Module 1 Flashcards
Distance of a Sphere from P to Q
|P-Q| = sqrt [ (a2-a1)^2 + (b2-b1)^2 + (c2-c1)^2 ]
SPhere of radius R centered at (a, b, c)
dist^2 = R^2
(x-a)^2 + (y-b)^2 + (z-c)^2 = R^2
Definition of a line in 3-D
vector r = < x(t), y(t), z(t) >
where x(t), y(t), and z(t) are linear functions
If you find the equation of a line with one point and then do the same process with a different point, you will get:
a different parameterization of the same line
r(t) =
= tv = t< a, b, c > = < at, bt, ct >
How to find the vector parametric equations for a line?
Use the two points P and Q to find PQ (Q-P). That vector is then the vector parameteriaztion and one of the points P or Q is the point vector.
perpendicular to the xz plane:
< 0, 1, 0 >
perpendicular to the xy plane:
< 0, 0, 1 >
perpendicular to the yz plane:
< 1, 0, 0 >
How do you find if two lines intersect?
set one variable of t equal to s. set up two systems of equations and set two variables equal to each other. find t and s. plug t and s into the last variable equal equation.
if the final equations do not match the lines do not intersect.
If they do intersect, plug in t or s into the corresponding r1 or r2 equation to find the point.
Dot Product
V . W
= a1a2 + b1b2 + c1c2
V . V
||V||^2 = || < a, b, c > || = sqrt(a^2 + b^2 + c^2)
cos (pheta)
= ( V . W) / (||V|| ||W||)
to find the angle: multipluy this my the inverse cosine
two non-zero vectors are called:
normal/ perpendicular/ orthongonal
acute angle =
V . W > 0
cos (pheta) > 0