IB Maths AA HL
Topic 3 ā Geometry & Trigonometry
Paper 1 & 2
~7 min read
HL only
Intersections of Two Planes
Two non-parallel planes meet along a line (not a point) ā like two pages of an open book joining at the spine. Two methods to find it: algebra (set one variable as Ī», solve for the others) or the cross product (n1 Ć n2 gives the line’s direction).
š What you need to know
- Three cases: planes intersect at a line, are parallel and distinct (no intersection), or are coincident (same plane, infinite intersection).
- Parallel test: normal vectors are scalar multiples of each other.
- Same plane if both sides of the Cartesian equations scale by the same factor.
- Direction of intersection line: n1 Ć n2 (perpendicular to both normals).
- Algebra method: set one variable = Ī», solve the two equations for the other two variables in terms of Ī», write parametric form.
- Cross product method: take direction = n1 Ć n2; find a single shared point by setting one variable to 0 and solving the resulting 2Ć2 system.
- Multiple valid forms: the line equation can be written with different anchors and direction scalings ā all describe the same line.
The three cases
| Case | Normals | RHS scaling | Intersection |
|---|
| Intersect along a line | not scalar multiples | ā | a line |
| Parallel, distinct | scalar multiples | RHS doesn’t match | none |
| Same plane (coincident) | scalar multiples | RHS scales by the same factor | entire plane |
Two methods for the intersection line
Algebra method
set one variable = Ī»
solve the 2Ć2 system for the other two variables in terms of Ī»
Cross product method
direction = n1 Ć n2
find one shared point by setting a variable to 0; combine into r = a + Ī»b
Which to use: algebra is more reliable on the no-calculator paper. Cross product is faster if the resulting 2Ć2 system is easy to solve. Either method is correct.
š§ Recipe ā find line of intersection (algebra method)
- Confirm not parallel: check the normals aren’t scalar multiples.
- Set one variable = Ī» (any of x, y, z ā pick whichever simplifies).
- Solve the two equations simultaneously for the other two variables in terms of Ī».
- Write parametric form: x = ā¦, y = ā¦, z = ā¦ (each in terms of Ī»).
- Stack into r = a + Ī»b: read off the constants as the anchor and the Ī»-coefficients as the direction.
Worked examples
WE 1Find the line of intersection (algebra method)
Find a vector equation of the line of intersection of the planes Ī 1: 3x ā y + z = 7 and Ī 2: x + y + 3z = 5.
Step 1: Normals (3, ā1, 1) and (1, 1, 3) ā not scalar multiples ā not parallel ā
Step 2: Let z = Ī»
3x ā y = 7 ā Ī» … (1)
x + y = 5 ā 3Ī» … (2)
Step 3: Add (1) + (2)
4x = 12 ā 4Ī» ā x = 3 ā Ī»
Step 4: Sub into (2)
y = 5 ā 3Ī» ā x = 5 ā 3Ī» ā (3 ā Ī») = 2 ā 2Ī»
Step 5: Stack parametric ā vector form
r = (3, 2, 0) + Ī»(ā1, ā2, 1)
verify at Ī» = 1: (2, 0, 1) ā Ī ā: 6 ā 0 + 1 = 7 ā; Ī ā: 2 + 0 + 3 = 5 ā
WE 2Find the line of intersection (cross product method)
Find a vector equation of the line of intersection of the planes Ī 1: 2x + y ā z = 4 and Ī 2: x ā y + 2z = 5.
Step 1: Direction = nā Ć nā
i: (1)(2) ā (ā1)(ā1) = 1
j: ā[(2)(2) ā (ā1)(1)] = ā5
k: (2)(ā1) ā (1)(1) = ā3
Direction b = (1, ā5, ā3)
Step 2: Find a shared point ā set z = 0
2x + y = 4; x ā y = 5
Add: 3x = 9 ā x = 3, y = ā2
Step 3: Combine into vector form
r = (3, ā2, 0) + Ī»(1, ā5, ā3)
verify: at point, 2(3) + (ā2) ā 0 = 4 ā and 3 ā (ā2) + 0 = 5 ā
WE 3Show two planes are parallel and do not intersect
Show that the planes Ī 1: 2x ā y + 3z = 4 and Ī 2: 4x ā 2y + 6z = 5 are parallel and do not intersect.
Step 1: Compare normal vectors
nā = (2, ā1, 3); nā = (4, ā2, 6)
nā = 2 Ć nā ā ā normals scalar multiples ā parallel
Step 2: Compare RHS
For same plane, RHS would need to be 2 Ć 4 = 8, but Ī ā has 5
8 ā 5 ā planes are different
Parallel and distinct ā no intersection
two parallel non-coincident planes never meet, so the line of intersection doesn’t exist
WE 4Show two equations represent the same plane
Show that the equations 3x + 2y ā z = 5 and 6x + 4y ā 2z = 10 represent the same plane.
Step 1: Compare normals
(6, 4, ā2) = 2 Ć (3, 2, ā1) ā ā parallel
Step 2: Compare RHS
2 Ć 5 = 10 ā ā matches
Both equations represent the same plane
infinitely many points of “intersection” ā every point of the plane satisfies both equations
WE 5Find a value to make two planes parallel
Find the value of k for which the planes Ī 1: 2x ā y + 3z = 5 and Ī 2: 4x + ky + 6z = 7 are parallel. State whether they represent the same plane.
Step 1: For parallel, normals are scalar multiples
(4, k, 6) = c Ć (2, ā1, 3)
Step 2: Use known components to find c
From x: 4 = 2c ā c = 2
From z: 6 = 3c ā c = 2 ā
Step 3: Apply c to y-component
k = ā1 Ć c = ā2
Step 4: Same plane check
RHS would need to be 2 Ć 5 = 10, but Ī ā has 7
k = ā2; planes parallel but NOT the same plane
parallel and distinct ā no line of intersection
WE 6Find the line of intersection, then verify a point lies on it
(a) Find a vector equation of the line of intersection of Ī 1: 2x + y + z = 5 and Ī 2: x ā y + 2z = 4. (b) Show that the point P(2, 0, 1) lies on this line.
Part (a): Algebra method, let x = Ī»
y + z = 5 ā 2Ī» … (1)
āy + 2z = 4 ā Ī» … (2)
Add: 3z = 9 ā 3Ī» ā z = 3 ā Ī»
Sub (1): y = 5 ā 2Ī» ā (3 ā Ī») = 2 ā Ī»
r = (0, 2, 3) + Ī»(1, ā1, ā1)
Part (b): Test P(2, 0, 1)
x: 0 + Ī» = 2 ā Ī» = 2
y: 2 ā Ī» = 0 ā Ī» = 2 ā
z: 3 ā Ī» = 1 ā Ī» = 2 ā
P lies on the line (Ī» = 2)
P also satisfies Ī ā: 2(2)+0+1 = 5 ā and Ī ā: 2ā0+2 = 4 ā
š” Top tips
- Check parallel first by comparing normals ā saves time if there’s no line to find.
- Choose your Ī»-variable wisely ā pick the one that gives the simplest 2Ć2 system.
- Cross product gives a quick check: the direction of the line should be perpendicular to both normals.
- Multiple valid answers ā different anchors and scaled directions all describe the same line.
- Always sanity check by plugging a point on your line back into both plane equations.
ā Common mistakes
- Forgetting to check parallel first ā you’ll waste time deriving “the line” before realising none exists.
- Treating “normals are scalar multiples” as “same plane” ā must also check the RHS scales by the same factor.
- Sign errors when adding or subtracting the two plane equations.
- Computing the cross product backwards (j-component sign).
- Accepting “no solution” when expecting a line ā re-pick the parameter variable; sometimes one choice leads to inconsistency.
Next: Angles Between a Line & a Plane. The angle between a line and a plane is the angle between the line and its projection onto the plane ā found via cosā1 on the direction and the normal, then subtracted from 90Ā° (or Ļ/2). Same dot-product idea, with one extra step.
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