From the applet above, can you derive how to find the angle between a line and a plane. Explain the key elements of this formula.
As we are finding the angle between the normal, and the direction vector, we have actually found [math]90-\theta[/math].So, we have [math]u\times v=\mid u\mid\mid v\mid cos\left(90-\theta\right)[/math] which is the same as [math]u\times v=\mid u\mid\mid v\mid sin\theta[/math].
Derive the equation for the angle between two planes.
As both normals are perpendicular to the plane, the angle between these is the same as the angle between the two planes. Therefore we have:[br][math]u\times v=\mid u\mid\mid v\mid cos\theta[/math]
You can vary the sliders for [math]d[/math] and [math]f[/math] (the constants in the equation of the plane), what does this do in order to explain why we consider the normals of planes.
This just varies to position of the plane, which is arbitrary, the key aspect is always the normal.
The applet has tools there to help you with questions a-c, for the rest, you will need to add other elements. The following commands will help, as you start to type, press the correct option:[br]- Vector (point)[br]- Line (Point, Direction)[br]- Cross (vector, vector)[br][br]You can also use the tools at the top to find angles between vectors, or lines
In part d) you discussed the direction of the line of intersection between two planes, can you explain why the line of intersection moves in the direction of the cross product of the normals to each of the planes?
The cross product finds the vector perpendicular to two vectors. The line of intersection is in the direction of a vector parallel to both lines, so one way to find this is to find a vector perpendicular to two vectors perpendicular that are perpendicular.