line in question: $$-x - 1 = \frac 12y - \frac12 = \frac 12z + 1$$ I wasn't really sure how to go about this one since it's not in the exact general form that I was taught, but I went along and tried to do it anyway: $$-x-1=t \rightarrow x = -t-1$$ $$\frac 12y - \frac 12 =t \rightarrow y = \frac t2+1$$ $$\frac 12z + 1 = t\rightarrow z = \frac t2 - \frac 12$$ and got: $$(-1, 1, -\frac 12)+t(-1, \frac 12, \frac 12)$$ Is this correct? If not, how do I do this problem?
PythonNewb asked Apr 1, 2015 at 10:08 PythonNewb PythonNewb 417 3 3 gold badges 7 7 silver badges 19 19 bronze badges$\begingroup$ That "vector in question" should probably be changed to "straight line in space in question" . Besides this, it looks fine to me. $\endgroup$
Commented Apr 1, 2015 at 10:09$\begingroup$ Maybe a typo it should be that $y=2t+1$ and $z=2t-2$ With that your solution is $(-1, 1, -2)+t(-1, 2, 2)$ $\endgroup$