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!


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).

| cite | improve this answer | |
  • $\begingroup$ Wait, what is the vector m and n? I didn't get that part. $\endgroup$ – You Xiao Ruan Oct 15 '17 at 23:08
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.