# We have four possibiliɵes ℓ 1 and ℓ 2 are the same

• 368
• 100% (1) 1 out of 1 people found this document helpful

This preview shows page 127 - 130 out of 368 pages.

we have four possibiliƟes:1and2arethe same linethey share all points;intersecƟng linesshare only 1 point;parallel linesd1d2, no points in common; orskew linesd1d2, no points in common.Notes:607
Figure 10.50: Sketching the lines from Ex-ample 346.Chapter 10VectorsThe next two examples invesƟgate these possibiliƟes.Example 346Comparing linesConsider lines1and2, given in parametric equaƟon form:1:x=1+3ty=2-tz=t2:x=-2+4sy=3+sz=5+2s.Determine whether1and2are the same line, intersect, are parallel, or skew.SÊ½çã®ÊÄWe start by looking at the direcƟons of each line. Line1has the direcƟon given byd1=3,-1,1and line2has the direcƟon given byd2=4,1,2. It should be clear thatd1andd2are not parallel, hence1and2are not the same line, nor are they parallel. Figure 10.50 verifies this fact (wherethe points and direcƟons indicated by the equaƟons of each line are idenƟfied).We next check to see if they intersect (if they do not, they are skew lines).To find if they intersect, we look fortandsvalues such that the respecƟvex,yandzvalues are the same. That is, we wantsandtsuch that:1+3t=-2+4s2-t=3+st=5+2s.This is a relaƟvely simple system of linear equaƟons. Since the last equaƟon isalready solved fort, subsƟtute that value oftinto the equaƟon above it:2-(5+2s) =3+ss=-2,t=1.A key to remember is that we havethreeequaƟons; we need to check ifs=-2,t=1 saƟsfies the first equaƟon as well:1+3(1)̸=-2+4(-2).It does not. Therefore, we conclude that the lines1and2are skew.Example 347Comparing linesConsider lines1and2, given in parametric equaƟon form:1:x=-0.7+1.6ty=4.2+2.72tz=2.3-3.36t2:x=2.8-2.9sy=10.15-4.93sz=-5.05+6.09s.Determine whether1and2are the same line, intersect, are parallel, or skew.Notes:608
Figure 10.51: Graphing the lines in Exam-ple 347.10.5LinesSÊ½çã®ÊÄIt is obviously very diﬃcult to simply look at these equaƟonsand discern anything. This is done intenƟonally. In the “real world,” most equa-Ɵons that are used do not have nice, integer coeﬃcients. Rather, there are lotsof digits aŌer the decimal and the equaƟons can look “messy.”We again start by deciding whether or not each line has the same direcƟon.The direcƟon of1is given byd1=1.6,2.72,-3.36and the direcƟon of2is given byd2=⟨-2.9,-4.93,6.09. When it is not clear through observaƟonwhether two vectors are parallel or not, the standard way of determining this isby comparing their respecƟve unit vectors. Using a calculator, we find:u1=d1||d1||=0.3471,0.5901,-0.7289u2=d2||d2||=⟨-0.3471,-0.5901,0.7289.

Upload your study docs or become a

Course Hero member to access this document

Upload your study docs or become a

Course Hero member to access this document

End of preview. Want to read all 368 pages?

Upload your study docs or become a

Course Hero member to access this document

Term
Spring
Professor
Unknown
Tags
hyperbolas, parametric equa ons
• • • 