Review of some of the linear algebra concepts that we need in this course.
Here are some questions you might think about:
Given two points in 3-space, write the parametric equations of the line passing through them. Find the points at 1/3 intervals across the segment connecting the points.
Find the angle (or at least the cosine of the angle) between two vectors.
Given a triangle, state a mathematical relationship which determines whether a given point is in the interior of the triangle.
In what sense can points be added in an affine space?How does an affine space differ from a vector space?
Find the intersection of the line passing through the points (1,1,1) and (2,3,4) with
(a) the x-y plane
(b) the y-z plane
(c) the x-z plane
As one can determine by looking at the two points and the way the x,y, and z are set for each, the line segment between the two does not have any intersection, but we have said the line passing through the points. So we need to find where (somewhere outside the line segment) the line that goes through these two points crossecs the planes.
If v = <1,2,3> is the vector from (1,1,1) to (2,3,4),
then the equation of the line is:
P = (1,1,1) + t <1,2,3>.
this vector equation into its x, y, and z components, we have:
x = 1 + t, y = 1 + 2t, z = 1
To find, for example, the intersection with the x-z plane, we solve
y = 0 for t and substitute for
x and z. y = 0 gives t = -1/2, so x =
1/2 and z = -1/2. The point of intersection is thus (1/2, 0,
-1/2). This point is obviously outside the segment but falls on
A related but more general is to find the intersection, if any, of a
line with a rectangular solid (like a cubical).
For example consider the line through (-2,4,6) and (-1,3.6,4.4) and
the rectangular solid with opposite points (0,0,0) and (6,4,-2) (where
the faces of the solid are parallel to the primary coordinate planes).
One of the faces of the solid is in the plane z = 0 and satisfies the
boundary for x: 0
<= x <= 6, boundary
for y: 0 <= y <= 4.
To see if the line intersects the face we first find the parametric
equations of the line as before.
x = -2 + t
y = 4 - 0.4t
z = 6 -1.6 t
We solve z = 0 for the parameter and
substitute into the equations for x and y to find the intersection in
the plane z = 0. Then we check the x and y coordinates of the point so
see that they're in the region:
0 <= x <= 6, 0 <= y <=4.
z = 0; 6 -1.6 t = 0 , that means t = (6/1.6). This
x = -2 + (6/1.6) = 1.75
y = 4 - 0.4(6/1.6) = 2.5
is 0 <= 1.75 <= 6
is 0 <= 2.5 <= 4 ? YES
So there is an intersection between the line that goes through the
above two points and the x-y plane. Check for others.
A typical line (one which isn't parallel to a face of the solid and which doesn't pass through a vertex) will either miss the solid altogether, or intersect two faces exactly. We are normally interested in finding the closer of the two points, the one with the smaller (but positive) value of the parameter. If we're shooting a projectile at a building, we will hit a front face before we hit a back face.
We talked about normalizing vectors. Given an arbitrary (nonzero) vector v, the vector v/|v| is a unit vector in the same direction.
We talked about finding the projection of one vector onto another. The projection of v onto u is a vector in the direction of u, with length |v|cos(t), where t is the angle between u and v. As an illustration, we considered landing a plane with a 20 kt. wind which is 50 degrees to the runway heading (but as a headwind). The headwind component is 20cos(50 deg) and the crosswind component is 20sin(50 deg). (Most small planes can't compensate for a crosswind which is more than about 17 or 18 kts, so a direct crosswind is a bad thing.)
We looked at cross products. The cross product u x v is a vector which is perpendicular to both u and v (with direction determined by the right-hand rule: if you curl your hand from u to v, the cross product points in the direction of your thumb). The magnitude of this vector is |u| |v| |sin(t)|, where t is the angle between the vectors. Another way to determine the cross product is to symbolically take the determinant
i j k
u1 u2 u3
v1 v2 v3
which results in i(u2v3 - u3v2) - j(u1v3 - u3v1) + k(u1v2 - u2v1), or
< u2v3 - u3v2, u3v1 - u1v3, u1v2 - u2v1>.
We'll look at lots of applications of cross products throughout the term. Next class we'll use them to find equations of planes. We often need to find the normal vector to a surface, which we do by taking the cross product of two tangent vectors to the surface. The normal vector is required for shading calculations when rendering surfaces.