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thomsd
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have this question

The points A and B have position vectors \begin{pmatrix} 1 \\ 6\\ 4\end{pmatrix} and \begin{pmatrix} 5 \\ 0\\ -6 \end{pmatrix} respectively, relative to a fixed origin.

Find in vector form an equation of the line which passes through A and B

I found the directional vector for this equation which I took vector B away from A getting directional vector \begin{pmatrix} 4 \\ -6 \\ -10 \end{pmatrix} which simplifies to \begin{pmatrix} 2 \\ -3 \\ -5\end{pmatrix}

so is the equation r=\begin{pmatrix} 1\\ 6 \\ 4 \end{pmatrix} +s\begin{pmatrix} 2\\ -3 \\ -5 \end{pmatrix} is this correct as the second part asked for the intersection point between this line and another which I know how to do, but came out with odd values which didn't work when i solved the simultaneously.
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Zhen Lin
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Your equation appears to be correct. Perhaps there was a printing error?
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generalebriety
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Well, maybe you can post your working for the second half of the question, since that's where you're having trouble?
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thomsd
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okay using the line I have found which we think is right
r=\begin{pmatrix} 1 \\ 6 \\ 4\end{pmatrix}+s\begin{pmatrix} 2 \\ -3 \\ -5\end{pmatrix} and line m from question which is \begin{pmatrix} 5 \\ -5 \\ 3 \end{pmatrix}+t\begin{pmatrix} 1\ -4 \\ 2\end{pmatrix}

x=1+2s=5+t
y=6-3s=-5-4t
z=4-5s=3+2t

which after so
s=39/23
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clear.twilight
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I get s = 1.

Try doubling the first equation. Take the third away from it. You get -2 + 9s = 7.
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Candescence
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How did you get to s=39/23, you should get s=1 and t =-2
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