How do I find the orthogonal basis for this plane?

Question

Question:

$P$ is a plane through the origin given by

$x + y + 2z = 0$.

Find an orthogonal basis v₁, v₂ ∈ $P$.

My answer:

I'm assuming the question asks for two vectors that span this plane $P$. But the chapter that this problem is for doesn't say anything about the $x,y,z$ equation of a plane that was given here...so I did some searching online and learned that this helps find the "normal vector".

In this case it would be $n = (1,1,2)$, right?

Then if all the vectors that span this plane are orthogonal to the normal vector, I can use the dot product.

I chose the following two vectors:

v₁ = $(1,1,-1)$

v₂ = $(3,3,-3)$

Was this question answered correctly?

Your vectors $v_1$ and $v_2$ are parallel which is not correct for this question. Try to find another $v_2$ which is orthogonal to both $n$ and $v_1$. — Spencer, Jan 19 '14 at 03:42

score 4 · Answer 1 · answered Jan 19 '14 at 03:53

First, find a basis for the plane as

$$ x + y + 2z = 0 \implies x=-y-2z \implies (x,y,z)=(-y-2z,y,z)$$

$$ \implies (-y,y,0)+(-2z,0,z)\implies y(-1,1,0)+z(-2,0,1). $$

So, our basis is $\left\{(-1,1,0),(-2,1,0)\right\}$. Now, you need to apply Gram-Schmidt process to the basis set to get the orthogonal basis.

score 1 · Accepted Answer · answered Jan 19 '14 at 03:43

1

Your solution is not correct as $v_1$ and $v_2$ are linearly dependent. However, your approach is correct. We need to find just another vector orthogonal to $n$ which is not a scalar multiple of $v_1$. Note that $(0,1,-2)$ is such a vector.

answered Jan 19 '14 at 03:43

voldemort

13,182

Note that $(0,1,-2)$ is indeed linearly independent to $(1,1,-1)$, but not orthogonal to it. – Olivier Jan 19 '14 at 03:59
Yah- but then you can apply Gram Schmidt. – voldemort Jan 19 '14 at 04:20
Haven't yet learned about Gram Schmidt, but if I'm looking for a vector orthogonal to both n and v1, can't I set up an equation like:
n dot v2 = v1 dot v2 = 0, and solve for the components?
– Bobby Lee Jan 19 '14 at 18:52
1

yah- or you can just take the vector $n$ x $v_1$ – voldemort Jan 19 '14 at 19:02

Olivier · Answer 3 · 2014-01-19T04:02:50.260

1

The question asks for two vectors in the plane $P$ that are orthogonal to each other. You can choose any vector $\vec{v_1}$ as the first vector, but not any vector $\vec{v_2}$. In fact, there are two vectors that are in the plane $P$ and that are orthogonal to $\vec{v_1}$ (try to find out why).

Furthermore, any vector in $P$ will be orthogonal to $\vec{n}$, by definition. We are thus looking for a vector $\vec{v_2}$ that is orthogonal to both $\vec{v_1}$ and $\vec{n}$.

One of the ways to find $\vec{v_2}$ is to use the vectorial product...

edited Jan 19 '14 at 04:02

answered Jan 19 '14 at 03:54

Olivier

3,893

Doesn't $\vec v_1$ have to be in the span? – Ahmed S. Attaalla Oct 08 '16 at 16:43

score 1 · Answer 4 · answered Jan 19 '14 at 07:29

1

Chose $v_1=(-1,1,0)$ and $v_2=v_1\times n$. No Gram-Schmidt needed in this case.

answered Jan 19 '14 at 07:29

Michael Hoppe

18,103
3
32
49

1

Is it necessary that $v_1$ is in the span? If yes, how do you know $<-1,1,0>$ is in the span? – Ahmed S. Attaalla Oct 08 '16 at 16:52

score 0 · Answer 5 · edited Oct 08 '16 at 16:42

0

$n = (1, 1, 2)$ - normal to plane
$v_1 = (1,1,-1)$ - first base vector on plane
$v_2 = n \times v_1 = (-3,3,0)$ - second base vector on plane

edited Oct 08 '16 at 16:42

Eman Yalpsid

3,044

answered Oct 08 '16 at 16:27

leonid

11

How do I find the orthogonal basis for this plane?

5 Answers5