Vectors, Lines & Planes: one small step at a time

Five topics, matching the five questions on the test (plane, parallel line, volume, proof, intersect/skew/parallel). Tap a card to open it. Tick the box when you're done with it.

First, a breath. Nothing here is new magic. Almost every problem on the list uses the same two moves: subtract points to get a vector, and take a cross product to get a perpendicular vector. If you know those two, you already know most of the test.

Short on time? For each topic, read only the one-sentence idea and the steps. Open the example only if you want to see it done.

1 Proofs 2 Area & volume 3 Lines 4 Skew or not 5 Planes

Inside each card: colored = the main idea and steps, yellow = watch out, gray = optional examples.

0 of 5 topics done
1Proving dot and cross product propertiesAbout 5 minutes▸
The idea: don't be clever. Write every vector in components, expand both sides, and show they match.

Know these two cold

a · b = a₁b₁ + a₂b₂ + a₃b₃ (a number)
a × b = ⟨ a₂b₃ − a₃b₂ , a₃b₁ − a₁b₃ , a₁b₂ − a₂b₁ ⟩ (a vector)

Steps for any proof

  1. Write each vector in components.
  2. Expand the left side using the definition.
  3. Expand the right side separately.
  4. If they match, write ∎ (it just means "done").
  5. If they don't look alike, regroup terms (for example, collect by c₁, c₂, c₃) and compare again.
Example: a · (b + c) = a · b + a · c
  1. Add inside: b + c = ⟨b₁+c₁, b₂+c₂, b₃+c₃⟩
  2. Dot product: a₁(b₁+c₁) + a₂(b₂+c₂) + a₃(b₃+c₃)
  3. Distribute: a₁b₁ + a₁c₁ + a₂b₂ + a₂c₂ + a₃b₃ + a₃c₃
  4. Regroup: (a₁b₁ + a₂b₂ + a₃b₃) + (a₁c₁ + a₂c₂ + a₃c₃) = a · b + a · c ∎
Example: a · (b × c) = (a × b) · c
  1. Expand the left: a₁(b₂c₃ − b₃c₂) + a₂(b₃c₁ − b₁c₃) + a₃(b₁c₂ − b₂c₁)
  2. Regroup by c₁, c₂, c₃: (a₂b₃ − a₃b₂)c₁ + (a₃b₁ − a₁b₃)c₂ + (a₁b₂ − a₂b₁)c₃
  3. That is ⟨components of a × b⟩ · ⟨c₁, c₂, c₃⟩ = (a × b) · c ∎
Test proof (fastest way): (a − b) × (a + b) = 2(a × b)
  1. Distribute: a × (a + b) − b × (a + b)
  2. Distribute again: a × a + a × b − b × a − b × b
  3. Drop the self-products (a × a = 0, b × b = 0): a × b − b × a
  4. Flip: b × a = −(a × b), so we get a × b + a × b = 2(a × b) ∎

Components backup: each component of the left side works out to 2 times the matching component of a × b. For example, the first is (a₂−b₂)(a₃+b₃) − (a₃−b₃)(a₂+b₂) = 2(a₂b₃ − a₃b₂).

⚠ Sign trap: (a − b) × (a + b) = +2(a × b), but (a + b) × (a − b) = −2(a × b). Check which factor comes first. Also, cross product is not commutative: a × b = −(b × a).
2Area of a triangle or rectangle, volume of a parallelepipedAbout 5 minutes▸
The idea: the length of a cross product is an area, and the triple product is a volume.
Parallelogram|a × b|
Triangle½ |a × b|
Rectangle (perpendicular sides)|a| |b|
Parallelepiped|a · (b × c)|

Steps: triangle from three points

  1. Edge vectors from P: PQ = Q − P, PR = R − P.
  2. Cross product PQ × PR.
  3. Find its length.
  4. Multiply by ½.

Steps: volume

  1. Put the three vectors in the rows of a 3×3 determinant.
  2. Expand along row 1 with signs + − +.
  3. Take the absolute value. If it's 0, the vectors are coplanar.
Example: triangle P(1,4,6), Q(−2,5,−1), R(1,−1,1)
  1. PQ = ⟨−3, 1, −7⟩, PR = ⟨0, −5, −5⟩
  2. PQ × PR = ⟨(1)(−5) − (−7)(−5), (−7)(0) − (−3)(−5), (−3)(−5) − (1)(0)⟩ = ⟨−40, −15, 15⟩
  3. Length: √(1600 + 225 + 225) = √2050 = 5√82
  4. Triangle: (5/2)√82
Example: are ⟨1,4,−7⟩, ⟨2,−1,4⟩, ⟨0,−9,18⟩ coplanar?
  1. 1[(−1)(18) − (4)(−9)] = 1(18) = 18
  2. −4[(2)(18) − (4)(0)] = −4(36) = −144
  3. −7[(2)(−9) − (−1)(0)] = −7(−18) = 126
  4. 18 − 144 + 126 = 0, so they are coplanar.
⚠ Don't forget the ½ for triangles and the absolute value for volume.
3Equation of a lineAbout 4 minutes▸
The idea: a line needs a point and a direction. Finding those two things is the whole problem.
x = x₀ + at, y = y₀ + bt, z = z₀ + ct (point (x₀,y₀,z₀), direction ⟨a,b,c⟩)

If you're given two points P and Q

  1. Direction = PQ = Q − P.
  2. Use P as the point.
  3. Fill in the template.

If it must be perpendicular to two vectors a and b

  1. Direction = a × b. (The cross product is perpendicular to both.)
  2. Use the point you were given.
  3. Check: direction · a = 0 and direction · b = 0.
Example: through P(1,2,3) and Q(4,0,5)
  1. Direction: ⟨3, −2, 2⟩
  2. x = 1 + 3t, y = 2 − 2t, z = 3 + 2t
  3. Check t = 1: (4, 0, 5) = Q ✓
Example: through (2,1,−1), perpendicular to ⟨1,2,0⟩ and ⟨0,1,3⟩
  1. Cross: ⟨(2)(3) − (0)(1), (0)(0) − (1)(3), (1)(1) − (2)(0)⟩ = ⟨6, −3, 1⟩
  2. x = 2 + 6t, y = 1 − 3t, z = −1 + t
  3. Check: 6 − 6 + 0 = 0 ✓ and 0 − 3 + 3 = 0 ✓

If it must be parallel to a line in symmetric form

  1. Read the direction from the denominators of (x − x₀)/a = (y − y₀)/b = (z − z₀)/c. Direction = ⟨a, b, c⟩.
  2. Use the new point you were given, not the old line's point.
  3. Fill in the template.
Example: through (11,22,33), parallel to (x−1)/2 = (y+3)/(−4) = (z−5)/6
  1. Direction from the denominators: ⟨2, −4, 6⟩
  2. Point: (11, 22, 33)
  3. x = 11 + 2t, y = 22 − 4t, z = 33 + 6t
⚠ Subtract endpoint minus start. A flipped sign flips the whole direction. For symmetric form, use the denominators (not the numerators) as the direction.
4Do two lines intersect or are they skew?About 5 minutes▸
The idea: ask "is there one point on both lines?" Yes means intersecting. No (and not parallel) means skew.

Steps

  1. Rename the parameters: t for line 1, s for line 2.
  2. Parallel? Is one direction a multiple of the other? If yes, stop: parallel.
  3. Set x = x, y = y, z = z. That gives three equations.
  4. Solve two of them for t and s.
  5. Plug t and s into the third. Works? Intersect. Fails? Skew.
Example (skew): L₁ = (1+t, −2+3t, 4−t), L₂ = (2s, 3+s, −3+4s)
  1. Directions ⟨1,3,−1⟩ and ⟨2,1,4⟩ are not multiples.
  2. 1 + t = 2s, and −2 + 3t = 3 + s
  3. From the first, t = 2s − 1. Into the second: 5s = 8, so s = 8/5, t = 11/5.
  4. Third equation: 4 − t = 9/5 but −3 + 4s = 17/5. Not equal, so skew.
Example (intersect): L₁ = (1+t, 2+t, 3+2t), L₂ = (2s, 5−2s, 4+s)
  1. Not parallel.
  2. 1 + t = 2s, and 2 + t = 5 − 2s. Subtract: 1 = 5 − 4s, so s = 1, t = 1.
  3. Third: 3 + 2(1) = 5 and 4 + 1 = 5 ✓
  4. Intersect at (2, 3, 5).
Example (parallel): L₁ = (1+2t, 2−t, 3+4t), L₂ = (5−4s, 1+2s, −5−8s)
  1. Directions: ⟨2, −1, 4⟩ and ⟨−4, 2, −8⟩. Since ⟨−4, 2, −8⟩ = −2⟨2, −1, 4⟩, the lines are parallel. Stop here.
  2. Only if asked "same line or different?": test L₁'s point (1, 2, 3) in L₂. x: 1 = 5 − 4s gives s = 1, then y = 3, but L₁'s y is 2. Not equal, so they are distinct parallel lines.
⚠ Do the parallel test first (even a negative multiple counts). Use two different letters (t and s), and always test the third equation.
5Equation of a plane through three pointsAbout 4 minutes▸
The idea: a plane needs a point and a normal (a vector sticking straight out of it). The cross product of two edges gives the normal.
a(x − x₀) + b(y − y₀) + c(z − z₀) = 0 with normal ⟨a, b, c⟩

Steps

  1. PQ = Q − P and PR = R − P.
  2. Normal n = PQ × PR. (Simplify by dividing out a common factor if you like.)
  3. Plug n and P into the formula above.
  4. Expand to ax + by + cz = d.
  5. Check all three points in your answer.
Example: P(1,0,0), Q(0,2,0), R(0,0,3)
  1. PQ = ⟨−1, 2, 0⟩, PR = ⟨−1, 0, 3⟩
  2. Normal = ⟨6, 3, 2⟩
  3. 6(x − 1) + 3y + 2z = 0, so 6x + 3y + 2z = 6
  4. Check: 6 ✓, 3(2) = 6 ✓, 2(3) = 6 ✓
Example: your triangle P(1,4,6), Q(−2,5,−1), R(1,−1,1)
  1. PQ × PR = ⟨−40, −15, 15⟩. Divide by −5: n = ⟨8, 3, −3⟩
  2. 8(x − 1) + 3(y − 4) − 3(z − 6) = 0, so 8x + 3y − 3z = 2
  3. Check Q: −16 + 15 + 3 = 2 ✓ and R: 8 − 3 − 3 = 2 ✓
⚠ If the cross product comes out as ⟨0, 0, 0⟩, the three points are on one line, so no single plane exists.

One last thing. Making a small mistake on the test doesn't mean you don't know the material. If you freeze, write down the definition or the formula for what the problem asks about. That alone often unlocks the next step, and it usually earns partial credit too.

You've prepared. Good luck.