Porisma

Worked solution

Find a projection foot using two methods

Relative to O, A,B,C have position vectors −3i+8j+k-3\mathbf i+8\mathbf j+\mathbf k, 7i+3j+6k7\mathbf i+3\mathbf j+6\mathbf k, and αi+βj−6k\alpha\mathbf i+\beta\mathbf j-6\mathbf k respectively, where α,β are real.

(i) Given BC:r=(7+3λ)i+3j+(6+4λ)kBC:\mathbf r=(7+3\lambda)\mathbf i+3\mathbf j+(6+4\lambda)\mathbf k, find α,β. marks

(ii) Find the length of projection of AB onto BC. marks

(iii) Hence or otherwise find the position vector of the perpendicular foot from A to BC. marks

(iv) Find the exact area of triangle ABC. marks

[2008/NYJC/Promo/8]

CJC · Review problems · Question 5

Answer

α=−2,β=3; projection 10; OF=(1,3,−2)OF=(1,3,-2); area 752/275\sqrt2/2.

(i) Put C on the line

α=7+3λ\alpha=7+3\lambda, β=3 and −6=6+4λ-6=6+4\lambda. Thus λ=−3, α=−2 and β=3, giving C=(−2,3,−6).

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(ii) Project onto the line direction

Let F be the perpendicular foot. BA=(−10,5,−5)BA=(-10,5,-5), so the projection length is BF=∣BA⋅(3,0,4)∣/5=∣−30−20∣/5=10BF=|BA\cdot(3,0,4)|/5=|-30-20|/5=10. Reversing BA to the asked AB changes the sign but not this length.

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(iii) Hence method: retain the projection sign

BA⋅(3,0,4)/5=−10BA\cdot(3,0,4)/5=-10. Therefore BF→=−10(3,0,4)/5=(−6,0,−8)\overrightarrow{BF}=-10(3,0,4)/5=(-6,0,-8), and OF=OB+BF=(7,3,6)+(−6,0,−8)=(1,3,−2)OF=OB+BF=(7,3,6)+(-6,0,-8)=(1,3,-2). The source writes length BF before introducing −10; the sign comes from the signed dot product, not from a negative length.

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(iii) Otherwise method: a zero perpendicular dot product

OF=(7,3,6)+λ(3,0,4)OF=(7,3,6)+\lambda(3,0,4) gives AF=(10+3λ,−5,5+4λ)AF=(10+3\lambda,-5,5+4\lambda). Thus AF⋅(3,0,4)=30+9λ+20+16λ=0AF\cdot(3,0,4)=30+9\lambda+20+16\lambda=0, λ=−2 and OF=(1,3,−2)OF=(1,3,-2).

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(iv) Method 1: cross product

BC=(−9,0,−12)BC=(-9,0,-12) and BA×BC=(−60,−75,45)BA\times BC=(-60,-75,45). The area is 1211250=752/2\frac12\sqrt{11250}=75\sqrt2/2 square units.

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(iv) Method 2: base and perpendicular height

AF=BA2−BF2=150−100=50=52AF=\sqrt{BA^2-BF^2}=\sqrt{150-100}=\sqrt{50}=5\sqrt2 and BC=15. Hence area=12(52)(15)=752/2\frac12(5\sqrt2)(15)=75\sqrt2/2 square units.

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Compare your sketch

Expand diagram

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B,F,C occur in that order on a horizontal line; A is above F and AF is perpendicular to BC. The vector from B to A is drawn; the positive direction(3,0,4) points left, opposite to BC, because BC=−3(3,0,4). Diagram is schematic, preserving incidences and directions rather than spatial coordinates.
B,F,C occur in that order on a horizontal line; A is above F and AF is perpendicular to BC. The vector from B to A is drawn; the positive direction(3,0,4) points left, opposite to BC, because BC=−3(3,0,4). Diagram is schematic, preserving incidences and directions rather than spatial coordinates.
B,F,C occur in that order on a horizontal line; A is above F and AF is perpendicular to BC. The vector from B to A is drawn; the positive direction(3,0,4) points left, opposite to BC, because BC=−3(3,0,4). Diagram is schematic, preserving incidences and directions rather than spatial coordinates.

Common mistakes

The direction(3,0,4) points opposite to BC; a positive length alone cannot determine the foot’s displacement.

Solution notes

All asks, both foot methods and both area methods retained. The source switch from a length 10 to signed −10 is explicitly justified. Source diagram independently reconstructed.

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