Find intersection point of two lines in 3D

Question

I need to find the intersection for the following two lines:

• $[x,y,z] = [2,-1,3]+k_1[1,2,3]$ and $[x,y,z] = [5,1,4]+k_2[3,2,1]$

So my approach is to find the intersection using gaussian elimination:

• $$ \left\{ \begin{array}{c} 2+k_1=5+3k_2 \\ -1+2k_1=1+2k_2 \\ 3+3k_1=4+k_2 \end{array} \right. $$
• $$ \left\{ \begin{array}{c} 3+3k_2=k_1 \\ 2+2k_2=2k_1 \\ 1+k_2=3k_1 \end{array} \right. $$
• $$ \left[ \begin{array}{cc|c} 3&3&1\\ 2&2&2\\ 1&1&3 \end{array} \right] $$
• And the result matrix I obtained is: $$ \left[ \begin{array}{cc|c} 1&1&\frac{1}{3}\\ 0&0&1\\ 0&0&0 \end{array} \right] $$

This is where the problem appears, I am not sure what value should I give for $k_1$ and $k_2$.

However, the answer key says the intersection is $(2,-1,3)$. Can anyone tell me where I did wrong?

Answer 1

After you get the system$$\left\{\begin{array}{l}3+3k_2=3k_1\\2+2k_2=2k_1\\1+k_2=3k_1,\end{array}\right.$$you should turn it into$$\left\{\begin{array}{l}-3k_1+3k_2=-3\\-2k_1+2k_2=-2\\-3k_1+k_2=-1.\end{array}\right.$$The second equation is redundant, since it is the first one times $\frac23$. So, consider just the system$$\left\{\begin{array}{l}-3k_1+3k_2=-3\\-3k_1+k_2=-1.\end{array}\right.$$Its only solution is $(k_1,k_2)=(0,-1)$. And then you will get that the intersection point is $(2,-1,3)$ (and not $(2,1,3)$).

Answer 2

When you’ve performed Gaussian elimination in the past, did you create the augmented matrix from a system of equations that had variables on both sides of the equations? Move all of the variables to the same side first. The resulting reduced matrix should make more sense.