Engineering Calculus Notes 48

Engineering Calculus Notes 48 - 36 CHAPTER 1. COORDINATES...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

Ask a homework question - tutors are online