3D-Geometry: Calculate a distance by given angles

Question

I got the angle $\psi$ and the $\Delta x$ like you can see in the picture below. Now I calculated $\Delta r$ as $\Delta x \cdot \cos(\psi)$.

Switching to 3D I now want to introduce an elevation of the red line versus the x,y-plane by the angle $\phi$. The blue line shall still intersect the red line perpendicularily at $\Delta r_{3D}$ and the x-axis at $\Delta x$. I was told that I can calculate $\Delta r = \Delta x \cdot \cos(\psi)\cos(\phi)$.

I can understand that I can still calculate $\Delta r = \Delta x \cdot \cos(\Delta x)$ and I can see that I can calculate $\Delta r_{3D}=\Delta r \cdot \cos(\phi)$ if there is a right angle between $\Delta r_{3D}$ and the line connecting it to $\Delta r$. I don't understand how I can prove that this right angle exists if I only know that the blue line intersects the elevated red line perpendicularily. Can you explain this to me please?

Thanks!

Answer 1

"I can understand that I can still calculate $\Delta r = \Delta x \cdot cos(\Delta x)$."

I would say:

"I can understand that I can still calculate $\Delta r_{horizontal} = \Delta x \cdot cos(\psi)$."

Now, because of the elevation, we have a vertical component:

$\Delta r_{vertical} = \Delta x \cdot cos(\phi)$.

And finally, to obtain $\Delta r_{3D}$ use Pythagoras.

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Having done this for the red line, we can do the same for the blue line, call it $\Delta y$. We have $\Delta y_{3D}$

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The point is : you want to maintain perpendicularity of "red and blue projections" or "red and blue true values" ?

You say to maintain true values perpendicularity.

Apply Pythagoras backwards. Because your right triangle is now in 3D.

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Hope it is clear to you.

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Answer 2

Attention should be to how the right triangles are drawn, how the hypotenuse is placed.

There are three right angles in the oblique view. The gray and yellow planes make a right angle along fold $dm$. Also $O$ marked angles are right angles. Since it is 3D I am labelling biggest differential segment as $ds$.

First component is

$$ dm=ds\, \cos \psi$$

Again a second component projection gives required segment

$$ dm\, \cos \phi =ds\, \cos \psi \cos \phi$$