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.
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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]
