Two lines have equations
\(\mathbf{r} = \begin{pmatrix} 5 \\ 1 \\ -4 \end{pmatrix} + s \begin{pmatrix} 1 \\ -1 \\ 3 \end{pmatrix}\) and \(\mathbf{r} = \begin{pmatrix} p \\ 4 \\ -2 \end{pmatrix} + t \begin{pmatrix} 2 \\ 5 \\ -4 \end{pmatrix}\),
where \(p\) is a constant. It is given that the lines intersect.
Find the value of \(p\) and determine the coordinates of the point of intersection.
Solution
To find the intersection, equate the parametric equations of the lines:
\(\begin{pmatrix} 5 + s \\ 1 - s \\ -4 + 3s \end{pmatrix} = \begin{pmatrix} p + 2t \\ 4 + 5t \\ -2 - 4t \end{pmatrix}\).
Equating components, we get:
1. \(5 + s = p + 2t\)
2. \(1 - s = 4 + 5t\)
3. \(-4 + 3s = -2 - 4t\)
From equation 2: \(s = -3 - 5t\).
Substitute \(s = -3 - 5t\) into equation 3:
\(-4 + 3(-3 - 5t) = -2 - 4t\)
\(-4 - 9 - 15t = -2 - 4t\)
\(-13 - 15t = -2 - 4t\)
\(-11 = 11t\)
\(t = -1\).
Substitute \(t = -1\) into \(s = -3 - 5t\):
\(s = -3 - 5(-1) = 2\).
Substitute \(s = 2\) and \(t = -1\) into equation 1:
\(5 + 2 = p + 2(-1)\)
\(7 = p - 2\)
\(p = 9\).
Thus, the value of \(p\) is 9. The intersection point is:
\(\begin{pmatrix} 5 + 2 \\ 1 - 2 \\ -4 + 3(2) \end{pmatrix} = \begin{pmatrix} 7 \\ -1 \\ 2 \end{pmatrix}\).
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