# Proving perpendicular vectors

1. Aug 11, 2008

### nobahar

Hello, this is a question from a pure mathematics textbook:
(2D vectors)
The vectors a and b are of equal magnitute k (k does not equal 0), and the angle between a and b is 60 degrees. If c=3a-b and d=2a-10b
A) Show that c and d are perpendicular vectors.
(Sadler, A.J., Thorning, D.W.S (2007, pg.63). Understanding Pure Mathematics, Glasgow: Oxford Universty Press).

c.d=0
=|c||d|cos(90)
Therefore, c.d/|c||d|=cos(90)

'Normally' for the dot product, the coefficients of the 'i' component of the two vectors are multiplied togeather, and the same process is applied to the j components, then they are added togeather.
I would attempt the question, but to be honest I'm not sure what steps to take; since my base vectors a and b are not orthogonal. Am I using the correct method? An answer would be much appreciated, I apologise for the language used and the presentation, I hope its accurate.

2. Aug 11, 2008

### tiny-tim

Welcome to PF!

Hi nobahar! Welcome to PF!

ah … you are about to understand the beauty of vectors.

You don't need components to understand vectors!

Vectors have a life of their own … they "don't know" they have components … components are just things we use when all else has failed.

Hint: just expand (3a-b).(2a-10b) in terms of a.a b.b and a.b

3. Aug 11, 2008

### HallsofIvy

Staff Emeritus
Additionally you are told that "The vectors a and b are of equal magnitute k (k does not equal 0), and the angle between a and b is 60 degrees" so $\vec{a}\cdot\vec{b}= k^2 cos(60)= (1/2)k^2$ and, of course, $\vec{a}\cdot\vec{a}= \vec{b}\cdot\vec{b}= k^2$.

4. Aug 13, 2008

### nobahar

I appreciate the response very much, thankyou.
In the same textbook:
(x1i+y1j).(x2i+y2j)
a.b=x1x2i.i+x1y2i.j+y1x2j.i+y1y2j.j
i.i=j.j=1 and i.j=j.i=0
therefore, a.b=x1x2+y1y2
Just wondering if someone could expand on this for me. I tried applying this to the question that I opened this topic with, before expanding it completely as suggested very kindly above. Presumably it is to do with being unit vectors? An explanation would really help me to grasp the concepts alot better, thankyou in advance.

5. Aug 13, 2008

### HallsofIvy

Staff Emeritus
It's not clear what your question is. As far as this particular example is concerned, yes, the fact that i and j are unit vectors and are perpendicular to one another, is essential. That is what allows you to say that i.i= j.j= 1 and i.j= j.i= 0 and so "get rid" of i and j.

As for your original question, (3a-b).(2a-10b) is concerned, you can use the distributive law to argue that (3a-b).(2a-10b)=3a.(2a-10b)-b.(2a-10b) and then using the distributive law on each of those, =6a.a-30a.b-2b.a+10b.b. That was tiny-tim's point.

The difference is that you do not know that a and b are unit vectors nor that they are they perpendicular. But because you do know their lengths and angle between them, you can do the same thing. My point, that you are told that a.b= b.a= (1/2)k2 and that a.a= b.b= k2, you can replace those dot products with those numbers.

Last edited: Aug 14, 2008
6. Aug 13, 2008

### nobahar

Firstly, thankyou again for the response.
I already applied those processes (the expansion and then the appropriate substitution for a.a, b.b and a.b with k2 and (1/2)k2) to get the required result after the first responses. The solution, although appreciated, was not necessary; I was looking for something more in the way of an explanation as to why i.i=j.j=1 and i.j=j.i=0. The above response has gone some way to addressing this issue, which I am grateful for.

7. Aug 13, 2008

### Dick

Like Halls said, i and j are chosen to be unit length (so i.i=j.j=1) and perpendicular (i.j=0). In the usual x-y coordinates you would write i=(1,0), j=(0,1). Check these dot products with your x,y formula.

8. Aug 13, 2008

### nobahar

Of course! The scalar product is 0 for i.j and j.i because they are perpendicular! Thankyou Dick, Halls and Tiny-tim. I am immensely grateful, greatly appreciative, and slightly disappointed with myself.