# Formula for changing a plane intersection angle

• MHB
• LMHmedchem
In summary, the conversation discusses finding the angle of intersection between two planes in three-dimensional space, specifically ∠ ABC ∩ DBC. The formula for calculating this angle is given, and it involves rotating a vector around an axis. One suggestion is to use the formula for finding the minimum distance between a point and a line, which would result in two possible circles for the locus of points with the same distance from both the fixed tail and the axis of rotation. The conversation also clarifies that the "x" in the formula refers to the cross product of vectors.
LMHmedchem
Hello,

This is a bit more complex than my previous post, but I don't think it qualifies for university level difficulty. Please move this to the appropriate forum if my assessment is not correct.

I have 4 vectors in three space where each vector has its tail at the same point (point B).
Code:
id    x        y        z
B     0.000    0.000    0.000
vBA   0.183   -0.479    0.358
vBD   1.970    2.395   -0.524
vBE  -3.798   -1.214    0.168
vBC  -5.847    4.571   -1.137

I have the angle between each vector and vBC (in radians).
Code:
∠ ABC = 2.466
∠ DBC = 1.570
∠ EBC = 0.989

Each vector pair forming one of the angles above lies on a Euclidean plane. These planes intersect along vBC. I can calculate the plane intersection angles as follows.

For the intersection angle ∠ ABC ∩ DBC (angle of intersection between planes ABC and DBC, sorry I don't know the proper symbolic nomenclature for plane intersection angles).

1. take the vector rejection of vBA against vBC
2. take the vector rejection of vBD against vBC
3. calculate the angle between the two rejection vectors
Code:
∠ ABC ∩ DBC = 2.484
∠ ABC ∩ EBC = 0.674
∠ DBC ∩ EBC = 3.125

I need to change the angle of intersection between two planes without changing any of the angles between vectors ∠ABC, ∠DBC, or ∠EBC. Changing ∠ ABC ∩ DBC would involve essentially rotating the tip of vBA in a circle around the axis of vBC while the tail of vBA stays at point B, but I don't know how to do that.

The methods I have seem for rotating a vector in 3D involve rotating the vector about one axis.

From stackexchange, a 3D rotation around the Z-axis would be,
Code:
|cos θ   −sin θ   0| |x|   |x cos θ − y sin θ|   |x'|
|sin θ    cos θ   0| |y| = |x sin θ + y cos θ| = |y'|
|  0       0      1| |z|   |        z        |   |z'|

To use such a method, I would have to translate my current coordinates so that vBC is aligned with one axis, make the 3D vector rotation, and then translate back. That seems overly involved. It seems as if I should be able to use something like the above using the coordinates of vBC.

Any suggestions would be appreciated,

LMHmedchem

Last edited:
In this post:

http://mathhelpboards.com/calculus-10/12-3-63-determine-smallest-distance-between-point-line-22704-post102528.html#post102528

I showed that if we have a line described by the two points:

$$\displaystyle P_1\left(x_1,y_1,z_1\right),\,P_2\left(x_2,y_2,z_2\right)$$

And the point:

$$\displaystyle P_0\left(x_0,y_0,z_0\right)$$

Then the minimum distance between them is given by:

$$\displaystyle D_{\min}=\frac{\left|\left(\vec{P_2}-\vec{P_1}\right)\times\left(\vec{P_1}-\vec{P_0}\right)\right|}{\left|\vec{P_2}-\vec{P_1}\right|}$$

I would use this formula to determine the distance of the head of the vector to be rotated from the axis of rotation, and then use the formula again to determine the locus of points all having this same distance from both the fixed tail and the axis of rotation. You will likely get two circles, one of which would be discarded.

MarkFL said:
I showed that if we have a line described by the two points:

$$\displaystyle P_1\left(x_1,y_1,z_1\right),\,P_2\left(x_2,y_2,z_2\right)$$

And the point:

$$\displaystyle P_0\left(x_0,y_0,z_0\right)$$

Then the minimum distance between them is given by:

$$\displaystyle D_{\min}=\frac{\left|\left(\vec{P_2}-\vec{P_1}\right)\times\left(\vec{P_1}-\vec{P_0}\right)\right|}{\left|\vec{P_2}-\vec{P_1}\right|}$$

Looking at the equation above,

View attachment 7610

does the numerator refer to the cross product of vectors (p2-p1) and (p1-p0)? Since the resulting distance Dmin is a scalar, it seems as if this should be a simple product, but "x" often means cross product when vectors are concerned so I thought I should ask. Since the shortest distance between the line and a point is a vector perpendicular to the line, the cross product also makes sense.

Sorry about not being very good with latex.

LMHmedchem

#### Attachments

• point_line_distance.png
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LMHmedchem said:
Looking at the equation above,

does the numerator refer to the cross product of vectors (p2-p1) and (p1-p0)? Since the resulting distance Dmin is a scalar, it seems as if this should be a simple product, but "x" often means cross product when vectors are concerned so I thought I should ask. Since the shortest distance between the line and a point is a vector perpendicular to the line, the cross product also makes sense.

Sorry about not being very good with latex.

LMHmedchem

The numerator is the magnitude of the indicated vectorial cross product. :)

## 1. What is the formula for changing a plane intersection angle?

The formula for changing a plane intersection angle is known as the Law of Sines, which states that the ratio of the length of a side of a triangle to the sine of its opposite angle is always constant.

## 2. How do I use the Law of Sines to change a plane intersection angle?

To use the Law of Sines, you will need to know the measure of at least two angles and the length of one side of the triangle. Then, you can use the formula: (sin A) / a = (sin B) / b = (sin C) / c, where A, B, and C are the angles and a, b, and c are the corresponding sides.

## 3. Can the Law of Sines be used for any triangle?

Yes, the Law of Sines can be used for any triangle, as long as you have enough information about the angles and sides. However, it is most commonly used for non-right triangles.

## 4. What is the purpose of changing a plane intersection angle?

The purpose of changing a plane intersection angle is to adjust the direction of a plane's flight path. This can be useful in various scenarios, such as avoiding obstacles, changing course due to weather conditions, or making a smooth landing.

## 5. Are there any limitations to using the Law of Sines for changing a plane intersection angle?

One limitation of using the Law of Sines is that it only applies to triangles. Additionally, if the triangle is very large or very small, the measurements may not be accurate due to the curvature of the Earth. It is also important to note that the Law of Sines does not take into account other factors such as wind or air resistance, which can affect the plane's actual flight path.

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