01 Coordinates, bases and division Convert arrows to coordinates, recover unknown endpoints and test when a basis actually determines coordinates. ↗ Question directory1. Foundation · Find the vector AB and explain its signs. 2. Standard · Express v as ae₁+be₂. 3. Transfer · Prove A,B,C are collinear and locate C relative to AB. 4. Foundation · Find the midpoint x-coordinate. 5. Foundation · Find B from A and AB. 6. Foundation · Find the distance AB. 7. Foundation · Find k so that u and v are parallel. 8. Standard · P lies inside AB with AP:PB=2:1. Find P. 9. Standard · Does ae₁+be₂=v determine a,b uniquely? Describe all solutions. 10. Standard · Find the y-coordinate of the centroid. 11. Standard · For which real k do these vectors form a plane basis? 12. Transfer · A triangle has A,B and centroid G as shown. Recover C and verify 13. Transfer · A is the origin of a parallelogram with edge vectors u,v. Locate ← →
What you will be able to explain Subtract endpoints in the correct order. Use affine weights for a midpoint, division point and centroid. Distinguish an independent basis from parallel generating vectors. Coordinates require an origin and independent directions Recovering a missing vertex from a centroid is an inverse problem, so the weights must be interpreted before arithmetic.
An arrow is endpoint minus start. Affine weights whose sum is one describe a point independently of a translated origin. Two parallel directions cannot give unique plane coordinates.
P = ( 1 − t ) A + t B , det ( e 1 , e 2 ) ≠ 0 P=(1-t)A+tB,\quad\det(e_1,e_2)\ne0 P = ( 1 − t ) A + tB , det ( e 1 , e 2 ) = 0 Point division follows from moving a fraction t of AB from A.
P = A + t ( B − A ) P=A+t(B-A) P = A + t ( B − A ) Independent columns give a unique solution for the coefficients.
a e 1 + b e 2 = v ae_1+be_2=v a e 1 + b e 2 = v When it applies Internal division requires 0<t<1. A two-dimensional basis has nonzero determinant. Counterexample: The pair (1,2),(2,4) spans only a line; (3,6) has infinitely many coefficient pairs.
Endpoint diagram ↔ displacement equation ↔ two simultaneous scalar equations. Pause at these common mistakes AP:PB=2:1 places P two thirds of the way from A, not one third. Solving coordinates in a dependent pair may give infinitely many or no solutions. PREDICT → EXPLORE → EXPLAIN → TRANSFER
Make a prediction, then explore the relationship. Lesson question: Predict when the moving weighted point lies inside AB.
Model exploration: predict which displayed result changes with the parameters, check the values, and compare the observation with the lesson question.
Numerical data for the diagram; printable Quantity Value t 0.5 P 1, 2 AP 3.16227766 PB 3.16227766
Explain: P=A+t(B−A) works for internal and external division; only the interval for t changes.
Transfer: Find a weight giving AP twice PB externally, then distinguish it from the internal solution.
My prediction, exploration and explanation Try it. Leave your reasoning visible. Use one hint at a time. A correction explains what changed, not just the final answer.
Examples 1–3: study a full demonstration; finish a solution from its first step; then choose a method independently before opening the solution. The remaining exercises may continue after class.
Find the vector AB and explain its signs. A = ( − 1 , 2 ) , B = ( 3 , − 4 ) A=(-1,2),\ B=(3,-4) A = ( − 1 , 2 ) , B = ( 3 , − 4 ) Skills and prerequisite lessons Basis theorem and coordinate representation Concept of plane vectors Operations on plane vectors Coordinates, bases and division Scope: JM02
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Hint 1+ Subtract the starting coordinates.
Hint 2+ A downward displacement has negative y.
Worked solution+ Subtract component by component.
A B = ( 3 − ( − 1 ) , − 4 − 2 ) = ( 4 , − 6 ) AB=(3-(-1),-4-2)=(4,-6) A B = ( 3 − ( − 1 ) , − 4 − 2 ) = ( 4 , − 6 ) This moves right four and down six.
AB=(4,−6).
Checks and common pitfalls: A−B would give BA instead.
Reasoning checklist · self / teacher assessment AB=(4,−6). A−B would give BA instead. Think first. Reveal a hint when the class is ready.
Express v as ae₁+be₂. e 1 = ( 1 , 1 ) , e 2 = ( 1 , − 1 ) , v = ( 6 , 2 ) e_1=(1,1),\ e_2=(1,-1),\ v=(6,2) e 1 = ( 1 , 1 ) , e 2 = ( 1 , − 1 ) , v = ( 6 , 2 ) First step provided — complete the reasoning: The basis gives two simultaneous equations.
a + b = 6 , a − b = 2 a+b=6,\ a-b=2 a + b = 6 , a − b = 2 Skills and prerequisite lessons Basis theorem and coordinate representation Concept of plane vectors Operations on plane vectors Coordinates, bases and division Scope: JM02
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Hint 1+ Equate the two coordinates.
Hint 2+ Add and subtract the equations.
Worked solution+ The basis gives two simultaneous equations.
a + b = 6 , a − b = 2 a+b=6,\ a-b=2 a + b = 6 , a − b = 2 Solve and substitute back.
a = 4 , b = 2 ; 4 ( 1 , 1 ) + 2 ( 1 , − 1 ) = ( 6 , 2 ) a=4,\ b=2;\ 4(1,1)+2(1,-1)=(6,2) a = 4 , b = 2 ; 4 ( 1 , 1 ) + 2 ( 1 , − 1 ) = ( 6 , 2 ) v=4e₁+2e₂.
Checks and common pitfalls: Basis coefficients need not equal Cartesian coordinates.
Reasoning checklist · self / teacher assessment v=4e₁+2e₂. Basis coefficients need not equal Cartesian coordinates. Think first. Reveal a hint when the class is ready.
Prove A,B,C are collinear and locate C relative to AB. A = ( 0 , 0 ) , B = ( 2 , 3 ) , C = ( 4 , 6 ) A=(0,0),B=(2,3),C=(4,6) A = ( 0 , 0 ) , B = ( 2 , 3 ) , C = ( 4 , 6 ) Skills and prerequisite lessons Basis theorem and coordinate representation Concept of plane vectors Operations on plane vectors Coordinates, bases and division Scope: JM02
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Hint 1+ Compare AC with AB.
Hint 2+ A factor greater than one places C beyond B.
Worked solution+ The displacements are proportional.
A C = ( 4 , 6 ) = 2 A B AC=(4,6)=2AB A C = ( 4 , 6 ) = 2 A B Since 2>1, C lies on the ray AB beyond B.
Collinear; B is the midpoint of AC.
Checks and common pitfalls: Parallel displacement vectors through A imply collinearity; equal lengths alone do not.
Reasoning checklist · self / teacher assessment Collinear; B is the midpoint of AC. Parallel displacement vectors through A imply collinearity; equal lengths alone do not. Think first. Reveal a hint when the class is ready.
Find the midpoint x-coordinate. A = ( − 2 , 4 ) , B = ( 6 , − 2 ) A=(-2,4),\ B=(6,-2) A = ( − 2 , 4 ) , B = ( 6 , − 2 ) Skills and prerequisite lessons Basis theorem and coordinate representation Concept of plane vectors Operations on plane vectors Coordinates, bases and division Scope: JM02
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Hint 1+ Average the x coordinates.
Hint 2+ Midpoint means equal weights.
Worked solution+ Use the midpoint formula.
M = ( A + B ) / 2 M=(A+B)/2 M = ( A + B ) /2 Evaluate the first coordinate.
M x = ( − 2 + 6 ) / 2 = 2 M_x=(-2+6)/2=2 M x = ( − 2 + 6 ) /2 = 2 x=2.
Checks and common pitfalls: Do not average an x-coordinate with a y-coordinate.
Think first. Reveal a hint when the class is ready.
Find B from A and AB. A = ( 3 , − 1 ) , A B = ( − 5 , 4 ) A=(3,-1),\ AB=(-5,4) A = ( 3 , − 1 ) , A B = ( − 5 , 4 ) Skills and prerequisite lessons Basis theorem and coordinate representation Concept of plane vectors Operations on plane vectors Coordinates, bases and division Scope: JM02
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Hint 1+ Endpoint equals start plus displacement.
Hint 2+ Check by subtracting A again.
Worked solution+ Add the displacement.
B = ( 3 , − 1 ) + ( − 5 , 4 ) = ( − 2 , 3 ) B=(3,-1)+(-5,4)=(-2,3) B = ( 3 , − 1 ) + ( − 5 , 4 ) = ( − 2 , 3 ) Recover the given arrow.
B − A = ( − 5 , 4 ) B-A=(-5,4) B − A = ( − 5 , 4 ) B=(−2,3).
Checks and common pitfalls: AB is not itself the position of B unless A is the origin.
Reasoning checklist · self / teacher assessment B=(−2,3). AB is not itself the position of B unless A is the origin. Think first. Reveal a hint when the class is ready.
Find the distance AB. A = ( 1 , 1 ) , B = ( 4 , 5 ) A=(1,1),\ B=(4,5) A = ( 1 , 1 ) , B = ( 4 , 5 ) Skills and prerequisite lessons Basis theorem and coordinate representation Concept of plane vectors Operations on plane vectors Coordinates, bases and division Scope: JM02
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Hint 1+ Form the displacement first.
Hint 2+ Then take its length.
Worked solution+ Subtract the endpoints.
B − A = ( 3 , 4 ) B-A=(3,4) B − A = ( 3 , 4 ) Use the Euclidean norm.
A B = 9 + 16 = 5 AB=\sqrt{9+16}=5 A B = 9 + 16 = 5 AB=5.
Checks and common pitfalls: Distance is unchanged if the subtraction order is reversed.
Think first. Reveal a hint when the class is ready.
Find k so that u and v are parallel. u = ( 2 , 3 ) , v = ( 4 , k ) u=(2,3),\ v=(4,k) u = ( 2 , 3 ) , v = ( 4 , k ) Skills and prerequisite lessons Basis theorem and coordinate representation Concept of plane vectors Operations on plane vectors Coordinates, bases and division Scope: JM02
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Hint 1+ The x component has doubled.
Hint 2+ Both components need the same multiplier.
Worked solution+ The multiplier is fixed by the nonzero first component.
Double the second component.
k = 2 ⋅ 3 = 6 k=2\cdot3=6 k = 2 ⋅ 3 = 6 k=6.
Checks and common pitfalls: Matching only one coordinate is insufficient.
Think first. Reveal a hint when the class is ready.
P lies inside AB with AP:PB=2:1. Find P. A = ( 0 , 0 ) , B = ( 9 , 6 ) A=(0,0),\ B=(9,6) A = ( 0 , 0 ) , B = ( 9 , 6 ) Skills and prerequisite lessons Basis theorem and coordinate representation Concept of plane vectors Operations on plane vectors Coordinates, bases and division Scope: JM02
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Hint 1+ P is two-thirds of the way from A.
Hint 2+ The endpoint nearer P has the larger weight.
Worked solution+ Use affine interpolation.
P = A + 2 3 ( B − A ) P=A+\frac23(B-A) P = A + 3 2 ( B − A ) Evaluate and compare segment lengths.
P = ( 6 , 4 ) , A P = 2 P B P=(6,4),\ AP=2PB P = ( 6 , 4 ) , A P = 2 P B P=(6,4).
Checks and common pitfalls: Using one-third would reverse the ratio.
Reasoning checklist · self / teacher assessment P=(6,4). Using one-third would reverse the ratio. Think first. Reveal a hint when the class is ready.
Does ae₁+be₂=v determine a,b uniquely? Describe all solutions. e 1 = ( 1 , 2 ) , e 2 = ( 2 , 4 ) , v = ( 3 , 6 ) e_1=(1,2),\ e_2=(2,4),\ v=(3,6) e 1 = ( 1 , 2 ) , e 2 = ( 2 , 4 ) , v = ( 3 , 6 ) Skills and prerequisite lessons Basis theorem and coordinate representation Concept of plane vectors Operations on plane vectors Coordinates, bases and division Scope: JM02
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Hint 1+ The second vector is twice the first.
Hint 2+ Both coordinate equations give one condition.
Worked solution+ Reduce the dependent system.
One parameter remains free.
( a , b ) = ( 3 − 2 t , t ) , t ∈ R (a,b)=(3-2t,t),\ t\in\mathbb R ( a , b ) = ( 3 − 2 t , t ) , t ∈ R Infinitely many representations.
Checks and common pitfalls: Two vectors form a plane basis only when independent.
Reasoning checklist · self / teacher assessment Infinitely many representations. Two vectors form a plane basis only when independent. Think first. Reveal a hint when the class is ready.
Find the y-coordinate of the centroid. A = ( 0 , 0 ) , B = ( 6 , 0 ) , C = ( 0 , 9 ) A=(0,0),B=(6,0),C=(0,9) A = ( 0 , 0 ) , B = ( 6 , 0 ) , C = ( 0 , 9 ) Skills and prerequisite lessons Basis theorem and coordinate representation Concept of plane vectors Operations on plane vectors Coordinates, bases and division Scope: JM02
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Hint 1+ The three vertices have equal weights.
Hint 2+ Average the y coordinates only.
Worked solution+ Use the position-vector centroid relation.
G = ( A + B + C ) / 3 G=(A+B+C)/3 G = ( A + B + C ) /3 Evaluate its vertical coordinate.
G y = ( 0 + 0 + 9 ) / 3 = 3 G_y=(0+0+9)/3=3 G y = ( 0 + 0 + 9 ) /3 = 3 Gᵧ=3.
Checks and common pitfalls: The centroid is not generally the bounding-box centre.
Think first. Reveal a hint when the class is ready.
For which real k do these vectors form a plane basis? e 1 = ( 1 , 2 ) , e 2 = ( k , 4 ) e_1=(1,2),\ e_2=(k,4) e 1 = ( 1 , 2 ) , e 2 = ( k , 4 ) Skills and prerequisite lessons Basis theorem and coordinate representation Concept of plane vectors Operations on plane vectors Coordinates, bases and division Scope: JM02
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Hint 1+ A basis excludes parallel vectors.
Hint 2+ Test the two-by-two determinant.
Worked solution+ The determinant measures independence.
det ( e 1 , e 2 ) = 4 − 2 k \det(e_1,e_2)=4-2k det ( e 1 , e 2 ) = 4 − 2 k Exclude precisely its zero.
All real k except 2.
Checks and common pitfalls: Nonzero individual vectors can still be dependent.
Reasoning checklist · self / teacher assessment All real k except 2. Nonzero individual vectors can still be dependent. Think first. Reveal a hint when the class is ready.
A triangle has A,B and centroid G as shown. Recover C and verify that the triangle is nondegenerate. A = ( 0 , 1 ) , B = ( 4 , 0 ) , G = ( 2 , 3 ) A=(0,1),B=(4,0),G=(2,3) A = ( 0 , 1 ) , B = ( 4 , 0 ) , G = ( 2 , 3 ) Skills and prerequisite lessons Basis theorem and coordinate representation Concept of plane vectors Operations on plane vectors Coordinates, bases and division Scope: JM02
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Hint 1+ Undo averaging before subtracting vertices.
Hint 2+ Test whether AB and AC are parallel.
Worked solution+ Solve the centroid equation for the unknown vertex.
C = 3 G − A − B = ( 2 , 8 ) C=3G-A-B=(2,8) C = 3 G − A − B = ( 2 , 8 ) Check the area determinant.
det ( ( 4 , − 1 ) , ( 2 , 7 ) ) = 30 ≠ 0 \det((4,-1),(2,7))=30\ne0 det (( 4 , − 1 ) , ( 2 , 7 )) = 30 = 0 C=(2,8); area 15, so the triangle is nondegenerate.
Checks and common pitfalls: The centroid condition alone must still be checked against a valid triangle.
Reasoning checklist · self / teacher assessment C=(2,8); area 15, so the triangle is nondegenerate. The centroid condition alone must still be checked against a valid triangle. Think first. Reveal a hint when the class is ready.
A is the origin of a parallelogram with edge vectors u,v. Locate P and decide whether it is inside. u = ( 4 , 2 ) , v = ( − 1 , 3 ) , A P = 1 4 u + 2 3 v u=(4,2),v=(-1,3),\ AP=\frac14u+\frac23v u = ( 4 , 2 ) , v = ( − 1 , 3 ) , A P = 4 1 u + 3 2 v Skills and prerequisite lessons Basis theorem and coordinate representation Concept of plane vectors Operations on plane vectors Coordinates, bases and division Scope: JM02
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Hint 1+ Combine the two scaled edge vectors.
Hint 2+ For independent edges, interior coefficients lie strictly between 0 and 1.
Worked solution+ Evaluate the affine coordinates.
P = ( 1 , 1 / 2 ) + ( − 2 / 3 , 2 ) = ( 1 / 3 , 5 / 2 ) P=(1,1/2)+(-2/3,2)=(1/3,5/2) P = ( 1 , 1/2 ) + ( − 2/3 , 2 ) = ( 1/3 , 5/2 ) Check independence and both coefficients.
det ( u , v ) = 14 ≠ 0 ; 0 < 1 / 4 , 2 / 3 < 1 \det(u,v)=14\ne0;\ 0<1/4,2/3<1 det ( u , v ) = 14 = 0 ; 0 < 1/4 , 2/3 < 1 P=(1/3,5/2), strictly inside.
Checks and common pitfalls: The sum of the two coefficients need not equal one for a parallelogram interior.
Reasoning checklist · self / teacher assessment P=(1/3,5/2), strictly inside. The sum of the two coefficients need not equal one for a parallelogram interior. Think first. Reveal a hint when the class is ready.
