Finding the Cartesian Equation of a Hyperplane

[jhg12345]

Homework Statement

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)

Homework Equations

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.

 

[Mark44]

Homework Statement

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)

Homework Equations

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.

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?

 

[jhg12345]

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.

 

[Mark44]

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.

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?

 

[jhg12345]

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

 

[Mark44]

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?

 

[jhg12345]

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?

 

[Mark44]

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?

? 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?

 

[jhg12345]

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

 

[Mark44]

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.

 

[jhg12345]

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

 

[jhg12345]

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