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I was given a multiple choice practice question by my lecturer on finding a vector orthogonal to two given lines. I do understand that I have to use the cross product to do this. I was initially given these two lines represented as parametric equations.

$x=-1+3t,y=3-2t,z=1+t$

$x=4+5t,y=2-t,z=-1-2t$

I have been given the answer which is: $-5i-11j-7k$
However, I do not know how to reach this answer. I have attempted to let t=0 and then cross-product the values which gave me $-5i+3j+10k$ where all three values were present somewhere in one of the given 4 choices, however, I could not reach the actual answer.

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You should do the next procedure:

To get the vectors for the lines, substitute $t$ in the equations for 0 and 1, so that you get the directional vectors for each line: $\bar{v_1}=v_1(t=1)-v_1(t=0)=(3,-2,1)$ and $\bar{v_2}=v_2(t=1)-v_2(t=0)=(5,-1,-2)$

Then if you do the cross product between those two vectors, you will obtain =

$\bar{v_1}\times\bar{v_2} = 5i+11j+7k $

Which is the negative of the vector that you give as an answer, so it is also orthogonal to the lines you specified.

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You have to visualize what a parametrization of a line looks like and what parts goes into making it:

A line $l$ can be represented by a vector $$\vec{l}=\vec{r}_0+t\cdot \vec{r}$$ where $\vec{r}_0$ is vector that goes from the origin to anywhere on the line, and $\vec{r}$ is a vector in the direction of the line (try drawing this if in doubt; any point on the line will be reached if you just vary $t$ and use the usual vector addition).

So it is only $\vec{r}$ that is parallel to the line. Thus the cross product should be $$\begin{pmatrix}3 \\ -2 \\ 1 \end{pmatrix} \times \begin{pmatrix}5 \\ -1 \\ -2 \end{pmatrix}$$

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