Distinguish no intersection from containment

A plane π1\pi_1 has equation r⋅(2,−1,3)=9\mathbf r\cdot(2,-1,3)=9 and a line l1l_1 has equation r=(3,0,0)+λ(4,1,−2)\mathbf r=(3,0,0)+\lambda(4,1,-2).

(i) Find the coordinates of PP, the point of intersection of l1l_1 and π1\pi_1. marks

Hence, or otherwise, find the shortest distance from point A(3,0,0)A(3,0,0) to π1\pi_1. marks

The equations of planes π2,π3\pi_2,\pi_3 are given as π2:r=(1,1,1)+s(2,0,−3)+t(2,2,1)\pi_2:\mathbf r=(1,1,1)+s(2,0,-3)+t(2,2,1), and π3:αx+y−z=β\pi_3:\alpha x+y-z=\beta, where α,β∈R\alpha,\beta\in\mathbb R.

(ii) Find an equation of plane π2\pi_2 in the form r⋅n=d\mathbf r\cdot\mathbf n=d. Explain why the planes π1,π2\pi_1,\pi_2 intersect. marks

(iii) The line of intersection of planes π1,π2\pi_1,\pi_2 is l2l_2. The line l2l_2 has equation r=(3,3,2)+μ(2,1,−1)\mathbf r=(3,3,2)+\mu(2,1,-1). Given that l2,π3l_2,\pi_3 do not have any points in common, find the conditions satisfied by α,β\alpha,\beta. marks

[modified 2010/IJC/Prelim/I/12]

Total marks: 13

Solution

(i) intersection

Line meeting a plane

π1:r⋅(2,−1,3)=9\pi_1:\mathbf r\cdot(2,-1,3)=9 and l1:r=(3,0,0)+λ(4,1,−2)l_1:\mathbf r=(3,0,0)+\lambda(4,1,-2). Find where they meet.

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Hint

  1. Substitute the line into the plane equation.

Answer · (i) intersection

P=(15,3,−6)P=(15,3,-6).

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The line pierces the plane

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(i) distance

Distance from a point to a plane

Hence find the shortest distance from A(3,0,0)A(3,0,0) to π1\pi_1.

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Hint

  1. Use AP→⋅n^\overrightarrow{AP}\cdot\hat{\mathbf n}.

Answer · (i) distance

3/143/\sqrt{14}.

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Find the perpendicular distance from A to π₁

N is the perpendicular foot on 2x − y + 3z = 9, not the oblique intersection P from the previous part. Move A to explore; Reset restores A = (3, 0, 0) and AN = 3/√14.

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(ii)

Plane equation and intersection

π2:r=(1,1,1)+s(2,0,−3)+t(2,2,1)\pi_2:\mathbf r=(1,1,1)+s(2,0,-3)+t(2,2,1). Find r⋅n=d\mathbf r\cdot\mathbf n=d for π2\pi_2 and explain why π1\pi_1 and π2\pi_2 intersect.

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Hint

  1. Normal =(2,0,−3)×(2,2,1)=(2,0,-3)\times(2,2,1).

Answer · (ii)

r⋅(3,−4,2)=1r\cdot(3,-4,2)=1; nonparallel normals imply an intersection line.

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Two planes meeting along l₂

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(iii)

No common points

π3:αx+y−z=β\pi_3:\alpha x+y-z=\beta and l2l_2 have no points in common. Find the conditions on α\alpha and β\beta.

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Hint

  1. No intersection means l2l_2 is parallel to π3\pi_3 but not in it.

  2. Parallel: d⋅n=0\mathbf d\cdot\mathbf n=0. Not contained: a point of l2l_2 fails π3\pi_3.

Answer · (iii)

α=−1\alpha=-1, β≠−2\beta\ne-2.

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A plane parallel to l₂

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