10.7#16

 I can't seem to solve question 16 on 10.7 and need help. I got part A, but my method doesn't work for parts B or C, so I think I just got lucky on part A. Thanks.

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One parameterization of the line will be
r(t)=u+tv,
where u=<1,-1,2> and v=<4,1,-4>, but v can be replaced by α<4,1,-4> for any nonzero constant α and u can be replaced by any other vector with time on the line,  i.e. u can be replace by
<1,-1,2> +β<4,1,-4> for any real number β.  Since u+0v = <1,-1,2> and u+1v =<5,0,-2>, all we have to do to get the answer to (a) is replace v by v*=v/5, so now we have the equations
u+0v* = <1,-1,2> and u+5v* =<5,0,-2>, and your answer <1,-1,2>+t<4/5,1/5,-4/5>  is the correct one.  In more generality, you would need to look at something like
r(t)=(<1,-1,2> +β<4,1,-4>)+tα<4,1,-4> 
and choose α,β to fit your parameters.  For example in case (b) r(5)= (<1,-1,2> +β<4,1,-4>)+5α<4,1,-4>=<1,-1,2>  and
r(7)= (<1,-1,2> +β<4,1,-4>)+7α<4,1,-4>=<5,0,2>  so we get equations
5α+β=0 and 7α+β=1 from which 2α=1=>α=1/2, and β=-5/2,
r(t)=(<1,-1,2> -5/2<4,1,-4>)+t/2<4,1,-4>
    = <-9,-7/2,12> +t<2,1/2,-2>