📝 Unit 6: Introduction to Vectors — Unit Test

R², R³, Operations, Magnitude
✅ Graded
⏱️ 75 min  |  Total: /60 marks
K/U
/15
Thinking
/15
Comm.
/15
Applic.
/15
Part A: K/U [15]
1 [2]
Find the magnitude of \(\vec u=(2,-3,6)\).
2 [2]
Compute \(3\vec a-2\vec b\) where \(\vec a=(1,4,-2),\,\vec b=(3,0,5)\).
3 [3]
Find the unit vector in the direction of \((4,-3)\).
4 [3]
A vector \(\vec v\) has magnitude 20 and makes an angle of 150° with the positive x-axis. Find its Cartesian components.
5 [2]
MC: Two vectors \((2,k,1)\) and \((4,-2,2)\) are parallel iff \(k=\)
6 [3]
Find vector \(\vec{AB}\) and \(|\vec{AB}|\) for \(A(1,2,3)\) and \(B(-2,5,7)\).
Part B: Thinking [15]
7 [5]
Two forces of magnitudes 12 N and 16 N act on a body. The angle between them is 60°. Find the magnitude of the resultant and the angle the resultant makes with the 12 N force.
8 [5]
Find scalars \(s,t\) such that \(s(1,2)+t(3,-1)=(11,3)\).
9 [5]
Show that \(\vec u=(2,-1,3)\), \(\vec v=(4,-2,6)\), and \(\vec w=(0,1,1)\) — determine if \(\vec u\) and \(\vec v\) are collinear, and whether \(\vec u\) and \(\vec w\) are coplanar (as position vectors from origin, three points + origin always coplanar with one vector — discuss with sketch).
Part C: Communication [15]
10 [4]
Define and contrast: position vector, displacement vector, free vector. Use a diagram-description.
11 [4]
Describe the parallelogram law and triangle law for vector addition. Give a real-world physics example for each.
12 [4]
Explain how to convert a polar form (magnitude, direction angle) to Cartesian form, and vice versa, in R². Include the formulas.
13 [3]
Justify why \(|\vec u+\vec v|\le |\vec u|+|\vec v|\) (the triangle inequality). When is equality achieved?
Part D: Application [15]
14 [5]
An airplane flies on a bearing of N40°E with airspeed 500 km/h. The wind blows from the west (from 270°) at 80 km/h. Find the ground velocity (speed and bearing).
15 [5]
A 100 kg crate is suspended by two cables: one at 30° above horizontal (left) and one at 45° above horizontal (right). Find the tension in each cable. (Use \(g=9.8\) m/s²; resolve forces.)
16 [5]
A river flows east at 3 m/s. A boat heads N30°E at 8 m/s relative to the water. Find the boat's actual ground velocity (magnitude and direction).

Evaluation Rubric

LevelDescription%
4Thorough80–100
3Considerable70–79
2Some60–69
1Limited50–59
RInsufficient<50
📕 Answer Key

1. \(\sqrt{4+9+36}=7\).

2. \((3,12,-6)-(6,0,10)=(-3,12,-16)\).

3. \(|\vec v|=5\); \(\hat v=(0.8,-0.6)\).

4. \(v_x=20\cos 150°=-10\sqrt 3\); \(v_y=20\sin 150°=10\); \(\vec v=(-10\sqrt 3,10)\).

5. Parallel iff scalar multiple. Ratios: \(4/2=2,\,2/1=2\), so middle: \(-2/k=2\) → \(k=-1\) → (b).

6. \(\vec{AB}=(-3,3,4)\); \(|\vec{AB}|=\sqrt{9+9+16}=\sqrt{34}\approx 5.83\).

7. \(|\vec R|^2=12^2+16^2+2(12)(16)\cos 60°=144+256+192=592\); \(|\vec R|\approx 24.33\) N. Angle with 12 N: \(\sin\alpha/16=\sin 60°/24.33\) → \(\alpha\approx 34.7°\).

8. System: \(s+3t=11,\ 2s-t=3\). From first: \(s=11-3t\); sub: \(22-6t-t=3\) → \(t=19/7\)? Re-check: actually \(2(11-3t)-t=3\) → \(22-6t-t=3\) → \(7t=19\) → \(t=19/7\approx 2.71\); \(s=11-3(19/7)=(77-57)/7=20/7\approx 2.86\).

9. \(\vec v=2\vec u\) → collinear. Three position vectors plus origin lie in 3-D; \(\vec u\) and \(\vec w\) span a plane through origin (always coplanar as 2 vectors).

10. Position: tail at origin. Displacement: tail at one point, head at another (specific). Free: equivalence class — same magnitude/direction regardless of location.

11. Parallelogram: place tails together; resultant = diagonal. Triangle: head-to-tail; resultant from first tail to last head. Examples: forces (parallelogram), successive displacements (triangle).

12. Polar → Cartesian: \((|\vec v|\cos\theta,|\vec v|\sin\theta)\). Cartesian → polar: \(|\vec v|=\sqrt{x^2+y^2}\), \(\theta=\arctan(y/x)\) with quadrant adjust.

13. Geometrically: shortest path between two points is the direct one (the third side of a triangle is at most the sum of the other two). Equality when \(\vec u,\vec v\) are parallel and same direction.

14. Plane velocity: \(500(\sin 40°,\cos 40°)\approx(321.4,383.0)\). Wind from west = blowing east: \((80,0)\). Ground = sum: \((401.4,383.0)\); \(|v|\approx \sqrt{161122+146689}\approx 555\) km/h. Bearing \(\arctan(401.4/383.0)\approx 46.4°\) → N46.4°E.

15. Weight = 980 N down. Equilibrium: \(T_1\cos 30°=T_2\cos 45°\) (horizontal); \(T_1\sin 30°+T_2\sin 45°=980\). From first: \(T_1=T_2\cos 45°/\cos 30°=T_2\sqrt{2/3}\). Sub: \(T_2\sqrt{2/3}(0.5)+T_2(0.707)=980\) → \(T_2(0.408+0.707)=980\) → \(T_2\approx 879\) N; \(T_1\approx 718\) N.

16. Boat: \(8(\sin 30°,\cos 30°)=(4,6.93)\) (E,N). + River \((3,0)\): \((7,6.93)\). Speed: \(\sqrt{49+48}\approx 9.85\) m/s. Bearing: \(\arctan(7/6.93)\approx 45.3°\) → N45.3°E.