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Original question & working

Skew lines and a perpendicular foot · source reference

[modified 2009/NYJC/Prelim/I/3] · Question pp. 1 · Working pp. 4

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Question

Lines l1,l2l_1,l_2 are r=(1,0,1)+λ(1,1,−2)\mathbf r=(1,0,1)+\lambda(1,1,-2) and r=(1,2,−1)+μ(7,5,1)\mathbf r=(1,2,-1)+\mu(7,5,1) respectively.

(i) Show they are skew lines. [3]

(ii) Find the position vector of the foot of perpendicular from A(1,2,−1)(1,2,-1) to l1l_1. [3]

[modified 2009/NYJC/Prelim/I/3]

Source working

(i) The direction vectors (1,1,−2)(1,1,-2) and (7,5,1)(7,5,1) are not parallel, so the lines may intersect. Equating line coordinates:

λ−7μ=0(1),λ−5μ=2(2),2λ+μ=2(3).\lambda-7\mu=0\quad(1),\quad\lambda-5\mu=2\quad(2),\quad2\lambda+\mu=2\quad(3).

Solving (1),(2) using GC gives μ=1,λ=7\mu=1,\lambda=7. Substitute into unused third equation: LHS =2(7)+1=15≠2==2(7)+1=15\ne2= RHS. The source concludes: “Since there is no unique solution, the lines are skew lines.”

Qualification: The computed inconsistency shows there is no solution, rather than merely “no unique solution”. Nonparallel lines that do not intersect are skew.

(ii) Let N be the foot. N on l1l_1:

ON→=(1,0,1)+λ(1,1,−2),λ∈R.\overrightarrow{ON}=(1,0,1)+\lambda(1,1,-2),\quad\lambda\in\mathbb R.

AN⊥l1⇒AN→⋅(1,1,−2)=0AN\perp l_1\Rightarrow\overrightarrow{AN}\cdot(1,1,-2)=0:

[(1,0,1)+λ(1,1,−2)−(1,2,−1)]⋅(1,1,−2)=0,[(1,0,1)+\lambda(1,1,-2)-(1,2,-1)]\cdot(1,1,-2)=0, [(0,−2,2)+λ(1,1,−2)]⋅(1,1,−2)=0,[(0,-2,2)+\lambda(1,1,-2)]\cdot(1,1,-2)=0, (−2−4)+λ(1+1+4)=0⇒λ=1.(-2-4)+\lambda(1+1+4)=0\Rightarrow\lambda=1.

Thus ON→=(2,1,−1)\overrightarrow{ON}=(2,1,-1).

Teacher notes

In r=(1,0,1)+λ(1,1,−2)\mathbf r=(1,0,1)+\lambda(1,1,-2), the first vector is the position vector of a known point on the line; the second is its direction vector; λ∈R\lambda\in\mathbb R.

Solve for μ, λ using two coordinate equations and substitute into the third to check satisfaction.

For the perpendicular foot, use two conditions: N lies on l1l_1 so satisfies the line equation; AN⊥l1AN\perp l_1, so AN→⋅d=0\overrightarrow{AN}\cdot\mathbf d=0, with d its direction vector.

Native diagrams

The source shows a straight l1l_1 through N, a perpendicular AN with A(1,2,−1)(1,2,-1) above, a right-angle mark at N and direction arrow (1,1,−2)(1,1,-2) along the line.

Foot of perpendicular from A to l₁

Printed answer key

(ii) (2,1,−1)(2,1,-1).

Provenance

2025 TJC H2 Math (9758) JC2 MSM: VECTORS. Question document pages 1; solution document pages 4. Original page numbers are retained as text metadata.