Unit Vector question?

Question

How to find a unit vector perpendicular to a line?
Find two unit vectors perpendicular to the xy plane with equation 3x-4y=17.
I just need to know how the process works! Thanks

Answer 1

It is not clear if you mean find two unit vectors perpendicular to the xy plane (which would then be perpendicular to ALL lines in the xy plane)

OR find two unit vectors perpendicular to the PLANE 3x-4y=17 (=plane 3x-4y+0z=0)

OR find two unit vectors perpendicular to the LINE 3x-4y=17 (which is in the xy plane)



Since a plane has essentially only one unit vector perpendicular to it ((0,0,1 ) in the first case (0.6 , -0.8 ,0)) in the second case, I'll assume you meant the third possibility.



(This is not quite true. The vectors (0,0,-1 ) AND (0,0,1 ) are both unit vectors perpendicular to the xy plane but you'll note that they are "dependent" meaning that one is a multiple of the other,so if you've found one of them you've also found the other.The two are also on the same (perpendicular)line to the plane). The same applies to (0.6 , -0.8 ,0) and (-0.6 , 0.8 ,0). To see how I reached these vectors-see below.



Regarding the "unit vector" note that any vector can easily be transformed (or "normalised") into a unit vector by dividing its components by the sum of the squares of all the components.

For example the unit vector corresponding to the vector (1,2,3) (i.e. the vector that has the same direction as the original vector but has a length of 1) is (1/√14 , 2/√14 ,3/√14) the 14 coming from 1²+2²+3²=14 .



So the problem is reduced to finding ANY two vectors perpendicular to the line 3x-4y=17.You can do the normalising later.



There are an infinite number of vectors (and unit vectors) perpendicular to a line. Here's how to find some.The method I show below applies to ANY line in space not necessarily in the xy plane.



If your question is limited to a line in the xy plane and unit vectors inside this plane you can use the rule that the line ax+by+c=0 is always perpendicular to the vector (a,b) (and (-a,-b) )



A more general method that applies to any line in the plane is based on finding the "direction vector".



Find any two points on the line say (3,-2,0) (-1,-5,0) (the zero is because the line is in the xy plane. Finding the x and y is elementary -choose any x then find y or the other way around...)



The vector that connects these two points is (3,-2,0)- (-1,-5,0)=(4 , 3 ,0) so this is a "direction vector" of the line.

Any vector that is perpendicular to the line must be perpendicular to its "direction vector".



In order for vectors to be to be perpendicular their cross (scalar) product must be zero.



So we are looking for a,b,c such that: a*4+b*3+c*0=0



Obviously c can be anything so you can easily find as many solutions as you want to this equation.



Here's a few (found by choosing a convenient value for a then finding b from the equation. c can be any number..)



(3,-4,0) (3,-4,1) (3,-4,5) (3,-4,9)





Normalise any two of these and you have a solution.

Note:

1)The 3,-4 is the same as the coefficients of the line and of course this is no coincidence-see the rule I mentioned above.

If you choose any other value for a,b, after normalising you will reach the essentially the same unit vectors (I can't say the same since you can choose any number for c)

2) If you normalise (3,-4,0) you get the vector perpendicular to the PLANE 3x-4y+0z=17 that I mentioned above.

3)The vector (0,0,1) that I mentioned above is also perperndicular to this line since it is it perpendicular to the whole xy plane.)



Good luck!

Answer 2

In general, the hyperplane in n-dimensional space

c1 x1 + c2 x2 + ... + cn xn = d



has normal vector (c1, c2, ..., cn). For n=2, this turns into the equation for a line,

a x + b y = d



which has normal vector (i.e. vector perpendicular to the line) (a, b). Aside from using this very general formula, you can figure it out in this particular case. Suppose (x, y) is on your line, ax + by = d. Then (x+b, y-a) is also on the line: a(x+b) + b(y-a) = ax + ab + by - ab = ax + by = d. Thus you want a vector perpendicular to (x+b, y-a) - (x, y) = (b, -a). The usual 2D perpendicular vector formula says that (a, b) is such a vector. Alternatively, (b, -a) dot (a, b) = ab - ab = 0, so these are again perpendicular. This same style of reasoning works in general, though it gets a bit more complex.



So again, your line 3x - 4y = 17 has perpendicular vector (3, -4). This is not a unit vector, though. It has magnitude sqrt(3^2 + 4^2) = sqrt(25) = 5, so to make it a unit vector we have to divide by 5. In all, a unit vector perpendicular to 3x - 4y = 17 is (3/5, -4/5). We can also flip the vector to get a second unit vector perpendicular to 3x - 4y = 17 given by (-3/5, 4/5).

Answer 3

visualize a vector A from the beginning to (6, -4) A = 6i - 4j imagine the vector from (6, -4) to (-2, 8) as B and visualize vector C going from the beginning to (-2, 8) C = -2i + 8j From this "comic strip" it is sparkling that A + B = C and to that end B = C - A B = (-2i + 8j) - (6i - 4j) wish this receives you there...