Porisma

Worked solution

Build an intersection line and a perpendicular plane

Planes π1,π2\pi_1,\pi_2 have equations r⋅(0,1,2)=2\mathbf r\cdot(0,1,2)=2, r⋅(2,−1,4)=12\mathbf r\cdot(2,-1,4)=12. A=(7,2,0), O is the origin.

(i) Verify A lies in both planes. mark

(ii) They meet in line L. Show that a vector equation of L is r=(7,2,0)+λ(3,2,−1)\mathbf r=(7,2,0)+\lambda(3,2,-1), λ∈R\lambda\in\mathbb R. marks

(iii) B=(7,−4,9), and N is its perpendicular foot onto π2\pi_2. Find N. marks

(iv) Find the Cartesian equation of π3\pi_3 containing L and perpendicular to π2\pi_2. marks

[2008/SRJC/Promo/15]

CJC · Review problems · Question 1

Answer

L:r=(7,2,0)+λ(3,2,−1)L:r=(7,2,0)+\lambda(3,2,-1); N=(3,−2,1); π3:x−2y−z=3\pi_3:x-2y-z=3.

(i) Substitute A in both equations

(7,2,0)⋅(0,1,2)=2+0=2(7,2,0)\cdot(0,1,2)=2+0=2 and (7,2,0)⋅(2,−1,4)=14−2=12(7,2,0)\cdot(2,-1,4)=14-2=12. Thus A belongs to both planes.

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(ii) Cross the normals

(0,1,2)×(2,−1,4)=(6,4,−2)=2(3,2,−1)(0,1,2)\times(2,-1,4)=(6,4,-2)=2(3,2,-1). This nonzero vector is parallel to both planes. Through their common point A, L:r=(7,2,0)+λ(3,2,−1)L:\mathbf r=(7,2,0)+\lambda(3,2,-1), λ real.

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(iii) Follow a normal through B

BN:r=(7,−4,9)+μ(2,−1,4)BN:\mathbf r=(7,-4,9)+\mu(2,-1,4). At N, (7+2μ,−4−μ,9+4μ)⋅(2,−1,4)=12(7+2\mu,-4-\mu,9+4\mu)\cdot(2,-1,4)=12. Thus 14+4μ+4+μ+36+16μ=1214+4\mu+4+\mu+36+16\mu=12, 21μ=−4221\mu=-42, μ=−2 and N=(3,−2,1).

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(iv) Choose the required normal

The normal of π3 must be perpendicular to L’s direction and to π2’s normal. Their cross product is (3,2,−1)×(2,−1,4)=(7,−14,−7)=−7(−1,2,1)(3,2,-1)\times(2,-1,4)=(7,-14,-7)=-7(-1,2,1). Through A, r⋅(−1,2,1)=(7,2,0)⋅(−1,2,1)=−3\mathbf r\cdot(-1,2,1)=(7,2,0)\cdot(-1,2,1)=-3. Hence −x+2y+z=−3-x+2y+z=-3, or x−2y−z=3x-2y-z=3.

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Common mistakes

Use a cross product for the new normal; the direction of L is not itself generally a plane normal.

Solution notes

All 4 printed asks and complete working retained. No source diagram.

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