# vector problem!

• Jan 15th 2010, 10:32 PM
iExcavate
vector problem!
i apologise if this is the incorrect forum, i assume it would go under here or perhaps calculus forum - vector help!

find the formula for $||b -a||^2$ in terms of $||b||^2$, $||a||^2$ and $b·a$

• Jan 16th 2010, 12:13 AM
tonio
Quote:

Originally Posted by iExcavate
i apologise if this is the incorrect forum, i assume it would go under here or perhaps calculus forum - vector help!

find the formula for $||b -a||^2$ in terms of $||b||^2$, $||a||^2$ and $b·a$

This question belongs, me believes ,in trigonometry-analytic geometry. Anyway:

$b-a$ is an arrow beginning at $a$ and ending at $b$ and $\|b-a\|$ is its length, so using the cosines law (theorem) on the resulting triangle we get:

$\|b-a\|^2=\|a\|^2+\|b\|^2-2\|a\|\|b\|\cos \theta$

I'm not sure what is $ba$, but if you meant the dot product of these two vectors then just remember that $\cos \theta=\frac{b\cdot a}{\|b\|\|a\|}$ , so you can substitute above.

Tonio
• Jan 16th 2010, 12:42 AM
iExcavate
how do i deduce $\|b-a\|^2 = \|b\|^2 + \|a\|^2 - 2\|a\|\|b\|cos\theta$
to prove: $\cos \theta=\frac{b\cdot a}{\|b\|\|a\|}$
• Jan 16th 2010, 01:49 AM
HallsofIvy
tonio told you: apply the cosine law to the triangle with sides a, b, and a- b.