Find all unit vectors orthogonal to the line

[annpaulveal]

Homework Statement

Find all the unit vectors orthogonal on the line L.

Homework Equations

L passes through the vectors u = [9; 7] and v = [1; -5]

The Attempt at a Solution

I found the slope of L from the two vectors: 3/2. I know that to be orthogonal, the vector will have a slope of -2/3, and that to be a unit vector, the square root of (x^2 + y^2) will be 1. I tried to define the line using the slope and one of the vectors, coming up with y = -2/3x + 13, and then using this value for y in the square root equation to solve for x. However, the square root equation yields imaginary values. Could someone give me some guidance as to where I've gone wrong?

 

[mfb]

It just means that your arbitrary orthogonal line does not have to have points of unit length - and even if it had, those are irrelevant. You need a vector (x,y) which gives the calculated slope. You can care about x^2+y^2=1 afterwards with a prefactor.

 

[vela]

This plot may help you see where you've gone astray. The blue line is L. The red line is the other line you found. And then there's the unit circle. Unit vectors will extend from the origin to a point on the unit circle.

 

[LCKurtz]

Homework Statement

Find all the unit vectors orthogonal on the line L.

Homework Equations

L passes through the vectors u = [9; 7] and v = [1; -5]

The Attempt at a Solution

I found the slope of L from the two vectors: 3/2. I know that to be orthogonal, the vector will have a slope of -2/3, and that to be a unit vector, the square root of (x^2 + y^2) will be 1. I tried to define the line using the slope and one of the vectors, coming up with y = -2/3x + 13, and then using this value for y in the square root equation to solve for x. However, the square root equation yields imaginary values. Could someone give me some guidance as to where I've gone wrong?

You don't need the equation of any lines. A vector between those points is $\langle 8,12\rangle$, or a shorter one is $\langle 2,3\rangle$. A vector $\langle a,b\rangle$ perpendicular would satisfy $\langle a,b\rangle\cdot \langle 2,3\rangle=0$. Find any nonzero $a$ and $b$ that work, make a unit vector out of it and attach a $\pm$ sign and you will have it.