Intersecting Lines Solution
A line passing through the point $ (1,1,1) $ intersects the line defined by
\[\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + t \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix}\]at $ P, $ and intersects the line defined by
\[\begin{pmatrix} -2 \\ 3 \\ -1 \end{pmatrix} + s \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix}\]at $ Q $. Find point $ Q $.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
For the first line, $ P = (2t + 1, 3t + 2, 4t + 3) $. For the second line, $ Q = (s - 2, 2s + 3, 4s - 1) $.
Since $ (1,1,1), $ $ P, $ and $ Q $ are collinear, the vectors
\[\begin{pmatrix} 2t + 1 \\ 3t + 2 \\ 4t + 3 \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2t \\ 3t + 1 \\ 4t + 2 \end{pmatrix}\]and
\[\begin{pmatrix} s - 2 \\ 2s + 3 \\ 4s - 1 \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} s - 3 \\ 2s + 2 \\ 4s - 2 \end{pmatrix}\]will be proportional. Thus,
\[\frac{2t}{s - 3} = \frac{3t + 1}{2s + 2} = \frac{4t + 2}{4s - 2}.\]Let
\[k = \frac{2t}{s - 3} = \frac{3t + 1}{2s + 2} = \frac{4t + 2}{4s - 2}.\]Then
\begin{align*}
2t &= k(s - 3), \\
3t + 1 &= k(2s + 2), \\
4t + 2 &= k(4s - 2).\end{align*}From the first equation, $ 4t = k(2s - 6) $. Subtracting from the equation $ 4t + 2 = k(4s - 2), $ we get
\[2 = k(2s + 4).\]From the second equation, $ 6t + 2 = k(4s + 4) $. Subtracting the equation $ 4t + 2 = k(4s - 2), $ we get
\[2t = 6k,\]so $ t = 3k $. Substituting into the first equation, we get $ 6k = k(s - 3) $.
If $ k = 0, $ then from the equations above,
\[2t = 3t + 1 = 4t + 2 = 0,\]which is not possible. So $ k \neq 0, $ which gives us $ 6 = s - 3, $ and $ s = 9 $. Then $ Q = \boxed{(7,21,35)} $.
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