# More linear algebra - Perpindicular Line spanned by a vector

• succubus
In summary, the conversation is discussing a linear algebra problem involving finding a basis for a subspace that is perpendicular to a given vector. The main point of confusion is understanding the concept of the line spanned by a vector in R^3 and how to find a basis for the perpendicular subspace. The conversation also touches on the idea of linear combinations and scalar multiples.
succubus
I'm back, I'm just brushing up on some linear algebra for my test tomorrow and am doing some recommended questions. I have a bit of confusion on some of them, and some of them I just want to make sure my thinking is correct.

So on to the problem...

Consider the line spanned by

1
2
3

in $$\Re^{3}$$ Find a basis of $$\bot$$L

So the basis would be the set of all independent column? vectors... Since it's $$\Re^{3}$$ I would assume that it is at least n x 3? So would I find a vector that when dotted against this would equal 0? :/

Do you know what the phrase "line spanned by [1 2 3]^T in R^3" means?

So the basis would be the set of all independent column? vectors
No, I don't think so. I'm not sure what you mean by independent column vectors. I think you might be getting ahead of yourself by thinking in terms of a matrix before you fully comprehend the geometry here.

... Since it's $\Re^{3}$
I would assume that it is at least n x 3?
What is? Again, I think you're getting ahead of yourself by thinking about matrices. At least that's what it seems like.
So would I find a vector that when dotted against this would equal 0?
That's an excellent idea! The perpendicular subspace consists of vectors that are perpendicular to the line, so the dot product of anyone of them with the vector that generates the line should be zero.

Write an equation that expresses this idea and you'll be on your way to solving this problem.

Well, I've come up with $$\overline{x_{1}}$$ + 2$$\overline{v_{2}}$$ + 3$$\overline{v_{3}}$$ = 0

And since the first column vector has a 1 coefficient (or can be made that way in any other case), then I solve for

$$\overline{v_{1}}$$ = -2$$\overline{v_{2}}$$ - 3$$\overline{v_{3}}$$

Then where do I go?

"Do you know what the phrase "line spanned by [1 2 3]^T in R^3" means?"
Could you explain a little in more detail. I somewhat understand the concept of a span and basis. We just started getting into these concepts in class. I'm shaky in my confidence. The concept is testing me spatially :(

Last edited:
succubus said:
Well, I've come up with $$\overline{x_{1}}$$ + 2$$\overline{v_{2}}$$ + 3$$\overline{v_{3}}$$ = 0
Nope, that's not it at all. You shouldn't have an equation with vectors in it after you have taken a dot product. Also, you're mixing up x's and v's, which makes things only slightly worse.
succubus said:
And since the first column vector has a 1 coefficient (or can be made that way in any other case), then I solve for

$$\overline{v_{1}}$$ = -2$$\overline{v_{2}}$$ - 3$$\overline{v_{3}}$$

Then where do I go?

"Do you know what the phrase "line spanned by [1 2 3]^T in R^3" means?"
Could you explain a little in more detail. I somewhat understand the concept of a span and basis. We just started getting into these concepts in class. I'm shaky in my confidence. The concept is testing me spatially :(

When the text talks about a space spanned by a collection of vectors, it means the set of all linear combinations of those vectors. For example, the span of {v1, v2, v3} is the set of all possible sums c1 * v1 + c2 * v2 + c3 * v3, where the c's are scalars and the v's are the vectors. If we're talking about the space spanned by one vector, the set of all linear combinations of one thing is c1 * v1. In other words, scalar multiples of the vector v1. This is a line through the origin. The line spanned by [1 2 3] is the line through the origin that goes through (1, 2, 3), and (2, 4, 6), and (-3, -6, -9), and (sqrt(7), 2sqrt(7), 3sqrt(7), and ...

Back to the problem at hand. You have a vector, [1 2 3]. You want to find the subspace of R^3 consisting of vectors that are perpendicular (hint: dot product) to [1 2 3]. What does your arbitrary, garden-variety vector in R^3 look like?

## 1. What is a perpendicular line in linear algebra?

A perpendicular line in linear algebra is a line that intersects another line at a right angle, or 90 degrees. This means that the two lines are orthogonal to each other and do not share any common direction.

## 2. How do you determine if a line is perpendicular to another in linear algebra?

In order to determine if a line is perpendicular to another in linear algebra, you can use the dot product. If the dot product of the two lines is equal to zero, then they are perpendicular. Another way is to use the slopes of the lines and check if they are negative reciprocals of each other.

## 3. Can a vector span a perpendicular line in linear algebra?

Yes, a vector can span a perpendicular line in linear algebra. This means that the vector can be used to create a perpendicular line that is orthogonal to another line.

## 4. What is the relationship between a vector and a perpendicular line in linear algebra?

In linear algebra, a vector and a perpendicular line are related in that the vector can be used to create the perpendicular line. This is because the direction of the vector is perpendicular to the direction of the line it spans.

## 5. How does finding a perpendicular line help solve problems in linear algebra?

Finding a perpendicular line can help solve problems in linear algebra by providing a way to create a new line that is related to the original line. This can help with calculations and visualizing geometric concepts, such as finding the shortest distance between two lines or planes. It can also be used in applications such as creating a basis for a vector space.

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