IB Maths AI HL 3.13 Scalar and vector products

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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 AI HL Ahl 3 13 Hl Scalar and Vector Products Questions question 1 — IB Maths AI HL

(a)

In reality, these three cities lie on the curved surface of the Earth which will change the distances and directions found in part (a).

Now consider a curved model using a coordinate system (x, y, z) with its origin, O , at the centre of the Earth. The units of this system are thousands of kilometres and the Earth is modelled as a sphere with radius 6000 km . The North Pole, P, lies on the z-axis, and Nairobi, N, is modelled as being on the equator and lying on the y-axis.

Figure for Question (a) — IB Maths AI HL

P has position vector $\overrightarrow{\mathrm{OP}}=\boldsymbol{p}=\left(\begin{array}{l}0 \\ 0 \\ 6\end{array}\right)$ and N has position vector $\overrightarrow{\mathrm{ON}}=\boldsymbol{n}=\left(\begin{array}{l}0 \\ 6 \\ 0\end{array}\right)$ .

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(i)

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

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$\quad \boldsymbol{p} . \boldsymbol{n}=0$
so the angle is $90^{\circ}$ OR $\frac{\pi}{2}$ rads

Note: Award M0A0 for an answer of 90 degrees without any mention that the scalar product is zero.

(b)

Point A, which is also on the equator, has position vector $\boldsymbol{a}=\left(\begin{array}{l}6 \\ 0 \\ 0\end{array}\right)$ as shown in the
following diagram.

Figure for Question (b) — IB Maths AI HL

P, N and A , and the arcs connecting them, form a spherical triangle.
The angle at vertex A is defined as the angle between the vectors $\boldsymbol{a} \times \boldsymbol{p}$ and $\boldsymbol{a} \times \boldsymbol{n}$ .

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(i)

Find the vector $\boldsymbol{a} \times \boldsymbol{p}$ .

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attempt to find cross product (e.g. correct except for sign error)
$\boldsymbol{a} \times \boldsymbol{p}=\left(\begin{array}{c}0 \\ -36 \\ 0\end{array}\right)$

(ii)

Show that the angle at vertex A in the spherical triangle is $90^{\circ}$ .

Moscow, M , has position vector $\overrightarrow{\mathrm{OM}}=\boldsymbol{m}=\left(\begin{array}{c}0 \\ 6 \cos \theta \\ 6 \sin \theta\end{array}\right)$ , as shown in the following diagram.

Figure for Question (ii) — IB Maths AI HL

The shortest distance between two points on the sphere lies along an arc of a circle on the sphere with centre O . In this model the shortest distance from Moscow to Nairobi is 6000 km .

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METHOD 1
$\boldsymbol{a} \times \boldsymbol{n}=\left(\begin{array}{c}0 \\ 0 \\ 36\end{array}\right)$
choosing to use scalar product for their $\boldsymbol{a} \times \boldsymbol{p}$ and $\boldsymbol{a} \times \boldsymbol{n}$ $(\boldsymbol{a} \times \boldsymbol{p}) \cdot(\boldsymbol{a} \times \boldsymbol{n})=\mathbf{0}$

Note: The final A1 can be awarded for seeing the dot product equals zero without incorrect working. Follow through from (i) is permitted and if they explicitly state that they are using $\boldsymbol{a} \times \boldsymbol{n}$ then allow internal follow through.
so angle is $90^{\circ}$
AG

METHOD 2
$\boldsymbol{a} \times \boldsymbol{n}=\left(\begin{array}{c}0 \\ 0 \\ 36\end{array}\right)$
attempt to use cross product for their $\boldsymbol{a} \times \boldsymbol{p}$ and $\boldsymbol{a} \times \boldsymbol{n}$ . This must include both an attempt to find the cross product and knowledge that $\sin \theta=\frac{|x \times y|}{|x||y|}$ $(\boldsymbol{a} \times \boldsymbol{n}) \times(\boldsymbol{a} \times \boldsymbol{p})=\left(\begin{array}{c}1296 \\ 0 \\ 0\end{array}\right) \quad$ OR $(\boldsymbol{a} \times \boldsymbol{p}) \times(\boldsymbol{a} \times \boldsymbol{n})=\left(\begin{array}{c}-1296 \\ 0 \\ 0\end{array}\right)$ $\sin \theta=\frac{1296}{36 \times 36}=1$
so angle is $90^{\circ}$

(c)

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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(e) 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 $\cos (\mathrm{BOM})=\frac{\boldsymbol{b} \boldsymbol{.} \boldsymbol{m}}{|\boldsymbol{b}||\boldsymbol{m}|}$ .

=\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) \text { OR } 1.84 \text { radians }(1.84434 \ldots)

OR (Cosine Rule)

\begin{aligned} & |\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) \text { OR } 9560 \mathrm{~km} \end{aligned}

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}\left(=105.673 \ldots{ }^{\circ}\right) \text { OR } 1.84 \text { radians }(1.84434 \ldots)

THEN

\begin{aligned} & \text { attempt to use } \frac{\theta}{360} \times 2 \pi r \text { OR } s=r \theta \\ & =\frac{105.671^{\circ}}{360} \times 2 \pi \times 6 \\ & =11.1(11.0660 \ldots \text { thousand } \mathrm{km}) \end{aligned}

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

METHOD 2 (Cross product)

\text { 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 \end{aligned}

\begin{aligned} \text { BÔM } & =180-\arcsin 0.962818 \ldots \\ & =106^{\circ}(105.673 \ldots, 1.844346 \mathrm{rads}) \end{aligned}

\text { attempt to use } \frac{\theta}{360} \times 2 \pi r \quad \text { OR } \quad s=r \theta

=\frac{105.671}{360} \times 2 \pi \times 6

=11.1 (11.0660 thousand km)

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.

(d)

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

Note: Award (A1) for numerator, (A1) for denominator.
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 M1A0M1A1A1A0.

IB Maths AI HL Ahl 3 13 Hl Scalar and Vector Products Questions question 1

Question (a)

Question (i)

Question (b)

Question (i)

Question (ii)

Question (c)

Question (d)