- #1

- 908

- 0

The formula of angle between two lines in space is:

[tex]\vec{a}=(a_1,a_2,a_3)[/tex] ; [tex]\vec{b}=(b_1,b_2,b_3)[/tex]

[tex]cos\alpha=\frac{|\vec{a} \vec{b}|}{|\vec{a}||\vec{b}|}[/tex]

or out from there:

[tex]cos\alpha=\frac{|a_1b_1+a_2b_2+a_3b_3|}{\sqrt{a_1^2+a_2^2+a_3^2}\sqrt{b_1^2+b_2^2+b_3^2}}[/tex]

Why it is [itex]|\vec{a}\vec{b}|[/itex]? Why not [itex]\vec{a} \vec{b}[/itex]?

Scalar product of two vectors is [tex]\vec{a}\vec{b}=|\vec{a}||\vec{b}|cos(\vec{a},\vec{b})[/tex]

[tex]\vec{a}=(a_1,a_2,a_3)[/tex] ; [tex]\vec{b}=(b_1,b_2,b_3)[/tex]

[tex]cos\alpha=\frac{|\vec{a} \vec{b}|}{|\vec{a}||\vec{b}|}[/tex]

or out from there:

[tex]cos\alpha=\frac{|a_1b_1+a_2b_2+a_3b_3|}{\sqrt{a_1^2+a_2^2+a_3^2}\sqrt{b_1^2+b_2^2+b_3^2}}[/tex]

Why it is [itex]|\vec{a}\vec{b}|[/itex]? Why not [itex]\vec{a} \vec{b}[/itex]?

Scalar product of two vectors is [tex]\vec{a}\vec{b}=|\vec{a}||\vec{b}|cos(\vec{a},\vec{b})[/tex]

Last edited: