Unformatted text preview: ISM: Linear Algebra Section 1.1 113} Fa no.n... . Figure 13?: for Problem 1.1.39a. 4D. We can think of the line through the points [1.1.1] and {3. EM] as the intersection of an},r two planes through these two points; each of these planes will he deﬁned by an equation
ofthe formom+by+cz = d. It isreqiﬁredthat1n+1b+1c=dand 3n+5b+ﬂc=d. a. +b +c —d = [1
Now the system 30. +55 —d = [1 reduoes to
a. +§c 4s = t:
EI —§c +d = [1' We can choose arbitraryr real numbers for e and d; then o = —%c+ 2d and b = %C— d. For
example, if we choose e = 2 and d = ll, then a = —5 and b = 3, leading to the equation
—51r+3y+22 = I]. [fwe chooses: I] andd= 1, then a. = 2 andEl: —1, giving the
equation 21' — y = 1. —5:r +31: +22 We have found one possible answer:
2:: —y D
1 . 41. To eliminate the arbitraryr constant t, we can solve the last equation for t to give it = z — 2.
a: = B + 5E2 — 2} and subst1tute z — 2 for t 1n the ﬁrst two equat1onsT obta1n1ng y = 4 + 313 _ 2} or 11 ...
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 Spring '09
 Derksen
 Algebra

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