Porisma

Worked solution

Extend a segment and use two vectors based at the angle vertex

Relative to O, A,B,C have position vectors 7i+6j−k7\mathbf i+6\mathbf j-\mathbf k, −11i−3j+26k-11\mathbf i-3\mathbf j+26\mathbf k, and 4i+j+7k4\mathbf i+\mathbf j+7\mathbf k. P lies on BA produced such that 3AB=PB.

(i) Find the position vector of P. marks

(ii) Find angle ABC. marks

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

CJC · Review problems · Question 6

Answer

OP=(43,24,−55)OP=(43,24,-55); angle ABC≈0.132rad=7.6°; area=633/263\sqrt3/2.

(i) Resolve the external division

P,A,B occur in that order, with PA:AB=2:1. Thus OA=(OP+2OB)/3OA=(OP+2OB)/3, so OP=3OA−2OB=(21,18,−3)−(−22,−6,52)=(43,24,−55)OP=3OA-2OB=(21,18,-3)-(-22,-6,52)=(43,24,-55).

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(ii) Form both sides outward from B

BA=OA−OB=(18,9,−27)BA=OA-OB=(18,9,-27) and BC=OC−OB=(15,4,−19)BC=OC-OB=(15,4,-19). Therefore

θ=cos⁡−1(BA⋅BC∣BA∣∣BC∣)=cos⁡−1(8191134602).\begin{aligned}&\theta\\ & =\cos^{-1}\left(\frac{BA\cdot BC}{|BA||BC|}\right)\\ & =\cos^{-1}\left(\frac{819}{\sqrt{1134}\sqrt{602}}\right).\end{aligned}

This gives 0.132rad, or 7.6°, matching the printed rounding.

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(iii) Evaluate the cross product

BA×BC=(−63,−63,−63)BA\times BC=(-63,-63,-63). Hence

area⁡(ABC)=12∣BA×BC∣=1211907=6332.\begin{aligned}&\operatorname{area}(ABC)\\ & =\frac12|BA\times BC|\\ & =\frac12\sqrt{11907}\\ & =\frac{63\sqrt3}{2}.\end{aligned}

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Compare your sketch

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Upper diagram: P,A,B lie in that order on a line, PA:AB=2:1, with O off the line and position-vector arrows OP, OA, OB. Lower diagram: triangle ABC has the angle theta at B between the directed sides BA and BC. Both are schematic, not coordinate projections.
Upper diagram: P,A,B lie in that order on a line, PA:AB=2:1, with O off the line and position-vector arrows OP, OA, OB. Lower diagram: triangle ABC has the angle theta at B between the directed sides BA and BC. Both are schematic, not coordinate projections.
Upper diagram: P,A,B lie in that order on a line, PA:AB=2:1, with O off the line and position-vector arrows OP, OA, OB. Lower diagram: triangle ABC has the angle theta at B between the directed sides BA and BC. Both are schematic, not coordinate projections.

Common mistakes

BA produced places P beyond A, not between A and B. Reversing just one angle vector gives the supplementary angle.

Solution notes

All three asks, calculations and both source figures retained through schematic reconstructions. The printed 3AB=PB relation concerns lengths.

Read the hint