# Thread: Proving Vector Dot Product Properties

1. ## Proving Vector Dot Product Properties

I have a big hole in my math skills in that I don't really no where to begin in proving properties. I can use the dot product to work out things just not sure how to prove the properties.

Let U = (Ux, Uy, Uz), Let V = (Vx, Vy, Vz), Let W = (Wx, Wy, Wz), Let c and k be scalars. Prove the following dot properties.

A. U . V = V . U
B. U . (V + W) = U . V + U . W
C. k (U . V) = (kU) . V = U . (kV)
D. V . V = || V ||^2
E. 0 . V = 0

I have worked out D as:-

V . V = Vx Vx + Vy Vy + Vz Vz
= Vx^2 + Vy^2 + Vz^2
= ( SQRT( Vx^2 + Vy^2 + Vz^2) )^2
= ( ||V|| )^2

With A I know that,

U . V = Ux Vx + Ux Vy + Uz Vz
V . U = Vx Ux + Vy Uy + Vz Uz

I just don't know how to prove U . V = V . U without using values. My problem is the same for the remaining questions.

What topic do I need to study further to improve this skill I'm lacking?

2. Originally Posted by RedKMan
I just don't know how to prove U . V = V . U without using values.
Just notice that $\displaystyle (U_x)(V_x)=(V_x)(U_x)$.
That is true for the other two components as well.
Multiplication is commutive.

3. So I could simply put:-

(Ux Vx + Uy Vy + Uz Vz) = (Vx Ux + Vy Uy + Vz Uz) = U . V = V . U

This would be enough to prove that U . V = V . U ?

4. Originally Posted by RedKMan
So I could simply put:-

(Ux Vx + Uy Vy + Uz Vz) = (Vx Ux + Vy Uy + Vz Uz) = U . V = V . U

This would be enough to prove that U . V = V . U ?
Yes it is exactly that.
U . V =(Ux Vx + Uy Vy + Uz Vz) = (Vx Ux + Vy Uy + Vz Uz) = V . U

5. For this property, U . (V + W) = U . V + U . W

I have,

U . ( V + W ) = ( Ux(Vx + Wx) + Uy (Vy + Wy) + Uz (Vz + Wz)

= (Ux Vx + Uy Vy + Uz + Vz) + (Ux Wx + Uy Wy + Uz Wz) = U . V + U.W

Do I have the right idea here?

6. Originally Posted by RedKMan
For this property, U . (V + W) = U . V + U . W

I have,

U . ( V + W ) = ( Ux(Vx + Wx) + Uy (Vy + Wy) + Uz (Vz + Wz)

= (Ux Vx + Uy Vy + Uz + Vz) + (Ux Wx + Uy Wy + Uz Wz) = U . V + U.W

Do I have the right idea here?
Yes, that works nicely.
If would be nice if you could learn to use LaTex.

$$(U_x V_x + U_y V_y + U_z V_z) + (U_x W_x + U_y W_y + U_z W_z) = U\cdot V +U\cdot W$$
gives
$\displaystyle (U_x V_x + U_y V_y + U_z V_z) + (U_x W_x + U_y W_y + U_z W_z) = U\cdot V +U\cdot W$

7. I just wanted to confirm this is right:-

$\displaystyle k(U \cdot V) = (kU) \cdot V = U \cdot (kV)$

$\displaystyle (kU_x kV_x + kU_y kV_y + kU_z kV_z) = k(U \cdot V)$ $\displaystyle = ((kU_x) V_x + (kU_y) V_y + (kU_z) V_z) = (kU) \cdot V$ $\displaystyle = (U_x (kV_x) + U_y (kV_y) + U_z (kV_z) = U \cdot (kV)$

8. Originally Posted by RedKMan
I just wanted to confirm this is right:-

$\displaystyle k(U \cdot V) = (kU) \cdot V = U \cdot (kV)$

$\displaystyle (kU_x kV_x + kU_y kV_y + kU_z kV_z) = k(U \cdot V) = ((kU_x) V_x + (kU_y) V_y + (kU_z) V_z) = (kU)$$\displaystyle \cdot V = (U_x (kV_x) + U_y (kV_y) + U_z (kU_z) = U \cdot (kV) This is correct. \displaystyle k(U \cdot V) = ((kU_x) V_x + (kU_y) V_y + (kU_z) V_z) = (kU)$$\displaystyle \cdot V = (U_x (kV_x) + U_y (kV_y) + U_z (kU_z) = U \cdot (kV)$
You have too many k's in $\displaystyle (kU_x kV_x + kU_y kV_y + kU_z kV_z)$

9. That has confused me a bit as I thought the scalar $\displaystyle k$ would apply to both the $\displaystyle U$ and the $\displaystyle V$ vectors. I thought both vectors would be multiplied by the scalar k.

I was working from the idea that 2(a + b) = (2 * a) + (2 * b).

10. Originally Posted by RedKMan
That has confused me a bit as I thought the scalar $\displaystyle k$ would apply to both the $\displaystyle U$ and the $\displaystyle V$ vectors. I thought both vectors would be multiplied by the scalar k.
I was working from the idea that 2(a + b) = (2 * a) + (2 * b).
$\displaystyle \begin{gathered} k\left( {U + V} \right) = kU + kV \hfill \\ k\left( {U \cdot V} \right) = \left( {kU} \right) \cdot V = U \cdot \left( {kV} \right) \hfill \\ \end{gathered}$