Porisma

Worked solution

Separate a trigonometric parameter from a later special case

With respect to O, A,B,C have position vectors a=(1,sin⁡t,0)\mathbf a=(1,\sin t,0), b=(cos⁡t,2,1)\mathbf b=(\cos t,2,1), c=(2,−1,1)\mathbf c=(2,-1,1), where 0≤t<π0\le t<\pi.

(a) Find the exact t given a is perpendicular to b. marks

(b) X is on AB produced with AB:BX=1:4, and ACXY is a parallelogram. Take t=π/2t=\pi/2 for the rest of the question.

(i) Find the position vectors of X,Y. marks

(ii) Find the area of ACXY. Hence find the shortest distance from X to line AC. marks

(iii) Line l1l_1 has equation (x−2)/6=(3−y)/h(x-2)/6=(3-y)/h, z=k, where h,k are constants. The line through A,B intersects l1l_1 at a right angle. Find h,k. marks

[2012/HCI/Promo/13]

CJC · Review problems · Question 8

Answer

(a)t=π−tan⁡−1(1/2)t=\pi-\tan^{-1}(1/2); (b)OX=(−4,6,5)OX=(-4,6,5),OY=(−5,8,4)OY=(-5,8,4); area 350\sqrt{350}; distance 175/3\sqrt{175/3}≈7.64; h=−6,k=1/2.

(a) Use the specified quadrant

a⋅b=cos⁡t+2sin⁡t=0\mathbf a\cdot\mathbf b=\cos t+2\sin t=0, so tan⁡t=−1/2\tan t=-1/2. The basic angle is tan⁡−1(1/2)\tan^{-1}(1/2). Since 0≤t<π and tan t<0, t lies in quadrant 2: t=π−tan⁡−1(1/2)t=\pi-\tan^{-1}(1/2). The case cos t=0 does not satisfy the original equation.

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(b)(i) Reset t and locate X,Y

At t=π/2, OA=(1,1,0)OA=(1,1,0), OB=(0,2,1)OB=(0,2,1), OC=(2,−1,1)OC=(2,-1,1) and AB=(−1,1,1)AB=(-1,1,1). Since AX=5AB, OX=5AB+OA=(−4,6,5)OX=5AB+OA=(-4,6,5). The order ACXY gives XY=CAXY=CA, so OY=OA−OC+OX=(−5,8,4)OY=OA-OC+OX=(-5,8,4).

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(b)(ii) Area and height to AC

AC=(1,−2,1)AC=(1,-2,1), AY=(−6,7,4)AY=(-6,7,4) and AC×AY=(−15,−10,−5)AC\times AY=(-15,-10,-5). Hence area (ACXY)=350\sqrt{350}. Taking AC as the base gives shortest distance XN=350/6=175/3\sqrt{350}/\sqrt6=\sqrt{175/3}, or 7.64 units(3s.f.).

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(b)(iii) Enforce orthogonality and intersection separately

l1:r=(2,3,k)+λ(6,−h,0)l_1:\mathbf r=(2,3,k)+\lambda(6,-h,0); line AB:r=(1,1,0)+μ(−1,1,1)\mathbf r=(1,1,0)+\mu(-1,1,1). The directions have dot product −6−h, giving h=−6. At an intersection, 2+6λ=1−μ2+6\lambda=1-\mu, 3+6λ=1+μ3+6\lambda=1+\mu, k=μk=\mu. Solving gives λ=−1/4, μ=1/2, hence k=1/2. The two parameters must be independent despite a reused λ in one printed vector-equation line.

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Upper sketch: A,B,X collinear with AB:BX=1:4. Lower sketch: parallelogram A,C,X,Y in that vertex order, with equal side vectors CA and XY. Both source sketches are independently reconstructed as labelled schematics, not coordinate projections.
Upper sketch: A,B,X collinear with AB:BX=1:4. Lower sketch: parallelogram A,C,X,Y in that vertex order, with equal side vectors CA and XY. Both source sketches are independently reconstructed as labelled schematics, not coordinate projections.
Upper sketch: A,B,X collinear with AB:BX=1:4. Lower sketch: parallelogram A,C,X,Y in that vertex order, with equal side vectors CA and XY. Both source sketches are independently reconstructed as labelled schematics, not coordinate projections.

Common mistakes

Orthogonality gives h, while intersection supplies k. Neither condition alone is sufficient.

Solution notes

Every printed ask, quadrant note, calculations and both sketches retained. The independent line parameters in the printed subsequent scalar equations resolve a parameter-letter reuse in its preceding vector line.

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