IB Maths AA HL 3.13 Scalar product

[Maximum number: 13]

The following question compares the distance and direction between cities on a flat surface to the distance and direction between cities on a sphere.
Consider a model where the cities of Bogotá, Moscow, and Nairobi lie on a flat surface. In this model, Nairobi is 6000 km due south of Moscow and Bogotá is 12500 km due west of Nairobi, as shown in the following diagram.

Figure for Question IB Maths AA HL Ahl 3 13 Hl Scalar Product Questions question 1 — IB Maths AA HL

(a)

Use the scalar product to find the angle between p and n.

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$\mathbf{p}\cdot\mathbf{n}=0$ , so the angle is $90^\circ$ .

(b)

Find the shortest distance from Bogotá to Moscow on the sphere.

The bearing from B to M is defined as the angle at vertex B in the spherical triangle containing B, M and P . It is given that $\boldsymbol{b} \times \boldsymbol{p}=\left(\begin{array}{c}36 \cos 120^{\circ} \\ -36 \sin 120^{\circ} \\ 0\end{array}\right)$ .

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METHOD 1

EITHER (Scalar product)
attempt to use scalar product to find BÔM $=\arccos \left(\frac{\boldsymbol{b} . \boldsymbol{m}}{|\boldsymbol{b}||\boldsymbol{m}|}\right)$

Note: This may be written as $\cos ($ BÔM $)=\frac{\boldsymbol{b} . \boldsymbol{m}}{|\boldsymbol{b}||\boldsymbol{m}|}$ .
$=\arccos \left(\frac{\left(6 \cos 120^{\circ}\right)\left(6 \cos 57.3^{\circ}\right)}{6^{2}}\right)$

Note: This (A1) can be awarded for $\cos \theta=\frac{6 \cos 120^{\circ} \times 6 \cos 57.3^{\circ}}{6^{2}}$ .
$=106^{\circ}\left(=105.673 \ldots{ }^{\circ}\right)$ OR 1.84 radians $(1.84434 \ldots)$

OR (Cosine Rule)
$|\mathrm{BM}|=\sqrt{(6 \sin 120-0)^{2}+(6 \cos 120-6 \sin 57.3)^{2}+(0-6 \sin 57.3)^{2}}$ $=9.56(9.56299 \ldots)$ OR 9560 km
attempt to use the cosine rule to find angle BOM
$|\mathrm{BM}|^{2}=6^{2}+6^{2}-2 \times 6 \times 6 \times \cos \mathrm{BOM}$ $\mathrm{BOM}=106^{\circ}(=105.673 \ldots)$ OR 1.84 radians $(1.84434 \ldots)$

THEN
attempt to use $\frac{\theta}{360} \times 2 \pi r$ OR $s=r \theta$ $=\frac{105.671^{\circ}}{360} \times 2 \pi \times 6$ $=11.1(11.0660 \ldots$ thousand km)

Note: Accept an answer of 11100 OR $11066.0 \ldots$ without units (i.e. "km" is implicit).

METHOD 2 (Cross product)
attempt to use cross product to find BÔM $=\arcsin \frac{|\boldsymbol{b} \times \boldsymbol{m}|}{|\boldsymbol{b}||\boldsymbol{m}|}$

Note: This may be written as $\sin$ BÔM $=\frac{|\boldsymbol{b} \times \boldsymbol{m}|}{|\boldsymbol{b}||\boldsymbol{m}|}$ .

\begin{aligned} & =\frac{\left|\left(\begin{array}{c} 36 \cos 120 \sin \theta \\ -36 \sin 120 \sin \theta \\ 36 \sin 120 \cos \theta \end{array}\right)\right|}{6 \times 6} \\ & =\frac{\sqrt{\sin ^{2} \theta+3}}{2}=0.962818 \ldots \\ & \text { BÔM }=180-\arcsin 0.962818 \ldots \\ & \quad=106^{\circ}(105.673 \ldots, 1.844346 \mathrm{rads}) \end{aligned}

Note: Award at most (M1)(A1)(A0)(M1)A0 for an answer of 7.78 thousand km from finding the principal root of the arcsin.

(c)

Using the method from part (c), find the bearing from Bogotá to Moscow.
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attempt to calculate $\boldsymbol{b} \times \boldsymbol{m}$ $\boldsymbol{b} \times \boldsymbol{m}=\left(\begin{array}{c}36 \cos 120^{\circ} \sin 57.3^{\circ} \\ -36 \sin 120^{\circ} \sin 57.3^{\circ} \\ 36 \sin 120^{\circ} \cos 57.3^{\circ}\end{array}\right)\left(=\left(\begin{array}{c}-15.1464 \ldots \\ -26.2344 \ldots \\ 16.8449 \ldots\end{array}\right)\right)$
attempt to use scalar product to find angle between vectors

\begin{aligned} & (\cos \theta=)\left(\frac{(\boldsymbol{b} \times \boldsymbol{m}) \cdot(\boldsymbol{b} \times \boldsymbol{p})}{|\boldsymbol{b} \times \boldsymbol{m} \| \boldsymbol{b} \times \boldsymbol{p}|}\right) \\ & (\cos \theta=)\left(\frac{1090.54 \ldots}{34.6614 \ldots \times 36}\right) \end{aligned}

Note: Award (A1) for numerator, (A1) for denominator.

\text { bearing is }(0) 29.1^{\circ}\left(29.0770 \ldots{ }^{\circ}\right)

Note: Do not penalise absence of degree symbol.
Award full marks for answer in radians =0.507.
Award full marks for $029^{\circ}$ .
Special Case: If a candidate calculates $\boldsymbol{m} \times \boldsymbol{b}$ instead of $\boldsymbol{b} \times \boldsymbol{m}$ then you can award at most (M1)(A0)(M1)(A1)(A1)A0.

IB Maths AA HL Ahl 3 13 Hl Scalar Product Questions question 1

Question (a)

Question (b)

Question (c)