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# Unit Normal Vectors watch

1. I need help with 2 questions in an exercise:
1) Find a unit normal vector for the line −4x − 6y − 5 = 0
2) Find the equation of a line passing through point A(−1, −1) and orthogonal to vector
v = [−2, −1]T

I honestly do not know where to start with this. I've found plenty of help for 3D planes, but very little on 2D.

Note: I have put in random numbers; so I can't just copy someone else's work.
Also a link to a helpful tutorial on the subject would be appreciated
2. Normal Vector to a line: http://www.math.washington.edu/~king...ls-planes.html

You can then divide by its length to make it a unit normal vector.

Still trying to find the links for the other bit.
3. (Original post by claret_n_blue)
Normal Vector to a line: http://www.math.washington.edu/~king...ls-planes.html

You can then divide by its length to make it a unit normal vector.

Still trying to find the links for the other bit.
I do not understand where you get the values for A from...
4. (Original post by matte30)
I do not understand where you get the values for A from...
A vector is defined using two things: 1) Size (magnitude), 2) Direction. In the case of your question, we only need to look at direction. Let's assume we have a vector

.

What this effectively tells us that the vector goes 'm' units along the 'm' axis for every 'b' units along the 'y' axis. From your example, we see that the vector travels -2 units along the 'x' axis for every -1 unit along the 'y' axis, so it is pointing towards the third quadrant (bottom left) of your graph.

To get an orthogonal vector, we just switch the two numbers and change the sign on the new 'x' "value". So a vector orthogonal to my vector , lets call it , would be

.

A line is of the form y = mx + c. In the question, they have given you the point they want the line to pass through, which we will say are (a,b). So what you do is write the equation as a parametric equation and then solve for your new variable. So your line will be

where we have

and .

You then let both equations equal 't' and then equate them giving you your line.
5. for q1 using −4x − 6y − 5 = 0:

A = [-4, -6]^T
|A| = sqrt(16 + 36)
= sqrt(52)
=> unit normal vector is [-4/(sqrt 52), -6/(sqrt 52) ]^T

for q2 using v = [-2, -1]^T; a = b = -1:

x = 2t - 1
=> t = 0.5 (x) - 0.5

y = -1 -t
=> t = -1 -y

0.5 (x) - 0.5 = -1 -y
=> y = -0.5x -0.5

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Updated: April 16, 2013
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