Proving Perpendicular Vectors in 2D with Scalar Product

[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.

 

[tiny-tim]

Welcome to PF!

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.
…
'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.

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

 

[HallsofIvy]

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, $\vec{a}\cdot\vec{a}= \vec{b}\cdot\vec{b}= k^2$.

 

[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 a lot better, thankyou in advance.

 

[HallsofIvy]

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 a lot better, thankyou in advance.

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.

 

[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.

 

[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.

 

[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.