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plane through p with normal vector n consists of
all points r such that r→ - p→ orthogonal to n

vector form of plane
n→ • (r→ - p→) = 0
scalar form of plane
ax + by + cz = d
line can be described by a point p on line and a normal vector n orthogonal to line; line consists of all r such that
n→ • (r→ - p→) = 0 i.e. ax + by = c
how to find parallel planes
find one point on each plane by moving 1 unit in direction ± n
normalize n, points are p ± normalized n
planes are n→ • ((x y z) - p1 or p2) = 0
cross product
vector orthogonal to u and v (that are not parallel); u x v = (a2b3 - a3b2 | -(a1b3 - a3b1) | a1b2 - a2b1)
-a x b =
b x a
right hand rule
curl fingers from a to b, thumb is direction of a x b
a x b =
det ( i j k in first column | components of u in second | components of v in third)
k (a x b) =
ka x b = a x kb
a x (b + c) =
a x b + a x c
(b + c) x a =
b x a + c x a
a x a =
0
length of cross product
||a x b|| = ||a||||b||sinθ
direction n of cross product
n is a unit vector perpendicular to both a and b by the right hand rule
a • (a x b) =
0
b • (a x b) =
0
area of a parallelogram
||a x b||
volume of parallelepiped
area of the base || a x b|| * ||c||cosθ where θ is the angle between c and a x b and c is the vector forming slanted height thing

determinant of 3 × 3 matrix =
|det [ a b c ]| = volume of parallelepiped
span of vectors
all possible linear combos of those vectors
span (a, b) =
{ ta + sb | t, s are real}
span of two vectors is the xy-plane if
they are NOT collinear
span of two vectors is a line if
they ARE collinear
span of two vectors is a point if
both are zero vector
plane containing a point and a line
find line’s direction vector and vector connecting given point to any point on line, then cross product is the normal vector of the plane
equation through point perpendicular to a line
direction vector of line becomes normal vector of plane, then just plug in point
vector parallel to plane intersection
cross product of two planes’ normal vectors
better volume of a parallelepiped
|a•(b x c)|