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I was wondering if someone would be able to help me with this question:

"Your friend asks you to help perform some design checks on a stadium's floodlights. He gives you a sketch of the proposed layout using a vector coordinate system where the third coordinate, z, represents the vertical direction.

The seating will be fixed to a planar concrete deck supporting structure with corners at A = (25,10,0), B = (-15,20,0), C= (-12,34,8) and D = (28,24,8). The floodlight is positioned 20m in the air at position F = (-35,0,20) and its spotlight is pointing in the direction u = (2,2,-1)

Your maths lecturer has been invited the attend the opening ceremony as a guest of honour and as such has been allocated a seat at the point directly in the middle of the seating deck at point Q.

a) Find the position vector of point P, the place on the seating deck at which the floodlight is exactly pointing"

I'm not really sure how to go about this question. I have found the equation of the line in which the spotlight is pointing:

r = (-35,0,20) + µ(2,2,-1)

However, I am not really sure how to find the equation of the plane and then how to find where the line and plane intersect

EDIT:

Okay I have managed to do most of it but am struggling with the last part

So far I have got:

AB = (-40,10,0) and DA = (-3,-14,-8)

This then gives me the equation of the plane as:

(25,10,0)+s(-40,10,0)+t(-3,-14,-8)

I then made this equal my line equation to get three equations:

1) 25-40s-3t = -35+2μ

2) 10+10s-14t = 2μ

3) 8t = 20-μ

However I'm now confused as to how to solve these

S Hayward
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1 Answers1

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The equation of a plane is given by $\mathbf r(s,t) = \mathbf r_0 + s\mathbf v_1 + t\mathbf v_2$, where $\mathbf r_0$ is a vector pointing to one point on the plane, and $\mathbf v_1$ and $\mathbf v_2$ are two noncollinear vectors parallel to the plane. This problem technically describes a planar segment as opposed to a plane -- but that just means that $s$ and $t$ are restricted to some subsets of the reals.

You have $4$ points in your plane: $A$, $B$, $C$, $D$. You can thus find vectors parallel to your plane by taking differences of these vectors. In particular, one vector parallel to your plane will be $\mathbf v = \vec{AB} = \vec{OA} - \vec{OB}$.

Now can you figure out how to find $\mathbf r_0$, $\mathbf v_1$, and $\mathbf v_2$?

Once you have equations for your plane and your line, finding the intersection is as simple as setting the two equations equal to each other and solving for $s$ and $t$ or $\mu$. Then plug those back into their respective equations to get $(x,y,z)$.

  • Thanks so much for your help, I've edited my question to show the next steps I've taken. However, I'm now confused how to solve the three equations. Also is what I have done right? – S Hayward May 27 '15 at 22:20
  • Where you're stuck is just a system of linear equations. I went over $3$ different methods on how to solve these types of problems in this answer. So take a look at it to see how to finish. –  May 27 '15 at 22:24
  • Okay great, thanks! I worked that out and I got s=43/21, t=100/21 and mu=-380/21. These seem like pretty horrible answers so I'm sure I must've done something wrong. Is there any way of checking the answers? – S Hayward May 27 '15 at 22:37
  • Uh oh. The point that corresponds to those values is not in the planar segment. Let me check back through work. –  May 27 '15 at 22:43
  • @SHayward In your third equation, it should be $-8t$, not $8t$. Using that you should get a bad looking fractional answer (but that's fine -- don't worry) that is approximately equal to $(x,y,z) = (-6.5,28.5,5.8)$. That fractional answer is the correct one -- this is just an approximation you can check your answer against to know if you got it right or not. –  May 27 '15 at 22:55
  • Yes that's great, thanks so much! – S Hayward May 28 '15 at 09:25