INTRODUCTION Flashcards
distance formula
d = sqrt((x2-x1)^2+(y2-y1)^2+(z2-z1)^2)
equation of a sphere
d = r
center = (h, j, k), @ surface = (x, y, z)
(x-h)^2+(y-j)^2+(z-k)^2 = r^2
magnitude of v = <dx, dy, dz>
|v| = sqrt(dx^2+dy^2+dz^2)
unit vector
magnitude = 1
getting there: v/|v|
standard basis vectors in R3
i = <1, 0, 0>
j = <0, 1, 0>
k = <0, 0, 1>
ie. v = <4, 5, 6> = 4i + 5j + 6k
dot product: a*b
a = <xa, ya, za>
b = <xb, yb, zb>
a * b = xaxb + yayb + zazb
dot product properties (5)
- a = <a1, a2, a3>
a * a = a1^2 + a2^2 + a3^2 = |a|^2 - a * b = b * a
- a * (b + c) = a * b + a * c
- s = scalar
(sa) * b = s(a * b) = a * (sb) - 0 * a = 0
cosine theorem
a * b =
|a| * |b| * cos(theta)
a and b are orthogonal if
a * b = 0
a and b are parallel if
same direction:
theta = 0 so cos(theta) = 1
opposite direction:
theta = pi so cos(theta) = -1
radians to degrees
*180/pi
scalar projection component of b along a
length or projection
projection = x = |b| cos(theta)
x = (a * b)/|a|
compa(b) = (a * b)/|a|
vector projection of b onto a
vector along a with length of projection
proja(b) = compa(b) * (a/|a|)
projection with an obtuse angle
b will project in the direction of -a to form its right triangle
compa(b) = will be negative as cos<0
multiply proja(b) by -1
(for cross product)
determinant of a 2 x 2 matrix
M = (a, b)
(c, d)
det(M) = ad - bc
|a b| = ad - bc
|c d|
cross product
a x b = i(a2b3 - a3b2) - j(a1b3 - a3b1) + k(a1b2 - a2b1)
cross product is _____ to both vectors
orthogonal,
a x b = -(b x a)
to find if a vector crosses front or back
follow the curve of your right hand
sin theorem
|a x b| = |a| * |b| * sin(theta)
area of a parallelogram w/ sides a and b
A = |a x b|
Atriangle = “ “ /2
volume of a parallelopiped w/ sides a and b
V = a * (b x c)
a line is determined by…
a point on the line
a direction vector(parallel and scalar multiple of v by t when v lies on the line)
parametric equations for the line
through P0(x0, y0, z0) and any P
parallel to <a, b, c> scalar multiple by t
P0P = <x - x0, y - y0, z - z0>
x - x0 = at, y - y0 = bt, z - z0 = ct
x = x0 + at
y = y0 + bt
z = z0 + ct
a plane is determined by…
a point on the plane
a normal vector(perpendicular and scalar multiple of v by t when v lies on the plane)
equation of a plane
through P0 = (x0, y0, z0) and any P
w/ normal vector n = <a, b, c> scalar multiple of v by t
P0P = <x - x0, y - y0, z - z0>
(orthogonal to n)
a(x - x0) + b(y - y0) + c(z - z0) = 0
angle between planes =
= angle between normal vectors
