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Unformatted text preview: 36 CHAPTER 1. COORDINATES AND VECTORS We read oﬀ its direction vector as
→
− = 3− + − + −
→→→
v3
ı k
which is clearly not a scalar multiple of the other two. This tells us
immediately that ℓ3 is diﬀerent from both ℓ1 and ℓ2 (it has a diﬀerent
direction). Let us ask whether ℓ2 intersects ℓ3 . It might be tempting to try
to answer this by looking for a solution of the vector equation
→
− (t) = − (t)
→
p2
p3
but this would be a mistake. Remember that these two parametrizations
describe the positions of two points—one moving along ℓ2 and the other
moving along ℓ3 —at time t. The equation above requires the two points to
be at the same place at the same time —in other words, it describes a
collision. But all we ask is that the two paths cross: it would suﬃce to
locate a place occupied by both moving points, but possibly at diﬀerent
times. This means we need to distinguish the parameters appearing in the
→
→
two functions − 2 (t) and − 3 (t), by renaming one of them (say the ﬁrst) as
p
p
(say) s: the vector equation we need to solve is
→
− ( s ) = − ( t) ,
→
p2
p3
which amounts to the three equations in two unknowns
1 +6s =
1 +3t
4s =
2 +t
1 −2s = −3 +t.
The ﬁrst equation can be reduced to the condition
2s = t
and substituting this into the other two equations
2t =
2 +t
1 −t = −3 +t
we end up with the solution
t=2
s=1 ...
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 Fall '08
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 Calculus, Vectors, Scalar

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