0
$\begingroup$

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?

$\endgroup$
4
  • 2
    $\begingroup$ There is always a plane that contains two vectors $\endgroup$ Jun 9, 2020 at 14:01
  • $\begingroup$ Ya got it,just needed to confirm it $\endgroup$
    – tejas yadavalli
    Jun 9, 2020 at 14:05
  • 1
    $\begingroup$ The angle is defined for the co-planar vectors. They also share the same point of application i.e. they are co-initial. $\endgroup$ Jun 9, 2020 at 16:35
  • $\begingroup$ Another topic for you to consider: direction cosines $\endgroup$
    – Bill N
    Jun 9, 2020 at 18:12

1 Answer 1

2
$\begingroup$

"Two vectors from a point" defines a plane.
So, $\vec A \cdot \vec B=AB\cos\theta$ applies, where $\theta$ is an angle in that plane.

The dot product of two vectors can also be defined
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 $\vec A \cdot \vec B=(\hat x\cdot \vec A)(\hat x\cdot \vec B) +(\hat y\cdot \vec A)(\hat y\cdot \vec B) +(\hat z\cdot \vec A)(\hat z\cdot \vec B).$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .