IB Mathematics AHL 3.13 Definition and calculation of the scalar product-AI HL Paper 1- Exam Style Questions- New Syllabus

(a)
Using the distance formula:
\(AB = \sqrt{(3-1)^2 + (5-2)^2 + (2-1)^2}\)
\(AB = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1}\)
\(AB = \sqrt{14} \approx 3.74\)

(b)
First, find the position vectors relative to \(B\):
\(\vec{BA} = \begin{pmatrix} 1-3 \\ 2-5 \\ 1-2 \end{pmatrix} = \begin{pmatrix} -2 \\ -3 \\ -1 \end{pmatrix}\)
\(\vec{BC} = \begin{pmatrix} 2-3 \\ 8-5 \\ k-2 \end{pmatrix} = \begin{pmatrix} -1 \\ 3 \\ k-2 \end{pmatrix}\)

Since the triangle is right-angled at \(B\), the scalar product \(\vec{BA} \cdot \vec{BC} = 0\):
\((-2)(-1) + (-3)(3) + (-1)(k-2) = 0\)
\(2 – 9 – k + 2 = 0\)
\(-5 – k = 0\)
\(k = -5\)

(c)
We calculate \(\cos(B\hat{A}C)\) using the right-angled triangle properties or vectors.
Using right-angled trigonometry:
\(AC\) hypotenuse is not needed if we use \(\tan A\) or we can find \(AC\).
Let’s use vectors \(\vec{AB} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}\) and \(\vec{AC} = \begin{pmatrix} 2-1 \\ 8-2 \\ -5-1 \end{pmatrix} = \begin{pmatrix} 1 \\ 6 \\ -6 \end{pmatrix}\).
\(|\vec{AC}| = \sqrt{1^2 + 6^2 + (-6)^2} = \sqrt{1+36+36} = \sqrt{73}\)
\(|\vec{AB}| = \sqrt{14}\)
\(\cos(B\hat{A}C) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{14}}{\sqrt{73}}\)
\(B\hat{A}C = \cos^{-1}\left(\frac{\sqrt{14}}{\sqrt{73}}\right)\)
\(B\hat{A}C \approx 64.0^{\circ}\)