Porisma

Worked solution

Reflect a point in a line and calculate the resulting area

Relative to O, A,B have position vectors 4i+3j4\mathbf i+3\mathbf j and 5i+7j5\mathbf i+7\mathbf j. Line l1l_1 passes through B parallel to i+2j+αk\mathbf i+2\mathbf j+\alpha\mathbf k. Line l2l_2 has Cartesian equation (1−x)/2=(z−2)/1(1-x)/2=(z-2)/1, y=4.

(i) Given l1,l2l_1,l_2 are perpendicular, show α=2. marks

(ii) Find the length of projection of BA onto l1l_1. Hence find vector BN, where N is the perpendicular foot from A onto l1l_1. marks

(iii) A′ is the reflection of A in l1l_1, and C makes AA′CB a parallelogram. Find its exact area. marks

[2012/MJC/Promo/6]

CJC · Review problems · Question 6

Answer

α=2; projection 3; BN=(−1,−2,−2)BN=(-1,-2,-2); area 12212\sqrt2.

(i) Compare the direction vectors

l2:r=(1,4,2)+λ(−2,0,1)l_2:\mathbf r=(1,4,2)+\lambda(-2,0,1), λ∈R\lambda\in\mathbb R. Orthogonal directions require (1,2,α)⋅(−2,0,1)=−2+α=0(1,2,\alpha)\cdot(-2,0,1)=-2+\alpha=0, so α=2. The printed perpendicular-lines wording is used as an angle-between-directions condition; these spatial lines do not actually intersect at this value.

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(ii) Retain length and direction separately

BA=(−1,−4,0)BA=(-1,-4,0). The unit direction of l1l_1 is (1,2,2)/3(1,2,2)/3. Projection length=∣BA⋅(1,2,2)/3∣=∣−9∣/3=3|BA\cdot(1,2,2)/3|=|-9|/3=3. The signed component is −3, so BN=−3(1,2,2)/3=(−1,−2,−2)BN=-3(1,2,2)/3=(-1,-2,-2).

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(iii) Method 1: four equal triangle areas

∣AB∣=17|AB|=\sqrt{17} and ∣BN∣=3|BN|=3, giving ∣AN∣=17−9=8|AN|=\sqrt{17-9}=\sqrt8. The reflection makes AA′=2AN and its midpoint N lie on the line through B. Thus area (AA′CB)=4area⁡(ANB)=4(12)(8)(3)=1224\operatorname{area}(ANB)=4(\frac12)(\sqrt8)(3)=12\sqrt2 square units.

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(iii) Method 2: base and height

Use the base AA′=2√8 and perpendicular height BN=3. The area is (28)(3)=122(2\sqrt8)(3)=12\sqrt2 square units.

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Common mistakes

Orthogonal direction vectors do not establish that two spatial lines intersect.

Solution notes

All printed asks and both area methods retained. Source calls l1,l2 perpendicular, although they are skew: l1 at y=4 has (7/2,4,−3), while l2 at z=−3 has (11,4,−3). The intended orthogonal-direction condition and all numeric answers remain valid; no intersection is used in the reflection calculation. No source figure.

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