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Plane vectors and applications: mixed assessment

Read the idea, work independently, then explain what changed.

TOPIC 01

Plane vectors and applications: mixed assessment

A mixed assessment: identify the method, justify it and revise your reasoning.

What you will be able to explain

  • Connect the chapter skills without relying on the order of the exercises.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Foundation#Your turn

Find the vector magnitude.

a=(21,28)\mathbf a=(21,28)
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Separate magnitude, direction and position; compare vectors by displacement.
Hint 2
Use the distance from the origin in coordinate space.
Worked solution
  1. Separate magnitude, direction and position; compare vectors by displacement.

  2. Calculate or simplify this relation.

    ∣a∣=(21)2+(28)2=35|\mathbf a|=\sqrt{(21)^2+(28)^2}=35
  3. The magnitude is a nonnegative scalar.

The requested value is 35.

Checks and common pitfalls: The magnitude is a nonnegative scalar.

Think first. Reveal a hint when the class is ready.

02 / Foundation#Your turn

Calculate the dot product.

a=(7,2),b=(3,7)\mathbf a=(7,2),\quad\mathbf b=(3,7)
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use the vector operation law and preserve vector versus scalar types.
Hint 2
Multiply corresponding coordinates and add.
Worked solution
  1. Use the vector operation law and preserve vector versus scalar types.

  2. Calculate or simplify this relation.

    a⋅b=3(7)+2(7)=35\mathbf a\cdot\mathbf b=3(7)+2(7)=35
  3. The result is a scalar, not a vector.

The requested value is 35.

Checks and common pitfalls: The result is a scalar, not a vector.

Think first. Reveal a hint when the class is ready.

03 / Foundation#Your turn

Find m for the two vectors to be parallel.

a=(2,3),b=(14,m)\mathbf a=(2,3),\quad\mathbf b=(14,m)
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use a nonparallel basis or Cartesian coordinates and solve component equations.
Hint 2
Compare the first coordinate to find the scalar multiple.
Worked solution
  1. Use a nonparallel basis or Cartesian coordinates and solve component equations.

  2. Calculate or simplify this relation.

    b=7a⇒m=21\mathbf b=7\mathbf a\Rightarrow m=21
  3. The second coordinate must use the same multiplier.

The requested value is 21.

Checks and common pitfalls: The second coordinate must use the same multiplier.

Think first. Reveal a hint when the class is ready.

04 / Foundation#Your turn

Find the x-coordinate of the centroid.

A=(0,0),B=(21,0),C=(0,3)A=(0,0),\quad B=(21,0),\quad C=(0,3)
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometry or physical quantity into vectors and check angle conventions.
Hint 2
Average the three vertex position vectors.
Worked solution
  1. Translate the geometry or physical quantity into vectors and check angle conventions.

  2. Calculate or simplify this relation.

    G=(A+B+C)/3=(7,1)G=(A+B+C)/3=(7,1)
  3. The centroid lies two-thirds along each median from the vertex.

The requested value is 7.

Checks and common pitfalls: The centroid lies two-thirds along each median from the vertex.

Think first. Reveal a hint when the class is ready.

05 / Foundation#Your turn

Give a unit vector in the same direction.

a=(24,32)\mathbf a=(24,32)
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Separate magnitude, direction and position; compare vectors by displacement.
Hint 2
Divide by the positive magnitude.
Worked solution
  1. Separate magnitude, direction and position; compare vectors by displacement.

  2. Calculate or simplify this relation.

    a∣a∣=(3/5,4/5)\frac{\mathbf a}{|\mathbf a|}=(3/5,4/5)
  3. Normalizing a zero vector would be undefined.

(3/5,4/5).

Checks and common pitfalls: Normalizing a zero vector would be undefined.

Reasoning checklist · self / teacher assessment
  • State the relevant definition, condition or model.
  • Show a valid calculation, proof or counterexample.
  • Interpret the conclusion with its restrictions.

Think first. Reveal a hint when the class is ready.

06 / Foundation#Your turn

Find the angle in degrees between these nonzero vectors.

a=(8,0),b=(0,9)\mathbf a=(8,0),\quad\mathbf b=(0,9)
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use the vector operation law and preserve vector versus scalar types.
Hint 2
A zero dot product gives perpendicular nonzero vectors.
Worked solution
  1. Use the vector operation law and preserve vector versus scalar types.

  2. Calculate or simplify this relation.

    a⋅b=0⇒cos⁡θ=0\mathbf a\cdot\mathbf b=0\Rightarrow\cos\theta=0
  3. Nonzero assumptions are required to define the angle.

The requested value is 90.

Checks and common pitfalls: Nonzero assumptions are required to define the angle.

Think first. Reveal a hint when the class is ready.

07 / Standard#Your turn

Find the signed scalar projection on the positive x-axis.

a=(−8,3)\mathbf a=(-8,3)
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use a nonparallel basis or Cartesian coordinates and solve component equations.
Hint 2
Dot with a unit vector along the target direction.
Worked solution
  1. Use a nonparallel basis or Cartesian coordinates and solve component equations.

  2. Calculate or simplify this relation.

    a⋅(1,0)=−8\mathbf a\cdot(1,0)=-8
  3. A signed projection may be negative even though a length is nonnegative.

The requested value is -8.

Checks and common pitfalls: A signed projection may be negative even though a length is nonnegative.

Think first. Reveal a hint when the class is ready.

08 / Standard#Your turn

Solve both parts and justify the conditions used. Part A: Explain why the sine rule in an SSA problem may give two triangles. Part B: Do equal magnitudes guarantee equal vectors? Give a counterexample.

A: sin⁡B=s,0<s<1B: ∣a∣=∣b∣=10\begin{gathered}\text{A: }\sin B=s,\quad0<s<1\\\text{B: }|\mathbf a|=|\mathbf b|=10\end{gathered}
  • Use the domain, units and sampling assumptions stated in the question.
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
A: Translate the geometry or physical quantity into vectors and check angle conventions.
Hint 2
B: Separate magnitude, direction and position; compare vectors by displacement.
Worked solution
  1. Part A reasoning

  2. Translate the geometry or physical quantity into vectors and check angle conventions.

  3. Calculate or simplify this relation.

    sin⁡B=sin⁡(180∘−B)\sin B=\sin(180^{\circ}-B)
  4. A supplementary candidate is discarded if the angle sum makes it impossible.

  5. Part B reasoning

  6. Separate magnitude, direction and position; compare vectors by displacement.

  7. Calculate or simplify this relation.

    ∣(10,0)∣=∣(0,10)∣=10|(10,0)|=|(0,10)|=10
  8. The two perpendicular nonzero vectors cannot be equal.

A: Both B and 180°−B have the same sine; each must be checked against the remaining angle and side conditions. B: No: (10,0) and (0,10) have different directions.

Checks and common pitfalls: A supplementary candidate is discarded if the angle sum makes it impossible. The two perpendicular nonzero vectors cannot be equal.

Reasoning checklist · self / teacher assessment
  • Solve part A with its stated restrictions.
  • Solve part B using an appropriate representation.
  • Give the reasoning and check conditions in both parts.

Think first. Reveal a hint when the class is ready.

09 / Standard#Your turn

A walker travels 9 m east then 9 m west. Find the magnitude of displacement.

  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Separate magnitude, direction and position; compare vectors by displacement.
Hint 2
Displacement depends only on start and finish.
Worked solution
  1. Separate magnitude, direction and position; compare vectors by displacement.

  2. Calculate or simplify this relation.

    (9,0)+(−9,0)=(0,0)(9,0)+(-9,0)=(0,0)
  3. The distance travelled is 18 m, although displacement is zero.

The requested value is 0.

Checks and common pitfalls: The distance travelled is 18 m, although displacement is zero.

Think first. Reveal a hint when the class is ready.

10 / Standard#Your turn

Find the x-coordinate of a+b.

a=(9,2),b=(3,−1)\mathbf a=(9,2),\quad\mathbf b=(3,-1)
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use the vector operation law and preserve vector versus scalar types.
Hint 2
Add corresponding coordinates.
Worked solution
  1. Use the vector operation law and preserve vector versus scalar types.

  2. Calculate or simplify this relation.

    a+b=(12,1)\mathbf a+\mathbf b=(12,1)
  3. Vector addition combines displacements component by component.

The requested value is 12.

Checks and common pitfalls: Vector addition combines displacements component by component.

Think first. Reveal a hint when the class is ready.

11 / Standard#Your turn

Find the midpoint x-coordinate.

A=(9,1),B=(13,5)A=(9,1),\quad B=(13,5)
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use a nonparallel basis or Cartesian coordinates and solve component equations.
Hint 2
Average the corresponding endpoint coordinates.
Worked solution
  1. Use a nonparallel basis or Cartesian coordinates and solve component equations.

  2. Calculate or simplify this relation.

    M=A+B2=(11,3)M=\frac{A+B}{2}=(11,3)
  3. A midpoint is an affine combination with coefficients summing to one.

The requested value is 11.

Checks and common pitfalls: A midpoint is an affine combination with coefficients summing to one.

Think first. Reveal a hint when the class is ready.

12 / Standard#Your turn

Find the triangle area.

a=18,b=3,C=30∘a=18,\quad b=3,\quad C=30^{\circ}
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometry or physical quantity into vectors and check angle conventions.
Hint 2
Use half the product of two sides and sine of their included angle.
Worked solution
  1. Translate the geometry or physical quantity into vectors and check angle conventions.

  2. Calculate or simplify this relation.

    S=12absin⁡C=12(18)(3)(1/2)=13.5S=\frac12ab\sin C=\frac12(18)(3)(1/2)=13.5
  3. The angle must be the included angle, not an arbitrary triangle angle.

The requested value is 13.5.

Checks and common pitfalls: The angle must be the included angle, not an arbitrary triangle angle.

Think first. Reveal a hint when the class is ready.

13 / Transfer#Your turn

Do equal magnitudes guarantee equal vectors? Give a counterexample.

∣a∣=∣b∣=10|\mathbf a|=|\mathbf b|=10
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Separate magnitude, direction and position; compare vectors by displacement.
Hint 2
Equality requires both length and direction.
Worked solution
  1. Separate magnitude, direction and position; compare vectors by displacement.

  2. Calculate or simplify this relation.

    ∣(10,0)∣=∣(0,10)∣=10|(10,0)|=|(0,10)|=10
  3. The two perpendicular nonzero vectors cannot be equal.

No: (10,0) and (0,10) have different directions.

Checks and common pitfalls: The two perpendicular nonzero vectors cannot be equal.

Reasoning checklist · self / teacher assessment
  • State the relevant definition, condition or model.
  • Show a valid calculation, proof or counterexample.
  • Interpret the conclusion with its restrictions.

Think first. Reveal a hint when the class is ready.

14 / Transfer#Your turn

Solve both parts and justify the conditions used. Part A: Simplify using the triangle rule. Part B: Find the x-coordinate of the centroid.

A: AB→+BC→+CA→B: A=(0,0),B=(21,0),C=(0,3)\begin{gathered}\text{A: }\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}\\\text{B: }A=(0,0),\quad B=(21,0),\quad C=(0,3)\end{gathered}
  • Use the domain, units and sampling assumptions stated in the question.
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
A: Use the vector operation law and preserve vector versus scalar types.
Hint 2
B: Translate the geometry or physical quantity into vectors and check angle conventions.
Worked solution
  1. Part A reasoning

  2. Use the vector operation law and preserve vector versus scalar types.

  3. Calculate or simplify this relation.

    AB→+BC→=AC→;AC→+CA→=0\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC};\quad\overrightarrow{AC}+\overrightarrow{CA}=\mathbf0
  4. Returning to the start gives zero total displacement.

  5. Part B reasoning

  6. Translate the geometry or physical quantity into vectors and check angle conventions.

  7. Calculate or simplify this relation.

    G=(A+B+C)/3=(7,1)G=(A+B+C)/3=(7,1)
  8. The centroid lies two-thirds along each median from the vertex.

A: The zero vector. B: The requested value is 7.

Checks and common pitfalls: Returning to the start gives zero total displacement. The centroid lies two-thirds along each median from the vertex.

Reasoning checklist · self / teacher assessment
  • Solve part A with its stated restrictions.
  • Solve part B using an appropriate representation.
  • Give the reasoning and check conditions in both parts.

Think first. Reveal a hint when the class is ready.

15 / Transfer#Your turn

Solve both parts and justify the conditions used. Part A: Explain why parallel vectors cannot form a basis for the plane. Part B: Give a unit vector in the same direction.

A: e2=2e1B: a=(24,32)\begin{gathered}\text{A: }\mathbf e_2=2\mathbf e_1\\\text{B: }\mathbf a=(24,32)\end{gathered}
  • Use the domain, units and sampling assumptions stated in the question.
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
A: Use a nonparallel basis or Cartesian coordinates and solve component equations.
Hint 2
B: Separate magnitude, direction and position; compare vectors by displacement.
Worked solution
  1. Part A reasoning

  2. Use a nonparallel basis or Cartesian coordinates and solve component equations.

  3. Calculate or simplify this relation.

    ae1+be2=(a+2b)e1a\mathbf e_1+b\mathbf e_2=(a+2b)\mathbf e_1
  4. They cannot represent a vector outside that line.

  5. Part B reasoning

  6. Separate magnitude, direction and position; compare vectors by displacement.

  7. Calculate or simplify this relation.

    a∣a∣=(3/5,4/5)\frac{\mathbf a}{|\mathbf a|}=(3/5,4/5)
  8. Normalizing a zero vector would be undefined.

A: All their linear combinations stay on one line. B: (3/5,4/5).

Checks and common pitfalls: They cannot represent a vector outside that line. Normalizing a zero vector would be undefined.

Reasoning checklist · self / teacher assessment
  • Solve part A with its stated restrictions.
  • Solve part B using an appropriate representation.
  • Give the reasoning and check conditions in both parts.

Think first. Reveal a hint when the class is ready.

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