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Original question & working

Area identities and a cross-product degeneracy · source reference

[2018/DHS/Prelim/II/1] · Question pp. 2 · Working pp. 7

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The working and teaching notes are included in the interactive lesson.

Printed source relation needs a qualification for the valid case 2b+3d=0; original answer is preserved in the source panel.

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Full original question, working & notes

Question

Relative to O, A and B have position vectors a,b, nonzero and nonparallel.

(i) State the shortest distance from B to the line through O,A. [1]

(ii) Given c=a+b\mathbf c=\mathbf a+\mathbf b, state the geometrical meaning of ∣b×c∣|\mathbf b\times\mathbf c| and show ∣b×c∣=∣a×b∣|\mathbf b\times\mathbf c|=|\mathbf a\times\mathbf b|. [2]

(iii) Given a×2b=d×3a\mathbf a\times2\mathbf b=\mathbf d\times3\mathbf a, find a linear relationship between a,b,d. [2]

[2018/DHS/Prelim/II/1]

Source working

(i) Shortest distance =∣b×a^∣=|\mathbf b\times\hat{\mathbf a}| or ∣b×a∣/∣a∣|\mathbf b\times\mathbf a|/|\mathbf a|.

(ii) ∣b×c∣|\mathbf b\times\mathbf c| is the area of a parallelogram with sides OB and OC.

∣b×c∣=∣b×(a+b)∣=∣(b×a)+(b×b)∣=∣b×a∣=∣−(a×b)∣=∣a×b∣.\begin{aligned}|\mathbf b\times\mathbf c|&=|\mathbf b\times(\mathbf a+\mathbf b)|\\&=|(\mathbf b\times\mathbf a)+(\mathbf b\times\mathbf b)|\\&=|\mathbf b\times\mathbf a|=|-(\mathbf a\times\mathbf b)|=|\mathbf a\times\mathbf b|.\end{aligned}

The parallelograms with sides OB,OC and OA,OB have the same base and vertical height.

(iii)

a×2b=d×3a,\mathbf a\times2\mathbf b=\mathbf d\times3\mathbf a, (a×2b)−(d×3a)=0,(\mathbf a\times2\mathbf b)-(\mathbf d\times3\mathbf a)=\mathbf0, a×(2b+3d)=0.\mathbf a\times(2\mathbf b+3\mathbf d)=\mathbf0.

Printed conclusion: a is parallel to 2b+3d2\mathbf b+3\mathbf d, so a=k(2b+3d)\mathbf a=k(2\mathbf b+3\mathbf d), k∈Rk\in\mathbb R.

Teacher notes

The shortest distance is perpendicular BN from B to OA.

For parallelogram OBDC,

∣b×c∣=∣b∣∣c∣sin⁡θ=(OB)(h)=2×Area(OBC)=Area(OBDC).|\mathbf b\times\mathbf c|=|\mathbf b||\mathbf c|\sin\theta=(OB)(h)=2\times\text{Area}(OBC)=\text{Area}(OBDC).

For (iii), the margin writes

a×(2b+3d)=(∣a∣∣2b+3d∣sin⁡θ)n^=0,\mathbf a\times(2\mathbf b+3\mathbf d)=(|\mathbf a||2\mathbf b+3\mathbf d|\sin\theta)\hat{\mathbf n}=\mathbf0,

where θ is between a and 2b+3d2\mathbf b+3\mathbf d, and n^\hat{\mathbf n} is perpendicular to both. It concludes sin⁡θ=0\sin\theta=0, θ=0∘\theta=0^\circ or 180∘180^\circ, so parallel.

Qualification of printed relationship

The source omits 2b+3d=02\mathbf b+3\mathbf d=\mathbf0, where that angle is undefined and a cannot equal a scalar times the zero vector. The inclusive relationship is 2b+3d=ta2\mathbf b+3\mathbf d=t\mathbf a for real tt, allowing t=0t=0.

Native diagrams

The distance panel has O,N,A on a line, B above N and right angle at N. The area panel shows A,C,D collinear, O,B on the parallel base, dotted AC extension and AB/CB diagonals, red h and its right angle; a separate margin parallelogram OBDC labels θ between OB,OC and the same height h.

Distance BN from B to line OA Parallelograms sharing base OB and height h Area of parallelogram OBDC

Printed answer key

(i) ∣b×a^∣|\mathbf b\times\hat{\mathbf a}| or ∣b×a∣/∣a∣|\mathbf b\times\mathbf a|/|\mathbf a|; (ii) area of a parallelogram with sides OB and OC; (iii) a=k(2b+3d)\mathbf a=k(2\mathbf b+3\mathbf d), k∈Rk\in\mathbb R. See the qualification above for the omitted zero-vector case.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: VECTORS. Question document pages 2; solution document pages 7. Original page numbers are retained as text metadata.