How to find three angles of a point in 3D.

Question

Let's imagine a point in 3D coordinate such that its distance to the origin is $1 \text{ unit}$.

The coordinates of that point have been given as $x = a$, $y = b$, and $z = c$.

How can we calculate the angles made by the vector with each of the axes?

Answer 1

The vector point coordinates are $OP=(a,b,c)$ then the angles with $x,y,z$ with unitary vectors $e_1=(1,0,0),e_2=(0,1,0),e_3(0,0,1)$ are given by the dot product

• $\cos \alpha = \frac{OP\cdot e_1}{|OP||e_1|}=OP\cdot e_1=a$
• $\cos \beta = \frac{OP\cdot e_2}{|OP||e_2|}=OP\cdot e_2=b$
• $\cos \gamma = \frac{OP\cdot e_3}{|OP||e_3|}=OP\cdot e_3=c$

Answer 2

Suppose angle of vector related to x axis is $\alpha$, related to y axis is $\beta$ and related to z axis is $\gamma$ then we have:

Due to presumption; $\sqrt {a^2+b^2+c^2}=1$

$1\times \ cos \alpha=a$

$1\times \ cos \beta=b$

$1 \times \ cos \gamma=c$

Answer 3

I used a way finding the angle between two vectors.

angle = aCos ( (V1.V2) / (|V1|.|V2|)  )

When I want to calculate the angle of a point to x axis.

x Axis is

V1 = 1.i + 0.j + 0.k  

The point is

V2 = a.i + b.j + c.k

This way I calculated correct solution.

Answer 4

Suppose you have a vector $\vec v = xi+yj+zk$ where $i,j,k $ are the basis unit vectors then the angles $\alpha,\beta, \gamma$ of the vector to the $x,y,z $ axes respectively is given by ;

$\alpha = \frac{x}{\sqrt{x^2+y^2+z^2}} = \cos(a)\\\beta = \frac{y}{\sqrt{x^2+y^2+z^2}}=\cos(b)\\\gamma = \frac{z}{\sqrt{x^2+y^2+z^2}}=\cos(c)$

It follows that squaring the 3 equations and adding them results in

$\cos^2(a)+\cos^2(b)+\cos^2(c) =\alpha^2+\beta^2+\gamma^2 = 1 $