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C4 two lines intersection with unknown

How do I find a in the equations

r=(426)+t(812)r = \begin{pmatrix} 4 \\ 2 \\ -6 \end{pmatrix} + t\begin{pmatrix} -8 \\ 1 \\ -2 \end{pmatrix} and r=(2a2)+s(925)r = \begin{pmatrix} -2 \\ a \\ -2 \end{pmatrix} + s\begin{pmatrix} -9 \\ 2 \\ -5 \end{pmatrix}

Never seen one like this before, it's 9 marks to find a and then the point of intersection. The intersection is easy once I find a, but i'm not sure how to find it.
Original post by Coda
How do I find a in the equations

r=(426)+t(812)r = \begin{pmatrix} 4 \\ 2 \\ -6 \end{pmatrix} + t\begin{pmatrix} -8 \\ 1 \\ -2 \end{pmatrix} and r=(2a2)+s(925)r = \begin{pmatrix} -2 \\ a \\ -2 \end{pmatrix} + s\begin{pmatrix} -9 \\ 2 \\ -5 \end{pmatrix}

Never seen one like this before, it's 9 marks to find a and then the point of intersection. The intersection is easy once I find a, but i'm not sure how to find it.


If the two lines intersect, there must have a point in common. So equate coefficients and solve for s,t,a. You have three equations in three unknowns.

Your first one will be:

4-8t = -2 -9s
(edited 13 years ago)

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