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Given: $$(l_1) =(x,y,z)=(2,1,2)+t(a,1,0)$$ $$(l_2) =(x,y,z)=(5,2,3)+t(2,3,1)$$

Find the value of "$a$", if any, so that $(l_1)$ and $(l_2)$ intersect

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For the two lines to intersect at a point we need all of their coordinates to be the same. For that reason we want to solve the following:

$$\begin{pmatrix}2\\1\\2\end{pmatrix} + t_1 \begin{pmatrix}a\\1\\0\end{pmatrix} = \begin{pmatrix}5\\2\\3\end{pmatrix} + t_2 \begin{pmatrix}2\\3\\1\end{pmatrix}$$

Which, if we split them up into the parametric equations for each axis, are the following three equations:

\begin{cases} 2 + a t_1 = 5 + 2 t_2 \\ 1 + t_1 = 2 + 3 t_2 \\ 2 = 3 + t_2 \end{cases}

Rearranging them somewhat we have the following:

\begin{cases} -3 = - a t_1 + 2 t_2 \\ -1 = - t_1 + 3 t_2 \\ -1 = t_2 \end{cases}

Substituting the third expression, $t_2 = -1$ into the second and we find that

\begin{align*} -1 &= - t_1 + 3 \cdot - 1 \\ -1 &= - t_1 - 3 \\ 2 &= -t_1 \\ \therefore t_1 &= -2 \end{align*}

Putting both $t_1$ and $t_2$ back into the first of the simultaneous equations above and solve for $a$:

\begin{align*} -3 &= - a \cdot -2 + 2 \cdot -1 \\ -3 &= 2 a - 2 \\ -1 &= 2a \\ \therefore a &= - \frac{1}{2} \end{align*}

Finally, to confirm our findings (and to double check for mistakes), we plug it all into the original equation:

\begin{align*} \begin{pmatrix}2\\1\\2\end{pmatrix} - 2 \begin{pmatrix}-1/2\\1\\0\end{pmatrix} &= \begin{pmatrix}5\\2\\3\end{pmatrix} - \begin{pmatrix}2\\3\\1\end{pmatrix} \\ \begin{pmatrix}2\\1\\2\end{pmatrix} - \begin{pmatrix}-1\\2\\0\end{pmatrix} &= \begin{pmatrix}3\\-1\\2\end{pmatrix} \\ \begin{pmatrix}3\\-1\\2\end{pmatrix} &= \begin{pmatrix}3\\-1\\2\end{pmatrix} \end{align*}

And thus it was shown!


Edit: Notice how I have used two different parameters for the two lines. This is because two lines in space aren't dependent upon each other; using the same parameter causes problems. Take for example two expressions of the very same line:

\begin{align*} l_1: (x,y,z)=(1,1,1) + t_1 (1,1,1) \\ l_2: (x,y,z)=(1,1,1) + t_2 (2,2,2) \end{align*}

If we use the same parameter, we have this:

\begin{align*} l_1: (x,y,z)=(1,1,1) + t (1,1,1) \\ l_2: (x,y,z)=(1,1,1) + t (2,2,2) \end{align*}

which is quite problematic if we are to solve for points of intersection (of which there are an infinite amount, of course, since the two lines are the same). Each parametric equation will look like $1 + t = 1 + 2 t$, which leads to $t = 2t$ and we certainly aren't happy about that.

If, on the other hand, we use two different parameters, we simply find that $t_1 = 2 t_2$, which does indeed give us our desired result (that is, the two lines being the same).

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