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Precalculus / LEVEL 4 · DIFFICULTY 4/5

Matrices and Space

Measure orientation and describe planes.

3 stages · 24 practice problems · two 6-question assessment forms

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  1. MINI QUEST PC 4.1Determinants and AreaRead the lesson
  2. MINI QUEST PC 4.2Cross Products in SpaceRead the lesson
  3. MINI QUEST PC 4.3Planes and Normal VectorsRead the lesson
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STAGE PC 4.1 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Determinants and Area

Useful preparation: Dot Products

Goal: Understand and apply determinants and area.

Before you begin: Dot Products

Understand the idea

A 2×2 matrix sends the coordinate basis vectors to its columns. The signed area of their parallelogram is its determinant, describing area scaling and orientation.

det[[a,b],[c,d]]=ad−bc

Choose and carry out a method

Compute ad−bc for rows (a,b) and (c,d). Use the absolute value for ordinary area and preserve the sign when orientation is requested.

Check the reasoning

Swapping columns changes the sign. Proportional columns give zero determinant because the parallelogram collapses onto a line.

WORKED EXAMPLE 1

Find the determinant of the 2×2 matrix with rows (3,2) and (2,5).

  1. The determinant measures oriented area of the column parallelogram.
  2. det=3·5-2·2.
  3. The signed scale factor is 11; ordinary area scales by its absolute value.

11

WORKED EXAMPLE 2

Find the determinant of the 2×2 matrix with rows (4,3) and (2,6).

  1. The determinant measures oriented area of the column parallelogram.
  2. det=4·6-3·2.
  3. The signed scale factor is 18; ordinary area scales by its absolute value.

18

Common pitfalls

Possible mix-up: The determinant always equals unsigned area.

Its sign records orientation; unsigned area uses its absolute value.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

What geometric information does a zero determinant reveal?

Preview the eight practice prompts
  1. Find the determinant of the 2×2 matrix with rows (6,5) and (2,8).
  2. Find the determinant of the 2×2 matrix with rows (7,6) and (2,9).
  3. Find the determinant of the 2×2 matrix with rows (8,7) and (2,10).
  4. Find the determinant of the 2×2 matrix with rows (9,8) and (2,11).
  5. Find the determinant of the 2×2 matrix with rows (10,9) and (2,12).
  6. Find the determinant of the 2×2 matrix with rows (11,10) and (2,13).
  7. A linear transformation of the plane sends (1,0) to (12,2) and (0,1) to (11,14). Find its signed area scale factor. New context
  8. A linear transformation of the plane sends (1,0) to (13,2) and (0,1) to (12,15). Find its signed area scale factor. New context
Open stage PC 4.1 in the student workspace →

STAGE PC 4.2 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Cross Products in Space

Useful preparation: Determinants and Area

Goal: Understand and apply cross products in space.

Before you begin: Determinants and Area

Understand the idea

The cross product in three dimensions is perpendicular to both input vectors. Its direction follows their order, while its magnitude gives the area of the spanned parallelogram.

(u×v)_z=uₓvᵧ−uᵧvₓ

Choose and carry out a method

For the z-component, use uₓvᵧ−uᵧvₓ. Keep u×v in that order and enter only the component requested.

Check the reasoning

Reversing the vector order negates every component. A full computed cross product should have zero dot product with each input.

WORKED EXAMPLE 1

For u=(3,3,1) and v=(2,6,-1), find the z-component of u×v.

  1. Expand the cross product in the stated order.
  2. (u×v)z=uₓvᵧ-uᵧvₓ=3·6-3·2.
  3. The z-component is 12. Reversing the vectors reverses its sign.

12

WORKED EXAMPLE 2

For u=(4,3,1) and v=(2,7,-1), find the z-component of u×v.

  1. Expand the cross product in the stated order.
  2. (u×v)z=uₓvᵧ-uᵧvₓ=4·7-3·2.
  3. The z-component is 22. Reversing the vectors reverses its sign.

22

Common pitfalls

Possible mix-up: Reversing u and v preserves the cross product.

It preserves magnitude but reverses direction.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

How could dot products check a complete cross-product calculation?

Preview the eight practice prompts
  1. For u=(6,3,1) and v=(2,9,-1), find the z-component of u×v.
  2. For u=(7,3,1) and v=(2,10,-1), find the z-component of u×v.
  3. For u=(8,3,1) and v=(2,11,-1), find the z-component of u×v.
  4. For u=(9,3,1) and v=(2,12,-1), find the z-component of u×v.
  5. For u=(10,3,1) and v=(2,13,-1), find the z-component of u×v.
  6. For u=(11,3,1) and v=(2,14,-1), find the z-component of u×v.
  7. Two vectors u=(12,3,1) and v=(2,15,-1) span an oriented surface. Find the z-component of its area vector u×v. New context
  8. Two vectors u=(13,3,1) and v=(2,16,-1) span an oriented surface. Find the z-component of its area vector u×v. New context
Open stage PC 4.2 in the student workspace →

STAGE PC 4.3 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Planes and Normal Vectors

Useful preparation: Cross Products in Space

Goal: Understand and apply planes and normal vectors.

Before you begin: Cross Products in Space

Understand the idea

A normal vector is perpendicular to every displacement lying in a plane. Dotting the normal with the displacement from one known point therefore gives zero.

a(x−x₀)+b(y−y₀)+c(z−z₀)=0

Choose and carry out a method

Form n·(x−x₀,y−y₀,z−z₀)=0. Expand to a linear equation and substitute the requested horizontal coordinates to solve for height.

Check the reasoning

The given point must satisfy the plane equation. A nonzero z coefficient allows a unique height for each specified x and y.

WORKED EXAMPLE 1

A plane with normal (2,-3,4) passes through (3,4,5). At x=6 and y=2, what is z on the plane?

  1. A normal vector is perpendicular to every displacement within the plane.
  2. 2x-3y+4z=14. Substitute x=6, y=2.
  3. z=2. This value satisfies the same plane equation.

2

WORKED EXAMPLE 2

A plane with normal (2,-3,4) passes through (4,5,6). At x=7 and y=3, what is z on the plane?

  1. A normal vector is perpendicular to every displacement within the plane.
  2. 2x-3y+4z=17. Substitute x=7, y=3.
  3. z=3. This value satisfies the same plane equation.

3

Common pitfalls

Possible mix-up: The normal vector lies along the plane.

It points perpendicular to every direction within the plane.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Why do multiplying all plane coefficients by the same nonzero number describe the same plane?

Preview the eight practice prompts
  1. A plane with normal (2,-3,4) passes through (6,7,8). At x=9 and y=5, what is z on the plane?
  2. A plane with normal (2,-3,4) passes through (7,8,9). At x=10 and y=6, what is z on the plane?
  3. A plane with normal (2,-3,4) passes through (8,9,10). At x=11 and y=7, what is z on the plane?
  4. A plane with normal (2,-3,4) passes through (9,10,11). At x=12 and y=8, what is z on the plane?
  5. A plane with normal (2,-3,4) passes through (10,11,12). At x=13 and y=9, what is z on the plane?
  6. A plane with normal (2,-3,4) passes through (11,12,13). At x=14 and y=10, what is z on the plane?
  7. A flat ramp has normal (2,-3,4) and contains (12,13,14). Find its height z above the horizontal location (15,11). New context
  8. A flat ramp has normal (2,-3,4) and contains (13,14,15). Find its height z above the horizontal location (16,12). New context
Open stage PC 4.3 in the student workspace →
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