# Does only one angle exists between two vectors?

We know that the dot product of two vectors in $$\mathbf{R}^3$$ is defined as

$$\vec{A} \cdot \vec{B} = a_x b_x +a_y b_y+ a_z b_z$$

Now, if we choose an x-axis such that the vector A lies on it; then, dot product is $$|A||B| \cos \theta_{AB}$$.

It is so only if the two vectors are co-planar. If they aren't then does the formula still hold? i.e. if a vector A is along x-axis and the other vector B is vector, say [1,1,1], then the dot product must be something like

$$|A| |B| \cos(\alpha) \cos(\beta)$$

where, $$\alpha$$ is the angle made by the projection of B in the x-y plane with the vector A along the x-axis, and $$\beta$$ is the angle between the B vector and its projection on the plane.

Is it a correct way to formulate? or did i make a mistake somewhere?

• There is always a plane that contains two vectors – user3518839 Jun 9 '20 at 14:01
• Ya got it,just needed to confirm it – tejas yadavalli Jun 9 '20 at 14:05
• The angle is defined for the co-planar vectors. They also share the same point of application i.e. they are co-initial. – Ishika_96_sparkle Jun 9 '20 at 16:35
• Another topic for you to consider: direction cosines – Bill N Jun 9 '20 at 18:12

So, $$\vec A \cdot \vec B=AB\cos\theta$$ applies, where $$\theta$$ is an angle in that plane.
using $$\vec A \cdot \vec B=A_xB_x +A_yB_y +A_z B_z$$.
One can express the vector components in terms of angles with the coordinate axes $$\vec A \cdot \vec B=(\hat x\cdot \vec A)(\hat x\cdot \vec B) +(\hat y\cdot \vec A)(\hat y\cdot \vec B) +(\hat z\cdot \vec A)(\hat z\cdot \vec B).$$