# [SOLVED] Angle between vectors

• Apr 8th 2009, 07:03 AM
Spec
[SOLVED] Angle between vectors
The vectors a, b and c are all unit vectors (length=1). The angle between all the pairs is $\pi/3$.

What is the angle between $u = a+b+c$ and $v=3c-a-b$
• Apr 8th 2009, 08:36 AM
running-gag
Quote:

Originally Posted by Spec
The vectors a, b and c are all unit vectors (length=1). The angle between all the pairs is $\pi/3$.

What is the angle between $u = a+b+c$ and $v=3c-a-b$

Hi

Compute the dot product of u and v
Compute the length of u and the length of v

$\cos(u,v) = \frac{u.v}{||u||.||v||}$
• Apr 8th 2009, 12:02 PM
Soroban
Hello, Spec!

I have a start on the problem . . . but that's all.

Quote:

The vectors $\vec a, \vec b, \vec c$ are all unit vectors (length=1).
The angle between all the pairs is $\tfrac{\pi}{3}$

Find the angle between $\vec u \:=\:\vec a+\vec b+\vec c$ and $\vec v\:=\:3\vec c-\vec a-\vec b$

The angle $\theta$ between vectors $\vec u\text{ and }\vec v$ is given by: . $\cos\theta \:=\:\frac{|\vec u\cdot\vec v|}{|\vec u||\vec v|}$

Let $\alpha$ be the angle between $\vec b\text{ and }\vec c\!:\;\;\cos\alpha \:=\:\frac{\vec b\cdot\vec c}{|\vec b||\vec c|}$

Since . $\alpha = \tfrac{\pi}{3},\;|\vec b| = |\vec c| = 1$, we have: . ${\color{blue}\vec b\cdot\vec c \:=\:\tfrac{1}{2}}$

. . Similarly, we have: . ${\color{blue}\vec a\cdot\vec b \:=\:\tfrac{1}{2}}\:\text{ and }\;{\color{blue}\vec a\cdot\vec c \:=\:\tfrac{1}{2}}$

And I'm certain those will be used in subsequent calculations.

• Apr 8th 2009, 01:56 PM
Spec
I solved it, but it wasn't pretty.

$(a+b+c)\cdot (a+b+c) = a \cdot (a+b+c) + b \cdot (a+b+c) + c \cdot (a+b+c) =$
$=a \cdot a + a \cdot b + a\cdot c +b \cdot a + b \cdot b + b\cdot c + c \cdot a + c \cdot b + c\cdot c=6 \implies |a+b+c| = \sqrt{6}$

I'm not going to post all calculations, but that's basically how I solved it (and using the second dot product formula above).
• Apr 9th 2009, 03:06 AM
running-gag
That's it ! (Clapping)