Angle between line and the axis

[karthikS]

Consider a 3D straight line passing through z-axis. Following two angles are given (available):

(1) Angle of the line with the z-axis in X-Z plane
(2) Angle of the line with the z-axis in Y-Z plane

I want to know the true angle of the line with the z-axis, i.e.,the angle of the line with z-axis in the plane passing through the line and z-axis.View attachment 7971

 

[I like Serena]

Consider a 3D straight line passing through z-axis. Following two angles are given (available):

(1) Angle of the line with the z-axis in X-Z plane
(2) Angle of the line with the z-axis in Y-Z plane

I want to know the true angle of the line with the z-axis, i.e.,the angle of the line with z-axis in the plane passing through the line and z-axis.

Hi karthikS, welcome to MHB!

Can it be that you mean the angle of the line with the X-Z plane?
Respectively the angle with the Y-Z plane?
Because we can take an angle of a line either with respect to another line, or with respect to a plane.

If so, then let $\angle X\text-Z$ and $\angle Y\text-Z$ be those angles.
And let $\angle Z$ be the angle with the z-axis.

Then:
$$\angle Z = \arccos \sqrt{1-\sin(\angle Y\text-Z)^2-\sin(\angle X\text-Z)^2} \tag 1$$

This follows from the vector project formula, and how angles of lines and planes are defined.
That is, let $\mathbf{\hat a}$ be a direction vector along the line with length 1.
And let $\mathbf{\hat x}$, $\mathbf{\hat y}$, and $\mathbf{\hat z}$ be the standard coordinate vectors (of length 1).
Then:
\begin{cases}
\sin\angle Y\text-Z = \mathbf{\hat a}\cdot \mathbf{\hat x} = a_x \\
\sin\angle X\text-Z = \mathbf{\hat a}\cdot \mathbf{\hat y} = a_y \\
\cos\angle Z = \mathbf{\hat a}\cdot \mathbf{\hat z} = a_z \\
\sqrt{a_x^2+a_y^2+a_z^2} = 1 \tag 2
\end{cases}
From (2) we can find (1).

Since I do not know how familiar you are with these kind of manipulations, please let us know if any step is unclear.

 

[karthikS]

Hi karthikS, welcome to MHB!

Can it be that you mean the angle of the line with the X-Z plane?
Respectively the angle with the Y-Z plane?
Because we can take an angle of a line either with respect to another line, or with respect to a plane.

If so, then let $\angle X\text-Z$ and $\angle Y\text-Z$ be those angles.
And let $\angle Z$ be the angle with the z-axis.

Then:
$$\angle Z = \arccos \sqrt{1-\arcsin(\angle Y\text-Z)^2-\arcsin(\angle X\text-Z)^2} \tag 1$$

This follows from the vector project formula, and how angles of lines and planes are defined.
That is, let $\mathbf{\hat a}$ be a direction vector along the line with length 1.
And let $\mathbf{\hat x}$, $\mathbf{\hat y}$, and $\mathbf{\hat z}$ be the standard coordinate vectors (of length 1).
Then:
\begin{cases}
\sin\angle Y\text-Z = \mathbf{\hat a}\cdot \mathbf{\hat x} = a_x \\
\sin\angle X\text-Z = \mathbf{\hat a}\cdot \mathbf{\hat y} = a_y \\
\cos\angle Z = \mathbf{\hat a}\cdot \mathbf{\hat z} = a_z \\
\sqrt{a_x^2+a_y^2+a_z^2} = 1 \tag 2
\end{cases}
From (2) we can find (1).

Since I do not know how familiar you are with these kind of manipulations, please let us know if any step is unclear.

Have you meant 'arcsin' inside the square root, or it is simply 'sin'?

 

[I like Serena]

Have you meant 'arcsin' inside the square root, or it is simply 'sin'?

Yes. Sorry. I meant sine.
Fixed in my previous post.

 

[karthikS]

Yes. Sorry. I meant sine.
Fixed in my previous post.

Can you confirm that you have meant the angle marked with a '?' in the diagram as angle Z?View attachment 7981

'h' is the space diagonal in the cube shown above. When the space diagonal is projected onto the XZ plane, projected line is at an angle Ø1 with the Z-axis. Similarly, when the space diagonal is projected onto the YZ plane, projected line is at angle Ø2 with the Z-axis.

Now, look at the right angled triangle formed by the sides 'a', 'd' and 'h'. What is the the true angle of the space diagonal with the Z-axis in the plane of the right angled triangle?

When I use the above formula, I am not getting correct results. In a cube, when the space diagonal is projected on to the two planes, I have the 45 degrees on both Ø1 and Ø2. With the formula, I am getting 90 degrees, which is not true.
View attachment 7982View attachment 7983