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Worked solution

Prove a rhombus using a perpendicular projection

Relative to O, A,B,C have position vectors a=3i−j\mathbf a=3\mathbf i-\mathbf j, b=3i+4j−12k\mathbf b=3\mathbf i+4\mathbf j-12\mathbf k, c=λj+12k\mathbf c=\lambda\mathbf j+12\mathbf k. Line l1l_1 through A,C has equation (x−3)/3=(y+1)/4=−z/12(x-3)/3=(y+1)/4=-z/12.

(i) Show λ=−5. marks

(ii) Find the position vector of P on OA with BP perpendicular to OA. Hence or otherwise prove OBAC is a rhombus. marks

(iii) Find the area of OBAC. marks

(iv) Line l2l_2 passes through O,Q, where Q lies on BA produced with BA/BQ=1/4. Find the acute angle between l1,l2l_1,l_2. marks

[2016/AJC/Promo/7]

CJC · Review problems · Question 9

Answer

λ=−5; OP=(3/2,−1/2,0)OP=(3/2,-1/2,0); OBAC is a rhombus; area 40.8 units²(3s.f.) or 31853\sqrt{185} exactly; angle 18.5°.

(i) Method 1: use a vector parameter

l1:r=(3,−1,0)+μ(3,4,−12)l_1:\mathbf r=(3,-1,0)+\mu(3,4,-12). At C=(0,λ,12), the coordinates give 0=3+3μ0=3+3\mu, λ=−1+4μ\lambda=-1+4\mu, 12=−12μ12=-12\mu. Thus μ=−1 and λ=−5.

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(i) Method 2: substitute in the Cartesian equation

At C=(0,λ,12), −1=(λ+1)/4=−1-1=(\lambda+1)/4=-1, giving λ=−5 directly.

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(ii) Method 1: vector projection

OP=(b⋅a^)a^=b⋅a∣a∣2a=9−4+010(3,−1,0)=12(3,−1,0)OP=(\mathbf b\cdot\widehat{\mathbf a})\widehat{\mathbf a}=\frac{\mathbf b\cdot\mathbf a}{|\mathbf a|^2}\mathbf a=\frac{9-4+0}{10}(3,-1,0)=\frac12(3,-1,0).

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(ii) Method 2: impose perpendicularity

Write OP=k(3,−1,0)OP=k(3,-1,0), so BP=(3k−3,−k−4,12)BP=(3k-3,-k-4,12). Then BP⋅a=3(3k−3)−(−k−4)=10k−5=0BP\cdot\mathbf a=3(3k-3)-(-k-4)=10k-5=0, hence k=1/2, giving the same OP=(3/2,−1/2,0)OP=(3/2,-1/2,0).

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(ii) Prove the specific quadrilateral is a rhombus

OB=CA=(3,4,−12)OB=CA=(3,4,-12), or equivalently OC=BA=(0,−5,12)OC=BA=(0,-5,12), so OBAC is a parallelogram. Its adjacent sides satisfy ∣OB∣=∣BA∣=13|OB|=|BA|=13, making it a rhombus. The source alternative uses the parallelogram together with OA⊥BCOA\perp BC; the perpendicular diagonals of a parallelogram also establish a rhombus. Here BC=(−3,−9,24)=2BPBC=(-3,-9,24)=2BP, so the perpendicularity follows from BP⊥OA.

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(iii) Use four right triangles

OP=(3/2,−1/2,0)OP=(3/2,-1/2,0) and BP=(−3/2,−9/2,12)BP=(-3/2,-9/2,12). Hence

area⁡(OBAC)=4(12)∣OP∣∣BP∣=2(3/2)2+(−1/2)2(−3/2)2+(−9/2)2+122=40.8\begin{aligned}&\operatorname{area}(OBAC)\\ & =4\left(\frac12\right)|OP||BP|\\ & =2\sqrt{(3/2)^2+(-1/2)^2}\sqrt{(-3/2)^2+(-9/2)^2+12^2}\\ & =40.8\end{aligned}

square units(3s.f.), as printed. The corresponding exact value is 31853\sqrt{185} square units.

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(iv) Find Q, then an acute angle

OA=(OQ+3OB)/4OA=(OQ+3OB)/4, hence OQ=4OA−3OB=(3,−16,36)OQ=4OA-3OB=(3,-16,36). The other direction is (3,4,−12)(3,4,-12). Thus

cos⁡θ=∣(3,−16,36)⋅(3,4,−12)∣1561169=4871561169,\cos\theta=\frac{|(3,-16,36)\cdot(3,4,-12)|}{\sqrt{1561}\sqrt{169}}=\frac{487}{\sqrt{1561}\sqrt{169}},

and θ=18.5°.

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Compare your sketch

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Schematic rhombus O,B,A,C in that vertex order. P is the common midpoint of diagonals OA and BC; BP is perpendicular to OA. This extends the source proof sketch into the proved rhombus. AC is a side lying on l1. Lengths are schematic, not a projection of the coordinate data.
Schematic rhombus O,B,A,C in that vertex order. P is the common midpoint of diagonals OA and BC; BP is perpendicular to OA. This extends the source proof sketch into the proved rhombus. AC is a side lying on l1. Lengths are schematic, not a projection of the coordinate data.
Schematic rhombus O,B,A,C in that vertex order. P is the common midpoint of diagonals OA and BC; BP is perpendicular to OA. This extends the source proof sketch into the proved rhombus. AC is a side lying on l1. Lengths are schematic, not a projection of the coordinate data.

Common mistakes

Perpendicular diagonals alone do not establish a rhombus for an arbitrary quadrilateral.

Solution notes

Both printed λ methods and both foot methods retained, with both rhombus criteria and source area/angle working. The reconstructed rhombus includes the incidences in the source proof sketch. Exact area is an explanatory addition to the printed 3s.f. answer.

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