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There exists a real number \(k\) such that the equation \(\begin{pmatrix} -1 \\ -2 \end{pmatrix} + t \begin{pmatrix} \phantom -3 \\ -2 \end{pmatrix} = \begin{pmatrix} 5 \\ 0 \end{pmatrix} + s \begin{pmatrix} 4 \\ k \end{pmatrix}\)
does not have any solutions in t and s. Find k.

 Feb 8, 2020
 #1
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In the left-hand side, the y-coordinate is -2t - 2, which can be any real number.  If k = 0, then in the right-hand side, the y-coordinate can only be 0, so the answer is k = 0.

 Feb 8, 2020
 #2
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-1+3t=5+4s

3t=4s+6

6t=8s+12 (1)

 

-2-2t=sk

-2t=sk+2

-6t=6+3sk  (2)

 

add

0=18+8s+3sk

-18=s(8+3k)

 

 

\(8+3k\ne0\\ k\ne \frac{-8}{3}\)

 Feb 9, 2022

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