Rules of Dot product with vectors

[Harpermk]

Let u and v be vectors where u≠v
In the equation
2u.(u-v)=u.(u-v)

(. represents dot product)

I am trying to find a reason why the (u-v) can't just be cancelled out on both sides. My reasoning is that dividing either side by (u-v) would be dividing a scalar by a vector which you cannot do. I am unsure of whether this is true or if there is another reason.

 

[Dr.Peterson]

Let u and v be vectors where u≠v
In the equation
2u.(u-v)=u.(u-v)

(. represents dot product)

I am trying to find a reason why the (u-v) can't just be cancelled out on both sides. My reasoning is that dividing either side by (u-v) would be dividing a scalar by a vector which you cannot do. I am unsure of whether this is true or if there is another reason.

That's certainly the main reason you can't do this: You can't divide when there is no division operation to do!

Similarly, but not quite as obviously, you can only cancel when there is a theorem showing that cancellation is valid. In canceling, you would be assuming that [imath]\overrightarrow{a}\cdot \overrightarrow{c}=\overrightarrow{b}\cdot \overrightarrow{c}\implies \overrightarrow{a}=\overrightarrow{b}[/imath].

You have not learned such a theorem; and in fact you should be able to see, using the definition of the dot product (in terms of angles and lengths), that it is not true.

Of course, there are actual properties that you can use to solve this.

 

[blamocur]

My $0.02 worth: from [imath]\mathbf a \cdot \mathbf b = \mathbf a \cdot \mathbf c[/imath] you can deduce that [imath]\mathbf a \cdot (\mathbf b - \mathbf c) = 0[/imath]. But this does not mean that [imath]\mathbf b - \mathbf c[/imath] is zero, only that [imath]\mathbf a[/imath] and [imath]\mathbf b - \mathbf c[/imath] are orthogonal.