📝 Unit 7: Vector Applications — Unit Test

Dot / Cross Product · Work · Forces · Projections
✅ Graded
⏱️ 75 min  |  Total: /60 marks
K/U
/15
Thinking
/15
Comm.
/15
Applic.
/15
Part A: K/U [15]
1 [2]
Compute \(\vec u\cdot\vec v\) for \(\vec u=(3,-2,4)\), \(\vec v=(1,5,-2)\).
2 [3]
Find the angle between \(\vec u=(2,1,2)\) and \(\vec v=(0,3,4)\) (degrees).
3 [3]
Compute \(\vec u\times\vec v\) for \(\vec u=(1,2,3)\), \(\vec v=(4,5,6)\).
4 [2]
MC: \(\vec u\times\vec u =\)
5 [2]
Find scalar projection of \((4,3)\) onto \((1,1)\).
6 [3]
Find a vector perpendicular to both \((1,1,0)\) and \((0,1,1)\).
Part B: Thinking [15]
7 [5]
Find the area of triangle with vertices \(A(1,2,3)\), \(B(4,1,5)\), \(C(2,5,7)\) using a cross product.
8 [5]
Decompose \(\vec u=(3,4,5)\) into components parallel and perpendicular to \(\vec v=(1,0,0)\).
9 [5]
Verify the property \((\vec u+\vec v)\cdot\vec w=\vec u\cdot\vec w+\vec v\cdot\vec w\) using \(\vec u=(1,2,3),\,\vec v=(4,5,6),\,\vec w=(7,8,9)\).
Part C: Communication [15]
10 [4]
Compare and contrast the dot product and cross product. Address: input/output type, geometric meaning, commutativity, and one application of each.
11 [4]
Explain why \(\vec u\cdot\vec v=|\vec u||\vec v|\cos\theta\) follows from the cosine law applied to the triangle formed by \(\vec u\), \(\vec v\), and \(\vec u-\vec v\).
12 [4]
Describe how to use a vector projection to find the shortest distance from a point to a line in R². Give the formula.
13 [3]
Justify why \(\vec u\times\vec v\) is perpendicular to both \(\vec u\) and \(\vec v\) using the dot-product test.
Part D: Application [15]
14 [5]
A 200 N force is applied at 40° above horizontal to drag a sled 30 m horizontally. Calculate the work done.
15 [5]
A wrench of length 25 cm has a 60 N force applied to its end perpendicular to the handle. Calculate the magnitude of the torque (\(\tau=|\vec r\times\vec F|\)).
16 [5]
A 500 N crate sits on a frictionless ramp at 25° above horizontal. Find the component of gravity along the ramp (force tending to slide the crate down).

Evaluation Rubric

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

1. \(3-10-8=-15\).

2. \(\vec u\cdot\vec v=0+3+8=11\); \(|\vec u|=3,\,|\vec v|=5\); \(\cos\theta=11/15\) → \(\theta\approx 42.83°\).

3. \((2\cdot 6-3\cdot 5,\,3\cdot 4-1\cdot 6,\,1\cdot 5-2\cdot 4)=(-3,6,-3)\).

4. \(\vec 0\) → (b) (parallel vectors give zero cross).

5. \((4+3)/\sqrt 2=7/\sqrt 2\approx 4.95\).

6. \((1,1,0)\times(0,1,1)=(1,-1,1)\).

7. \(\vec{AB}=(3,-1,2)\), \(\vec{AC}=(1,3,4)\). \(\vec{AB}\times\vec{AC}=(-1\cdot 4-2\cdot 3,\,2\cdot 1-3\cdot 4,\,3\cdot 3-(-1)(1))=(-10,-10,10)\). Magnitude \(=10\sqrt 3\). Area \(=5\sqrt 3\approx 8.66\).

8. Parallel: \(\frac{\vec u\cdot\vec v}{|\vec v|^2}\vec v=3(1,0,0)=(3,0,0)\). Perpendicular: \(\vec u-(3,0,0)=(0,4,5)\).

9. LHS: \((5,7,9)\cdot(7,8,9)=35+56+81=172\). RHS: \(\vec u\cdot\vec w=7+16+27=50\); \(\vec v\cdot\vec w=28+40+54=122\); 50+122=172. ✓

10. Dot: vec×vec→scalar; commutative; gives \(|\vec u||\vec v|\cos\theta\); apps: work, angle. Cross: vec×vec→vec; anti-commutative; gives \(|\vec u||\vec v|\sin\theta\) perpendicular; apps: torque, area.

11. \(|\vec u-\vec v|^2=|\vec u|^2+|\vec v|^2-2\vec u\cdot\vec v\) (algebraic). Cosine law: \(=|\vec u|^2+|\vec v|^2-2|\vec u||\vec v|\cos\theta\). Equate → \(\vec u\cdot\vec v=|\vec u||\vec v|\cos\theta\).

12. Distance \(=\frac{|\vec{AP}\times\vec d|}{|\vec d|}\) where \(\vec d\) is the line's direction and \(\vec{AP}\) goes from a line point to \(P\); in R², use the perpendicular projection: \(d=|\vec{AP}-\text{proj}_{\vec d}\vec{AP}|\).

13. By component check: \(\vec u\cdot(\vec u\times\vec v)=u_1(u_2v_3-u_3v_2)+u_2(u_3v_1-u_1v_3)+u_3(u_1v_2-u_2v_1)=0\) (terms cancel). Same for \(\vec v\).

14. \(W=200\cdot 30\cdot\cos 40°\approx 6000\cdot 0.766\approx 4596\) J.

15. \(\tau=rF\sin 90°=0.25\cdot 60\cdot 1=15\) N·m.

16. \(F_{\parallel}=W\sin\theta=500\sin 25°\approx 500\cdot 0.4226\approx 211.3\) N down the ramp.