Question

Calc III Tutorial: Basic Review of Vector Algebra

Answer 1

\({\bf{Definition:}}\) a mathematical designation of a quantity w/ direction and magnitude

\({\bf{Notation:}}\) in textbooks this will usually appear as a bold-font letter and/or with an arrow above it. this is kind of a pain w/ LaTeX so I'll probably just specify when certain variables represent vectors instead of scalars.
- can also use angle brackets (they're sort of like <> but less sharp)



bold letter w/ hat represents unit vector (get to this later but a unit vector is a vector w/ magnitude of 1)

number of components = number of spatial dimensions of the vector

so <u1, u2> is a 2 dimensional vector and <u1, u2, u3> is a 3 dimensional vector
when written with each of the x, y, z, etc. components, called component form

\({\bf{Magnitude:}}\) length of a vector



more-or-less derivable from the pythagorean theorem
similar process for a more-than-2-d vector, just include (z2-z1)^2 according to the same logic

Answer 2

\({\bf{Vector~Operations:}}\) most of these are pretty straightforward until we get to cross products/dot products. assume u and v are vectors, k, a, and b are scalars

u + v = <u1 + v1, u2 + v2, u3 + v3>
ku = <ku1, ku2, ku3>

visually speaking you can place the two sum vectors tail to tail, and treat them as two sides of a parallelogram, then draw (would recommend using dashed lines as to not confuse them with the original vectors) the rest of the paralellogram. a line from the tails of the original vectors to the opposite angle of the parallelogram will produce the sum vector.
(will just show a pic because this is hard to explain w/ words)



in this particular example vector R is the sum of vectors A and B

Answer 3

cont.
u + v = v + u (commutative)
u + 0 = u (zero additive identity)
0u = 0 (zero multiplicative property)
a(bu) = (ab)u (associative, multiplicative)
(a+b)u = au + bu (distributive)
(u+v) + w = u + (v+w) (associative, additive)
u + (-u) = 0 (additive inverse)
1u = u (multiplicative identity)
a(u+v) = au + av (distributive, except this time using two vectors and a scalar, whereas the other example was two scalars and a vector. similar logic, though)

Answer 4

\({\bf{Unit~Vectors:}}\) in general, any vector with length (magnitude) 1, but we also specify unit vectors originating from the origin as
i^ = <1,0,0>
j^ = <0,1,0>
k^ = <0,0,1>, etc.
all vectors can be written as a linear combination of these vectors

<v1, v2, v3> = v1<1,0,0> + v2<0,1,0> + v3<0,0,1>

in general to find the unit vector in the same direction as the vector, divide vector/magnitude

and all vectors can be written as a product of the magnitude * the unit vector in the same direction as the original vector, or in other words

v = |v| * (1/|v|)

Answer 5

\({\bf{Useful~Formulas:}}\) probably should have put this at the top but

the vector PQ btwn P(x1,y1,z1) and Q(x2,y2,z2) has components

x2 - x1, y2 - y1, z2 - z1

the order is important especially when it comes to calculating the angle/cross product/etc. between two vectors, as changing the order of subtraction will change the direction of the vector

midpoint of a vector: basically works just like the midpoint of a line segment,

(x2 - x1)/2 , (y2 - y1)/2, (z2 - z1)/2

Answer 6

Source material is section 12.2 of Thomas' Calculus, Early Transcendentals, 14th edition by Hass, Heil, Weir, et. al.