The vectors a, b and c are all unit vectors (length=1). The angle between all the pairs is $\displaystyle \pi/3$.
What is the angle between $\displaystyle u = a+b+c$ and $\displaystyle v=3c-a-b$
Hello, Spec!
I have a start on the problem . . . but that's all.
The angle $\displaystyle \theta$ between vectors $\displaystyle \vec u\text{ and }\vec v$ is given by: .$\displaystyle \cos\theta \:=\:\frac{|\vec u\cdot\vec v|}{|\vec u||\vec v|} $The vectors $\displaystyle \vec a, \vec b, \vec c$ are all unit vectors (length=1).
The angle between all the pairs is $\displaystyle \tfrac{\pi}{3}$
Find the angle between $\displaystyle \vec u \:=\:\vec a+\vec b+\vec c$ and $\displaystyle \vec v\:=\:3\vec c-\vec a-\vec b$
Let $\displaystyle \alpha$ be the angle between $\displaystyle \vec b\text{ and }\vec c\!:\;\;\cos\alpha \:=\:\frac{\vec b\cdot\vec c}{|\vec b||\vec c|}$
Since .$\displaystyle \alpha = \tfrac{\pi}{3},\;|\vec b| = |\vec c| = 1$, we have: .$\displaystyle {\color{blue}\vec b\cdot\vec c \:=\:\tfrac{1}{2}}$
. . Similarly, we have: .$\displaystyle {\color{blue}\vec a\cdot\vec b \:=\:\tfrac{1}{2}}\:\text{ and }\;{\color{blue}\vec a\cdot\vec c \:=\:\tfrac{1}{2}}$
And I'm certain those will be used in subsequent calculations.
I solved it, but it wasn't pretty.
$\displaystyle (a+b+c)\cdot (a+b+c) = a \cdot (a+b+c) + b \cdot (a+b+c) + c \cdot (a+b+c) = $
$\displaystyle =a \cdot a + a \cdot b + a\cdot c +b \cdot a + b \cdot b + b\cdot c + c \cdot a + c \cdot b + c\cdot c=6 \implies |a+b+c| = \sqrt{6}$
I'm not going to post all calculations, but that's basically how I solved it (and using the second dot product formula above).