ$
$\boxed{90^\circ}$

(d)
Method: Using arc length formula $s = r\theta$ with $s = 6000$ km, $r = 6000$ km
$6000 = 6000 \times \theta$ ⇒ $\theta = 1$ radian
Convert to degrees: $1 \times \frac{180}{\pi} = 57.2957…^\circ$
$\boxed{57.3^\circ}$ (to 3 significant figures)

(e)
Method 1 (Scalar product):
First find angle BÔM using scalar product:
$b = \begin{pmatrix} 6\sin120^\circ \\ 6\cos120^\circ \\ 0 \end{pmatrix} = \begin{pmatrix} 6 \times \frac{\sqrt{3}}{2} \\ 6 \times (-\frac{1}{2}) \\ 0 \end{pmatrix} = \begin{pmatrix} 3\sqrt{3} \\ -3 \\ 0 \end{pmatrix}$
$m = \begin{pmatrix} 0 \\ 6\cos57.3^\circ \\ 6\sin57.3^\circ \end{pmatrix}$
$b \cdot m = (3\sqrt{3})(0) + (-3)(6\cos57.3^\circ) + (0)(6\sin57.3^\circ) = -18\cos57.3^\circ$
$|b| = \sqrt{(3\sqrt{3})^2 + (-3)^2 + 0^2} = \sqrt{27 + 9} = \sqrt{36} = 6$
$|m| = \sqrt{0^2 + (6\cos57.3^\circ)^2 + (6\sin57.3^\circ)^2} = \sqrt{36(\cos^2 57.3^\circ + \sin^2 57.3^\circ)} = 6$
$\cos(\text{BÔM}) = \frac{b \cdot m}{|b||m|} = \frac{-18\cos57.3^\circ}{36} = -\frac{\cos57.3^\circ}{2}$
$\text{BÔM} = \arccos\left(-\frac{\cos57.3^\circ}{2}\right) = 105.673…^\circ$
Shortest distance = $\frac{105.673}{360} \times 2\pi \times 6000 = 11066.0… \text{ km}$
$\boxed{11100 \text{ km}}$ (to 3 significant figures)

(f)
Method: Using method from part (c) – find angle between $b \times m$ and $b \times p$
First find $b \times m$:
$b \times m = \begin{pmatrix} 36\cos120^\circ\sin57.3^\circ \\ -36\sin120^\circ\sin57.3^\circ \\ 36\sin120^\circ\cos57.3^\circ \end{pmatrix}$
Using technology with $\sin57.3^\circ \approx 0.8415$, $\cos57.3^\circ \approx 0.5403$, $\sin120^\circ = \frac{\sqrt{3}}{2} \approx 0.8660$, $\cos120^\circ = -0.5$:
$b \times m \approx \begin{pmatrix} 36 \times (-0.5) \times 0.8415 \\ -36 \times 0.8660 \times 0.8415 \\ 36 \times 0.8660 \times 0.5403 \end{pmatrix} = \begin{pmatrix} -15.147 \\ -26.234 \\ 16.845 \end{pmatrix}$
Given $b \times p = \begin{pmatrix} 36\cos120^\circ \\ -36\sin120^\circ \\ 0 \end{pmatrix} = \begin{pmatrix} -18 \\ -31.176 \\ 0 \end{pmatrix}$
Scalar product: $(b \times m) \cdot (b \times p) \approx (-15.147)(-18) + (-26.234)(-31.176) + (16.845)(0) = 272.646 + 817.896 = 1090.542$
$|b \times m| \approx \sqrt{(-15.147)^2 + (-26.234)^2 + (16.845)^2} = \sqrt{229.43 + 688.22 + 283.75} = \sqrt{1201.40} = 34.661$
$|b \times p| = \sqrt{(-18)^2 + (-31.176)^2 + 0^2} = \sqrt{324 + 972.0} = \sqrt{1296} = 36$
$\cos\phi = \frac{1090.542}{34.661 \times 36} = \frac{1090.542}{1247.796} = 0.8740$
$\phi = \arccos(0.8740) = 29.1^\circ$
$\boxed{29.1^\circ}$ or $\boxed{029^\circ}$

IB Mathematics AHL 3.7 area of sector, length of arc AI HL Paper 3- Exam Style Questions- New Syllabus

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