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This is for my linear algebra assignment: enter image description here

This is a hard question in the assignment for me because we never did planes and line in R^4 (yet). Thus, I managed to find the answer for a). But b) is pretty hard. Since we can't do the cross product to find the normal vector, I had to proceed through another way. I tried to arbitrarily assign a direction vector for the line in question, then try to find if it is equal to sU + tV of the plane that I founded in a). I am not entirely sure if this approach is correct to find the non parallel line to the plane in (a). Then I have no idea how to find a way to make that line not intersect the plane. Here is my attempt at this question so far:

enter image description here

enter image description here

Thanks everyone in advance!

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You have vectors $\vec{AB}$ and $\vec{AC}$. Try finding some other 2 vectors independent to them.

for example $\vec{AB}$(1,3,-2,2),$\vec{AC}$(0,1,0,2),$\vec{m}$(1,0,0,0),$\vec{n}$(0,0,1,0). It is really simple finding it, try some random vectors.

Now you move from A by vector $\vec{m}$ it is point (2,0,1,0) and put a line in direction of $\vec{n}$. So parametric expression for the line is (2,0,1+k,0).

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  • $\begingroup$ Wait, what is the vector m and n? I didn't get that part. $\endgroup$ – You Xiao Ruan Oct 15 '17 at 23:08
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    $\begingroup$ In this space you can have 4 independent vectors, like coordinate vectors. You pick any vectors, it is only matter that the determinante of the 4 vectors is regular. Now you know that the line will not intersect the plane cos of independency. And also line is not parallel to plane cos n is not formed from AB and AC $\endgroup$ – Djura Marinkov Oct 16 '17 at 5:01

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