Calculate the distance to a 3D point in space

Question

I'm trying to find the distance from the origin to a point in 3d space which is offset by a couple angles and a line. I'm fairly certain it can be found, I'm just really struggling to wrap my head around it. The shape is pretty ordinary, and the point I'm trying to find is the furthermost point away. I've done it up in CAD and you can see the point below as the most right hand point. .

Here are the relevant angles and measurements from a couple of perspectives.

Now I can measure that distance in CAD to double check my calculations, but I'm finding that I keep getting the wrong answer. Here's the distance I'm trying to find, as measured in CAD.

My approach to the problem was the find the right angle triangle in the top down view, use the long side (1500+400) and the vertical side (250) to solve for the hypotenuse, but that gives 1916 rather than the 1897 as measured by CAD.

Any help is greatly appreciated!

Answer 1

You have three line segments, and thus four points: starting point $\vec{p}_0$, $\vec{p}_1$, $\vec{p}_2$, and ending point $\vec{p}_3$.

The question is how far is $\vec{p}_3$ from $\vec{p}_0$, i.e.

$$L = \left \lVert \vec{p}_3 - \vec{p}_0 \right \rVert$$

If we have $\vec{p}_0 = ( x_0, y_0, z_0 )$ and $\vec{p}_3 = ( x_3, y_3, z_3 )$, then

$$L = \sqrt{ \left ( x_3 - x_0 \right )^2 + \left ( y_3 - y_0 \right )^2 + \left ( z_3 - z_0 \right )^2 }$$

Let's assume that $x$ axis increases right in the front and top views, $y$ axis increases down in the top view, and $z$ axis increases up in the front view.

The first line segment is $1500$ units long, and is tilted $15°$ from the $x$ axis towards the $z$ axis:

$$\vec{p}_1 = \vec{p}_0 + ( 1500 \cos 15°, 0, 1500 \sin 15° )$$

The second line segment is $250$ units along the $y$ axis:

$$\vec{p}_2 = \vec{p}_1 + ( 0, 250, 0 )$$

The third line segment is $400$ units long, and is tilted $-5°$ from the $x$ axis towards the $z$ axis (tilted the other way compared to the first line segment):

$$\vec{p}_3 = \vec{p}_2 + ( 400 \cos -5°, 0, 400 \sin -5° )$$

Combining the above, we have

$$\vec{p}_3 - \vec{p}_0 = \vec{p}_0 + ( 1500 \cos 15°, 0, 1500 \sin 15° ) + ( 0, 250, 0) + ( 400 \cos -5°, 0, 400 \sin -5° ) - \vec{p}_0$$

i.e.

$$\vec{p}_3 - \vec{p}_0 = ( 1500 \cos 15° + 400 \cos -5°, 250, 1500 \sin 15° + 400 \sin -5° )$$

and therefore

$$L = \sqrt{ (1500 \cos 15° + 400 \cos -5°)^2 + 250^2 + (1500 \sin 15° + 400 \sin -5°)^2 }$$

This is a straightforward calculation, but remember that $\sin -5° \lt 0$ while $\cos 15° \gt 0$, $\sin 15° \gt 0$, and $\cos -5° \gt 0$, and that $\cos 15° = \cos\left(\frac{15 \pi}{180}\right)$, $\sin -5° = \sin\left(\frac{-5 \pi}{180}\right)$. Indeed, we get

$$L = 1897.4011551 \approx 1897$$