Porisma

Worked solution

Find an axis crossing and the origin-to-line distance

Line l passes through A=(1,2,−1) and B=(2,a,1).

(i) Show that l cuts the x-axis when a=−2. marks

(ii) Find the length of projection of OA onto l and hence the shortest distance from the origin to l. marks

[2008/NYJC/Promo/5]

CJC · Review problems · Question 2

Answer

a=−2; projection length 9/219/\sqrt{21}; shortest distance 15/7\sqrt{15/7}.

(i) Set both non-x coordinates to zero

AB=(1,a−2,2)AB=(1,a-2,2), so r=(1,2,−1)+λ(1,a−2,2)\mathbf r=(1,2,-1)+\lambda(1,a-2,2). On the x-axis, −1+2λ=0-1+2\lambda=0, hence λ=1/2. Then 2+12(a−2)=02+\frac12(a-2)=0 gives a=−2. The crossing has coordinates (3/2,0,0).

How did this part go?

(ii) Calculate a projection length

Using a=−2 from (i), the direction is (1,−4,2)(1,-4,2) of magnitude 21\sqrt{21}. Thus projection length=∣(1,2,−1)⋅(1,−4,2)∣/21=9/21|(1,2,-1)\cdot(1,-4,2)|/\sqrt{21}=9/\sqrt{21}.

How did this part go?

(ii) Use Pythagoras for the perpendicular distance

∣OA∣=6|OA|=\sqrt6, so the distance is

(6)2−(9/21)2=6−81/21=45/21=15/7.\begin{aligned}&\sqrt{(\sqrt6)^2-(9/\sqrt{21})^2}\\ & =\sqrt{6-81/21}\\ & =\sqrt{45/21}\\ & =\sqrt{15/7}.\end{aligned}

How did this part go?

Common mistakes

Part (ii) uses a=−2; without that value the projection depends on a.

Solution notes

All source calculations retained. The use of a=−2 in(ii), implicit in the source, is made explicit. No figure.

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