IB Mathematics AHL 3.10 Concept of a vector AI HL Paper 1- Exam Style Questions- New Syllabus
A vertical pole stands on a sloped platform. The bottom of the pole is used as the origin, O, of a coordinate system in which the top, F, of the pole has coordinates (0, 0, 5.8). All units are in metres.
The pole is held in place by ropes attached at F. One of these ropes is attached to the platform at point A(3.2, 4.5, −0.3). The rope forms a straight line from A to F.
(a) Find \(\overrightarrow{AF}\).
(b) Find the length of the rope.
(c) Find \(F \widehat{A} O\), the angle the rope makes with the platform.
▶️ Answer/Explanation
(a)
Compute vector: \(\overrightarrow{AF} = F – A\), where \( F = (0, 0, 5.8) \), \( A = (3.2, 4.5, -0.3) \)
Calculate: \( (0 – 3.2, 0 – 4.5, 5.8 – (-0.3)) = (-3.2, -4.5, 6.1) \) (A1)
Result: \(\overrightarrow{AF} = \begin{pmatrix} -3.2 \\ -4.5 \\ 6.1 \end{pmatrix}\) (A1) [2]
(b)
Magnitude: \( |\overrightarrow{AF}| = \sqrt{(-3.2)^2 + (-4.5)^2 + (6.1)^2} = \sqrt{10.24 + 20.25 + 37.21} = \sqrt{67.7} \) (M1)
Calculate: \( \sqrt{67.7} \approx 8.228 \approx 8.23 \) m (A1)
Result: 8.23 m [2]
(c)
Compute: \(\overrightarrow{AO} = O – A = (0 – 3.2, 0 – 4.5, 0 – (-0.3)) = (-3.2, -4.5, 0.3) \) (A1)
Dot product: \(\overrightarrow{AF} \cdot \overrightarrow{AO} = (-3.2) \times (-3.2) + (-4.5) \times (-4.5) + 6.1 \times 0.3 = 10.24 + 20.25 + 1.83 = 32.32 \) (A1)
Magnitudes: \( |\overrightarrow{AF}| \approx 8.228 \), \( |\overrightarrow{AO}| = \sqrt{(-3.2)^2 + (-4.5)^2 + (0.3)^2} = \sqrt{10.24 + 20.25 + 0.09} = \sqrt{30.58} \approx 5.5298 \) (M1)
Angle: \( \cos \theta = \frac{32.32}{8.228 \times 5.5298} \approx 0.710326 \), \( \theta = \cos^{-1}(0.710326) \approx 44.7384^\circ \approx 44.7^\circ \) (A1)
Result: \( 44.7^\circ \) [3]