Solve vector line intersections and justify their ratios
Represent one intersection by two directed paths, solve exact parameters using nonparallel planar directions and distinguish a line crossing from a segment crossing.
Exam scope and assessment notes
All nine lessons are additional Higher only: section5.1 has seven H5.1A–G rows and no Foundation vector rows. Retain directed endpoint order, signed and zero-vector exceptions, exact components and final-only rounding. Formal lists of two to four coefficients are planar position-vector identities, not independent spatial axes. Complete geometric arguments need their hypotheses and human review; numerical agreement or a text-entry attempt does not approve proof, marks or mastery. Linear4MA1 Issue2 is checked against November2026/June2027 entries; June regional options depend on centre eligibility and modular awards remain separate. Literal Book2 Shape and Space10 H5.1A–G and Hubs167/168/169 links are navigation only. Map427–450 conflicts with proof426/summary449/next450. The Vectors chapter and Hub bodies are unavailable; no Book1 tag, full-book, specialist or formal-assessment approval is inferred.
Before you begin
- Divide a directed segment in a ratio.
- Solve two linear equations with fractions.
- Use nonparallel vectors and line-versus-segment domains.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Why may coefficients of a and b be equated in these two-path arguments?
What you will learn
- Write both position-vector expressions for the same unknown point.
- Compare coefficients only in a justified nonparallel pair and solve exact parameters.
- Interpret parameter domains and unsigned ratios before making a segment claim.
OAB is a nondegenerate triangle, OA=a, OB=b. D lies on AB with AD:DB=2:1, and OE=(3/4)OB. Lines OD and AE meet at P. Find OP in a,b and the ratios OP:PD and AP:PE, proving the parameters place P on both segments.
Draw every requested ordered arrow and enter your own quantities, methods and reasons. Only the stated prompt supplies numerical facts. A symbolic figure is a planar illustration, not extra coordinate givens.
The planar x and y directions use equal drawing scale. Blue arrows are supplied operands; amber arrows are actual learner records; dashed amber is unfinished draft work. A point with coincident tail and head represents the zero vector and has no direction.
This is an illustrative planar pose. Its numerical coordinates, lengths and apparent angles are not additional givens. Formal arrays of two to four symbol coefficients are algebraic identities, never three- or four-dimensional geometry. Only the question states the actual hypotheses.
Supplied boundaries: line O–D; line A–E. A ray continues from its first anchor through its second; a line continues in both directions. A segment retains its two original endpoints. The viewport does not supply additional endpoints.
Actual ordered arrows, zero points and retained drafts do not certify written proofs, marks, mastery or formal assessment.
Construct my actual ordered OD arrow; retain both endpoints, including a coincident zero result.
No actual arrow recorded for this action.
A named pick uses the exact current supplied point or your current clean recorded endpoint. A free click records its labelled micro-unit grid point; exact keyboard fractions remain exact. No endpoint is snapped to a requested result or intersection. A zero vector has coincident endpoints and no defined nonzero direction.
My tail: not placed.
My head: not placed.
Place both your actual ordered tail and head. No endpoint is supplied from a reference. All actual points and raw strings remain yours; no endpoint is invented.
Choose my own illustrative planar source pose
This changes only an illustrative coordinate witness. Enter every permitted origin and source-vector component explicitly before Apply. These coordinates do not become new problem givens or alter formal coefficients. Requested resultant or intersection points are never pose inputs.
My source origin O
My planar source vector a
My planar source vector b
My quantities, directed method and reasons
Use a supported exact rational value or finite sum of single-radical terms where exactness is requested.
Use a supported exact rational value or finite sum of single-radical terms where exactness is requested.
Enter exactly 2 coefficients in the labelled order a, b, separated by semicolons. These are planar position-vector identities; the list is not a spatial coordinate system.
Enter the two unsigned length-ratio entries in their stated order, separated by a colon.
Enter the two unsigned length-ratio entries in their stated order, separated by a colon.
Choose Yes or No after checking the stated conditions.
Write your actual directed construction, substitutions, hypotheses and reason. A text attempt is not an automatically verified proof.
Write your actual directed construction, substitutions, hypotheses and reason. A text attempt is not an automatically verified proof.
This fixed mathematical tool retains actual planar geometry and requires your complete current attempt. It does not award marks, mastery, automatic proof, physical drawing competence, specialist signoff or formal assessment readiness. Formal coefficients name planar position-vector identities; they do not assert a three- or four-dimensional independent basis.
Give the same point two path expressions
For a point P on line OD, write OP=λOD. For the same point on line AE, write OP=a+μ(OE−a). Derive OD from the original division of AB and OE from its stated multiple of OB before expanding either expression; neither an apparent midpoint nor a drawn crossing replaces the supplied data.
When a and b are nonparallel, their coefficients in two equal planar combinations must match: otherwise a nonzero multiple of one would equal a multiple of the other, contradicting nonparallel directions. This justification permits a pair of scalar equations. It does not extend to treating three arbitrary planar position vectors as independent axes.
Solve exactly and substitute into both paths
Solve the two scalar equations together, keeping fractions exact. Substituting the recovered λ and μ into both vector expressions checks that they identify one point rather than two unrelated locations. Attach each parameter to its own path; interchanging λ and μ usually changes both the point and the ratios.
A nonparallel pair of original direction vectors a,b does not by itself guarantee that every pair of constructed lines has a unique crossing. The derived scalar system must also be consistent and determine the two path parameters. Parallel or coincident constructed lines would need their own no-unique-intersection conclusion; these fixed tasks have a uniquely solved intersection.
Interpret domains and ratios after the algebra
The point lies on closed segment OD only for 0≤λ≤1, and on closed segment AE only for 0≤μ≤1. For an internal point, OP:PD=λ:(1−λ) and AP:PE=μ:(1−μ). If P is beyond D, use OP:PD=|λ|:|1−λ| and state the actual point order rather than silently treating PD as a forward segment.
A line intersection beyond D can still lie on segment AE; it is not then an intersection of the two closed segments. Retain external parameters and the actual arrows. These original Higher vector arguments require explained working and human review; the literal book-map and Hub links supply navigation only, and no unseen book chapter, specialist review or formal assessment is established by a fixed mathematical comparison.
Sources for this lesson
- Pearson Mathematics A linear specification · Issue2 · Vectors: Higher-only Foundation statement printed23/physical29; Higher rows printed38/physical44
- Pearson 2026/2027 International GCSE entry manual · regional centres physical3 and Mathematics A physical25
- Pearson literal second-edition book/Hub map · Book2 Shape and Space10 physical15 · navigation only
Pause and explain
A uniquely solved point has OP=(5/4)OD and AP=(2/3)AE. Which domain conclusion is valid?
Worked example
OAB is a nondegenerate triangle, OA=a, OB=b. D lies on AB with AD:DB=2:1, and OE=(3/4)OB. Lines OD and AE meet at P. Find OP in a,b and the ratios OP:PD and AP:PE, proving the parameters place P on both segments.
Show the worked solution
- AD:DB=2:1 gives OD=a+(2/3)(b−a)=(1/3)a+(2/3)b; OE=(3/4)b.
- Two paths to P are OP=(λ/3)a+(2λ/3)b and OP=(1−μ)a+(3μ/4)b.
- Nonparallel a,b permit coefficient comparison: λ/3=1−μ and 2λ/3=3μ/4. The second gives μ=8λ/9; substituting into the first yields 11λ/9=1, so λ=9/11 and μ=8/11.
- Both paths give OP=(3/11)a+(6/11)b. Since both parameters lie strictly between 0 and 1, P is internal to OD and AE.
- OP:PD=(9/11):(2/11)=9:2, and AP:PE=(8/11):(3/11)=8:3. The parameter-domain checks justify the internal ratio interpretation.
Answer OP=(3/11)a+(6/11)b; λ=9/11, μ=8/11. P lies on both segments, with OP:PD=9:2 and AP:PE=8:3.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
OAB is nondegenerate, OA=a, OB=b. D divides AB in AD:DB=3:2 and OE=(2/3)OB. Lines OD and AE meet at P. Find OP, OP:PD and AP:PE, checking segment domains.
Give me a hint
Translate the ratio into a fraction of AB, then keep λ and μ attached to their own paths.
Compare my reasoning
- AD:DB=3:2 gives OD=(2/5)a+(3/5)b, while OE=(2/3)b.
- Equating the two paths λOD and a+μ(OE−a) gives 2λ/5=1−μ and 3λ/5=2μ/3, using the nonparallel basis.
- The second equation gives μ=9λ/10. Then 2λ/5+9λ/10=1 gives λ=10/13 and μ=9/13.
- Substituting gives OP=(4/13)a+(6/13)b. Both parameters are internal: OP:PD=10:3 and AP:PE=9:4.
OP=(4/13)a+(6/13)b; λ=10/13, μ=9/13. P lies on both segments, with OP:PD=10:3 and AP:PE=9:4.
Look for these in your work
- Write both position-vector expressions for the same unknown point.
- Compare coefficients only in a justified nonparallel pair and solve exact parameters.
- Interpret parameter domains and unsigned ratios before making a segment claim.
OAB is nondegenerate, OA=a, OB=b. AD:DB=3:1 and OE=(1/2)OB. Lines OD and AE meet at P. Derive OP, OP:PD and AP:PE without using a diagram measurement.
Give me a hint
Solve the two exact equations and check both vector paths before interpreting the ratios.
Compare my reasoning
- OD=a+(3/4)(b−a)=(1/4)a+(3/4)b and OE=(1/2)b.
- The paths give λ/4=1−μ and 3λ/4=μ/2. Hence μ=3λ/2 and λ/4+3λ/2=1, so λ=4/7 and μ=6/7.
- Both substitutions give OP=(1/7)a+(3/7)b. The nonparallel a,b condition justifies the coefficient equations.
- The internal parameters give OP:PD=(4/7):(3/7)=4:3 and AP:PE=(6/7):(1/7)=6:1. Neither ratio was read from a drawing.
OP=(1/7)a+(3/7)b; λ=4/7, μ=6/7. P is internal to both segments; OP:PD=4:3 and AP:PE=6:1.
Look for these in your work
- Write both position-vector expressions for the same unknown point.
- Compare coefficients only in a justified nonparallel pair and solve exact parameters.
- Interpret parameter domains and unsigned ratios before making a segment claim.
OAB is nondegenerate. D is the midpoint of AB and E lies on the ray OB with OE=2OB. Infinite lines OD and AE meet at P. Derive OP in a,b and locate P relative to O,D. Does the segment OD meet segment AE at P? Give OP:PD and AP:PE.
Give me a hint
Preserve the external line parameter and test each segment separately; do not redraw P onto OD.
Compare my reasoning
- D is the midpoint, so OD=(a+b)/2. The given ray point has OE=2b.
- The line paths λOD and a+μ(OE−a) give λ/2=1−μ and λ/2=2μ. Thus λ=4μ and 2μ=1−μ, giving μ=1/3 and λ=4/3.
- Substitution gives OP=(2/3)a+(2/3)b. Since λ>1, the actual order is O–D–P; P is beyond D. Since 0<μ<1, P is on segment AE.
- OP:PD=(4/3):(1/3)=4:1 and AP:PE=(1/3):(2/3)=1:2 as unsigned lengths. The infinite lines meet at P, but the closed segments OD and AE do not meet there because P is outside OD. No parameter is clamped to 1.
OP=(2/3)a+(2/3)b; λ=4/3, μ=1/3. Order O–D–P, OP:PD=4:1 and AP:PE=1:2. This line crossing is not a crossing of the two closed segments.
Look for these in your work
- Write both position-vector expressions for the same unknown point.
- Compare coefficients only in a justified nonparallel pair and solve exact parameters.
- Interpret parameter domains and unsigned ratios before making a segment claim.
Common mistakes
- Comparing coefficients without a nonparallel basis condition.
- Swapping the two path parameters or rounding an exact ratio.
- Clamping an external λ to the segment interval and claiming a different intersection.
What can you explain now?
If a solved intersection has λ=4/3 on OP=λOD, may λ be replaced by 1 to make P lie on segment OD?
Compare with the explanation
No. The original intersection lies beyond D; replacing λ changes the point and breaks the solved path equations.
Line parameters are not confined to [0,1]. Segment membership must be checked after solving, rather than enforced by moving the answer.
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Make it stick
Derive two paths to the same point, justify coefficient comparison, substitute exact parameters and explain why a valid line crossing may fail a segment-domain test.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.