# Homework Help: I'll get by with a little help from my friends- another parametric equation problem

1. Sep 15, 2010

### jhg12345

1. The problem statement, all variables and given/known data
Give the Cartesian equation of the following hyperplane:
The plane spanned by (1,1,1) and (2,1,0) and passing through (1,1,2)

2. Relevant equations

3. The attempt at a solution
I keep getting 4 = -2x(sub1) + -2x(sub2) +x(sub3). However, the answer is x(sub1) - 2x(sub2) + x(sub3) = 1. Someone please help! What I am doing wrong? Gabba guy come back you seemed nice. You say (a,b,c) + t(d,e,f) is not a plane but a line. But using the other equation doesn't work either.

2. Sep 15, 2010

### Staff: Mentor

Re: I'll get by with a little help from my friends- another parametric equation probl

If you know the normal n = <a, b, c> to a plane and a point P(x0, x0, x0) on it, an equation for the plane is a(x - x0) + b(y - y0) + c(z - z0) = 0.

You have two vectors in the plane, <1, 1, 1> and <2, 1, 0>. Can you find a third vector that is perpendicular to these two vectors?

3. Sep 15, 2010

### jhg12345

Re: I'll get by with a little help from my friends- another parametric equation probl

Is the normal n = (-1, 0, 1)? I subtracted the two vectors and plugged the result into the dot product equation and set it equal to 0.

4. Sep 15, 2010

### Staff: Mentor

Re: I'll get by with a little help from my friends- another parametric equation probl

No. Subtracting the two given vectors does not give you a normal to the plane.

Any linear combination of the two given vectors (au + bv, including 1u + (-1)v) gives you another vector in the plane. That's what it means to say that the two vectors span the plane.

If you are given two vectors, what operation can you perform on them to get a third vector that is perpendicular to the first two vectors?

5. Sep 15, 2010

### jhg12345

Re: I'll get by with a little help from my friends- another parametric equation probl

Put both vectors in a matrix and set it equal to 0?

6. Sep 15, 2010

### Staff: Mentor

Re: I'll get by with a little help from my friends- another parametric equation probl

Now you're just guessing. The only way a matrix can be zero is if all its elements are zero.

Surely you have heard of the cross product?

7. Sep 15, 2010

### jhg12345

Re: I'll get by with a little help from my friends- another parametric equation probl

Yea, but the section we've done so far doesn't use cross product. And can't the matrix be zero if the column vectors (that you solve for) cause the left side of the matrix to equal 0?

Last edited: Sep 15, 2010
8. Sep 15, 2010

### Staff: Mentor

Re: I'll get by with a little help from my friends- another parametric equation probl

??? I have no idea what you're asking here.

Since you don't know about the cross product, you can still find a vector that is perpendicular to <1, 1, 1> and <2, 1, 0> by using the dot product.

Let n = <a, b, c>

You want n $\cdot$ <1, 1, 1> = 0 and n $\cdot$ <2, 1, 0> = 0.
Can you solve these equations for a, b, and c?

9. Sep 15, 2010

### jhg12345

Re: I'll get by with a little help from my friends- another parametric equation probl

I subtracted a + b + c = 0 from 2a + b = 0 and got a = c.

10. Sep 15, 2010

### Staff: Mentor

Re: I'll get by with a little help from my friends- another parametric equation probl

What about b? You need to find values for a, b, and c. Since there are two equations in three unknowns, you will have one arbitrary variable (that you can set to any value).

You'll have an infinite number of solutions for a, b, and c, since there are an infinite number of vectors that are perpendicular to <1, 1, 1> and <2, 1, 0>. They are all scalar multiples of one another, though. All you need is a single vector.

11. Sep 15, 2010

### jhg12345

Re: I'll get by with a little help from my friends- another parametric equation probl

So vector (1,-2,1) will be just one of the possible normals. Gotcha.

12. Sep 15, 2010

### jhg12345

Re: I'll get by with a little help from my friends- another parametric equation probl

Yea, I wasn't solving for a,b and c before. Thanks for clearing that up.