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## windsylph 4 years ago : Linear Algebra: Describe the intersection of the two hyperplanes x1 + 2x2 - x3 + x4 = 8 and -x1 + 3x2 + x3 - x4 = 2.

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1. victorarana

They intersecate on a plane

2. windsylph

Is there a way to get an equation for that plane?

3. tdabboud

You could set up the augmented matrix, row reduce to get the values and that should be the point of intersection

4. windsylph

yep I did that.. and haha seemingly im stuck on a dumb question: let a and b be free variables. my solution then is (a-b+4, 2, a, b)^T. How do I express that as a span, then as an equation?

5. tdabboud

i get x2 = 2; x3 = x1 + x4 - 4; So like you said x1 and x4 are free so the span is: =(2 , -4) + x1(1, 0, 1, 0) + x4(0, 0, 1, 1) (these are vertical vectors I just cant write it on here) =(2, -4) + x1u + x4v (u and v are vectors above)

6. tdabboud

understand?

7. tdabboud

I made a small mistake: It is x3 and x4 that are free so it becomes x1=4 +x3 - x4 x2 =2 which translates to... =(2 , -4) + x3(1, 0, 1, 0) + x4(-1, 0, 0, 1)

8. tdabboud

(4, 2, 0, 0) sorry

9. windsylph

thanks, i was looking for an answer of the form span(u,v), but if that's that then that's okay. how about an equation for the plane?

10. tdabboud

well span(u,v) is just the span of those vectors: span{ (1, 0, 1, 0) , (-1, 0, 0, 1)}

11. windsylph

oh, so (4,2,0,0)^T doesn't matter when expressing it in that form?

12. tdabboud

nope because it only depends on the free variables

13. windsylph

okay thanks. is there a way to express that as an equation?

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