IB DP Math AI: Topic : AHL 3.10 :Concept of a vector: IB Style Questions HL Paper 2

Detail Solution:

Let’s tackle this problem step by step, diving into the geometry of these mountain summits with enthusiasm. We’re working in a 3D coordinate system with distances in kilometers, so we’ll use vectors, dot products, and the sine rule to solve each part.

Part(a) (i) Find the vector \(\overrightarrow{AB}\):
The position of point \( A \) is \( (0, 0, 2.8) \) and point \( B \) is \( (7.2, 5.1, 2.4) \). The vector \(\overrightarrow{AB}\) is found by subtracting the coordinates of \( A \) from \( B \):
\[ \overrightarrow{AB} = B – A = (7.2 – 0, 5.1 – 0, 2.4 – 2.8) = (7.2, 5.1, -0.4) \]
So,
\[ \overrightarrow{AB} = \begin{pmatrix} 7.2 \\ 5.1 \\ -0.4 \end{pmatrix} \]

Part(a) (ii) Hence find \( AB \), the distance between \( A \) and \( B \):
The distance \( AB \) is the magnitude of \(\overrightarrow{AB}\). For a vector \( (x, y, z) \), the magnitude is:
\[ |\overrightarrow{AB}| = \sqrt{x^2 + y^2 + z^2} \]
Substitute the components:
\[ |\overrightarrow{AB}| = \sqrt{(7.2)^2 + (5.1)^2 + (-0.4)^2} = \sqrt{51.84 + 26.01 + 0.16} = \sqrt{78.01} \]
Calculate:
\[ \sqrt{78.01} \approx 8.832 \]
So, the distance \( AB \approx 8.83 \) km (rounded to 2 decimal places).

Part(b) Find the angle between \(\begin{pmatrix} 1.1 \\ 8.4 \\ 0.2 \end{pmatrix}\) and \(\overrightarrow{AB}\):
Let’s call the given vector \(\overrightarrow{v} = \begin{pmatrix} 1.1 \\ 8.4 \\ 0.2 \end{pmatrix}\), which is parallel to \(\overrightarrow{AC}\). We need the angle \(\theta\) between \(\overrightarrow{v}\) and \(\overrightarrow{AB} = \begin{pmatrix} 7.2 \\ 5.1 \\ -0.4 \end{pmatrix}\). The cosine of the angle between two vectors is given by:
\[ \cos\theta = \frac{\overrightarrow{v} \cdot \overrightarrow{AB}}{|\overrightarrow{v}| |\overrightarrow{AB}|} \]

 Dot product \(\overrightarrow{v} \cdot \overrightarrow{AB}\):
\[ \overrightarrow{v} \cdot \overrightarrow{AB} = (1.1 \times 7.2) + (8.4 \times 5.1) + (0.2 \times -0.4) = 7.92 + 42.84 – 0.08 = 50.68 \]

 Magnitude of \(\overrightarrow{v}\):
\[ |\overrightarrow{v}| = \sqrt{(1.1)^2 + (8.4)^2 + (0.2)^2} = \sqrt{1.21 + 70.56 + 0.04} = \sqrt{71.81} \approx 8.471 \]

 Magnitude of \(\overrightarrow{AB}\):
From (a)(ii), \( |\overrightarrow{AB}| \approx 8.832 \).

 Cosine of the angle:
\[ \cos\theta = \frac{50.68}{8.471 \times 8.832} = \frac{50.68}{74.814} \approx 0.6773 \]

 Angle \(\theta\):
\[ \theta = \cos^{-1}(0.6773) \approx 47.36^\circ \] (using a calculator, rounded to 2 decimal places).
So, the angle is approximately \( 47.36^\circ \).

Part(c) Use the sine rule to find \( AC \):
We’re given the angle between \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\) as \( 55.2^\circ \). In triangle \( ABC \), this is the angle at \( B \) (since \(\overrightarrow{BA}\) goes from \( B \) to \( A \), and \(\overrightarrow{BC}\) goes from \( B \) to \( C \), the angle between them is \(\angle ABC\)). We need \( AC \), opposite \(\angle AB\).

 Vectors and directions:
 \(\overrightarrow{BA} = -\overrightarrow{AB} = \begin{pmatrix} -7.2 \\ -5.1 \\ 0.4 \end{pmatrix}\)
 \(\overrightarrow{BC} = \overrightarrow{C} – \overrightarrow{B}\), but we don’t have \( C \)’s coordinates yet.
 \(\overrightarrow{AC} = \overrightarrow{C} – \overrightarrow{A}\) is parallel to \(\overrightarrow{v} = \begin{pmatrix} 1.1 \\ 8.4 \\ 0.2 \end{pmatrix}\).

 Triangle setup:
 Side \( AB = 8.83 \) km (from (a)(ii)), opposite \(\angle AC\).
 Side \( AC \) is opposite \(\angle AB\), which is the angle between \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\), given as \( 55.2^\circ \).
 We need \(\angle BAC\) (angle at \( A \)) between \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\). Since \(\overrightarrow{AC} \parallel \overrightarrow{v}\), this is the angle from (b): \( 47.36^\circ \).

 Sine rule in \(\triangle ABC\):
\[ \frac{AC}{\sin(\angle AB)} = \frac{AB}{\sin(\angle AC)} \]
Here, \(\angle AB = 55.2^\circ\), \( AB = 8.83 \), and \(\angle AC\) is the angle at \( C \). Sum of angles in a triangle is \( 180^\circ \):
\[ \angle AC = 180^\circ – 47.36^\circ – 55.2^\circ = 77.44^\circ \]

 Apply sine rule:
\[ \frac{AC}{\sin 55.2^\circ} = \frac{8.83}{\sin 77.44^\circ} \]
\[ AC = 8.83 \times \frac{\sin 55.2^\circ}{\sin 77.44^\circ} \]
Using approximate values:
 \(\sin 55.2^\circ \approx 0.8192\)
 \(\sin 77.44^\circ \approx 0.9744\)
\[ AC = 8.83 \times \frac{0.8192}{0.9744} \approx 8.83 \times 0.8407 \approx 7.423 \]
So, \( AC \approx 7.42 \) km (rounded to 2 decimal places).

————Markscheme—————–

(a) (i) \begin{pmatrix}
7.2 \\
5.1 \\
2.4
\end{pmatrix} – \begin{pmatrix}
0 \\
0 \\
2.8
\end{pmatrix} = \begin{pmatrix}
7.2 \\
5.1 \\
-0.4
\end{pmatrix}

(ii) $$|\overrightarrow{AB}| = \sqrt{7.2^2 + 5.1^2 + (-0.4)^2} = 8.83 (km) (8.83232…)$$

(b) magnitude of $\begin{pmatrix}
1.1 \\
8.4 \\
0.2
\end{pmatrix}$ is $\sqrt{1.1^2 + 8.4^2 + 0.2^2} = 8.48 (8.47407…)$

EITHER
$\begin{pmatrix}
1.1 \\
8.4 \\
0.2
\end{pmatrix} \cdot \begin{pmatrix}
7.2 \\
5.1 \\
-0.4
\end{pmatrix} = 1.1 \times 7.2 + 8.4 \times 5.1 + 0.2 \times (-0.4) = 50.68$

$angle = arccos(\frac{50.68}{8.83232…\times8.47407…})$

OR

Attempt to find
$\begin{pmatrix}
1.1 \\
8.4 \\
0.2
\end{pmatrix} \times \begin{pmatrix}
7.2 \\
5.1 \\
-0.4
\end{pmatrix}$

$\begin{pmatrix}
1.1 \\
8.4 \\
0.2
\end{pmatrix} \times \begin{pmatrix}
7.2 \\
5.1 \\
-0.4
\end{pmatrix} = \sqrt{4.38^2 + 1.88^2 + 54.87^2} (= 55.0766…)$

$angle = arcsin(\frac{55.0766…}{8.83232…\times8.47407…})$

THEN

47.4° (47.3805…) OR 0.827 (0.826947…)

(c) using sum of angles in a triangle equals 180
$\hat{ACB} = 180 – 47.3805 – 55.2 = 77.4194…°$

$\frac{AC}{\sin(55.2)} = \frac{8.83232…}{\sin(77.4194…)}$

7.43 (km) (7.43107…)