1
$\begingroup$

This is the first year linear algebra question. Which of the following points is on the line containing the points $(1,0,5)$ and $(3,1,-2)$?

a). $( 0,-3,17)$

b). $(9,4,-23)$

c). $(-9,-4,23)$

d). $(0,3,-17)$

e). $(-1,-1,-12)$

The answer is b).

  • 1
    http://mathworld.wolfram.com/Collinear.html2012-12-07

1 Answers 1

3

The line $L$ through the points $(1,0,5)$ and $(3,1,-2)$ is parallel to the line $L'$ through the origin and

$(3,1,-2)-(1,0,5)=(2,1,-7)\;.$

Because it does pass through the origin, $L'$ is just the set of scalar multiples of $(2,0,-7)$, i.e.,

$L'=\big\{t(2,1,-7):t\in\Bbb R\big\}\;.$

$L$ is parallel to this and passes through $(1,0,5)$, so it’s just $L'$ shifted by $(1,0,5)$:

$L=\big\{(1,0,5)+t(2,1,-7):t\in\Bbb R\big\}\;.$

Or you can simplify $(1,0,5)+t(2,1,-7)$ to get

$L=\big\{(1+2t,t,5-7t):t\in\Bbb R\big\}\;.$

Now just check to see which of your points actually fit this description. Take the point $(0,3,-17)$, for instance: if it’s on $L$, the second coordinate tells you that $t$ must be $3$, but then the first coordinate ought to be $1+2\cdot3=7$. It isn’t, so this point isn’t on $L$.

  • 0
    @SusumuMaxYakushijin: You’re very welcome.2012-12-07