# Perpendicular Unit Vectors

1. Jul 30, 2015

### PhotonSSBM

1. The problem statement, all variables and given/known data
From Kleppner and Kolenkow Chapter 1 (Just checking to see if I'm right)

Given vector A=<3, 4, -4>
a) Find a unit vector B that lies in the x-y plane and is perpendicular to A.
b) Find a unit vector C that is perpendicular to both A and B.
c)Show that A is perpendicular to the plane defined by B and C.

2. Relevant equations
Cross Product
Dot Product
Knowing how to find a plane from two vectors

3. The attempt at a solution
a) Using the following system...
3x+4y=0 (using dot product and z=0)
x+y=1 (For it to be a unit vector the sum of the two components must be 1)
The perendicular vector is
$<4, -3, 0>$
and the unit vector after dividing by the magnitude is...
$B=<4/5, -3/5, 0>$

b)Finding the cross product of A and B and dividing by the magnitude...
$C=<\frac{-12,}{5\sqrt{41}}, \frac{-16}{5\sqrt{41}}, \frac{-5}{\sqrt{41}}>$

c)$B \times C=<75, 100, -100>$
Therefore the plane is defined by...
$3x+4y-4z=0$
Therefore A is perpendicular since its components equal the constants that define the plane.

I'm pretty sure I did everything right I just wanted to make sure and see if any of my wording is dumb or if I made a mistake. I don't know how to write the last part formally. Thanks all!

2. Jul 30, 2015

### Dr. Courtney

You can show things are perpendicular by computing the dot product between two vectors.

3. Jul 31, 2015

### HallsofIvy

Staff Emeritus
NO!! For it to be a unit vector we must have $\sqrt{x^2+ y^2}= 1$, not x+ y!

So x+ y is NOT 1!

4. Jul 31, 2015

### PhotonSSBM

I see, but since the unit vector I found follows that condition...
$\sqrt{(\frac{4}{5})^2 + (\frac{-3}{5})^2} = 1$
is my solution still right?