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Angle between line and its shadow on a plane |
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| Jun8-12, 10:41 AM | #1 |
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Angle between line and its shadow on a plane
1. The problem statement, all variables and given/known data
A line intersects a vertical wall at 40.78° (that is, 49.22° with respect to the normal vector of the wall). The line is contained within a vertical plane perpendicular to the wall such that the aforementioned angle is measured with respect to the plumb line dropped from the point of intersection. The line casts a shadow on the wall that is 68.04° from the plumb line. What is the angle between the line and its shadow? 2. Relevant equations I haven't the foggiest. There's got to be a simple expression out there somewhere for this, but I've never encountered it before. Thank you very much for your help! added After talking with a math professor from the College of Wooster this afternoon, I found that a better way to describe this would be in spherical coordinates. So, I guess my question now would be: how does one go about finding the angle between to rays given in polar coordinates? Taking the normal vector to the wall as (radius, azimuth, zenith)=(1,0,0), the two rays (unit length for simplicity) would be (1, 0°, 49.22°) and (1, 68.04°, 90°). |
| Jun8-12, 02:30 PM | #2 |
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Personally, I would be inclined to use "vectors" to do this. Setting up a coordinate system so that the z-axis is up the wall is the yz-plane, and the line is in the xz-plane, we have a vector, of length 1, pointing along the line, given by sin(40.78)i+ cos(40.78)k. The shadow, making an ange of 68.04 with the z- axis, would be along the unit vector sin(68.04)j+ cos(68.04)k. Now, you can find the angle between those vectors by using the dot product and the fact that [itex]u\cdot v= |u||v|cos(\theta)[/itex] where [itex]\theta[/itex] is the angle you want to find.
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| shadow, shadow problems, tetrahedron |
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