Find Norm of Vector Subtraction w/o Using Formula

[cscott]

How can I find the norm of (\vec{v} - \vec{w}) without using ||\vec{v} - \vec{w}||^2 = ||\vec{v}||^2 + ||\vec{w}||^2 - 2 ||\vec{v}|| \cdot ||\vec{w}|| \cos \theta?

 

[marlon]

How can I find the norm of (\vec{v} - \vec{w}) without using ||\vec{v} - \vec{w}||^2 = ||\vec{v}||^2 + ||\vec{w}||^2 - 2 ||\vec{v}|| \cdot ||\vec{w}|| \cos \theta?

Do you know the components of these vectors ?
Do you have a base in which you can write them down ?

Do not say NO, because you must have this
marlon

 

[cscott]

I know that the terminal point of the two vectors are v = (1, 6, 2) and w = (3, 1, 7)

 

[marlon]

I know that the terminal point of the two vectors are v = (1, 6, 2) and w = (3, 1, 7)

Well then, in components the subtraction is just (a,b,c) - (a',b',c') = (a-a', b-b', c-c') and the magnitude of a vector with components a, b and c is \sqrt {a^2 + b^2 + c^2}

So you have everything to calculate the norm of a vector with given components.

enjoy

marlon

 

[cscott]

Thanks a lot!