Question

Calc III Tutorial: Dot Products & Cross Products

Answer 1

\({\bf{Definition:}}\) u ⋅ v = u1v1 + u2v2 + u3v3, also the projection of vector u onto vector v, times the length of v

Angle between two vectors: cos(theta) = (u ⋅ v) / (|u| |v|), which is derived from performing law of cosines on the desired angle and basically factoring out the dot product (not really that important to know for the scope of this course but can provide if anyone's curious)

Orthogonality: iff u ⋅ v = 0

\({\bf{Properties:}}\)
u ⋅ v = v ⋅ u (commutative, don't confuse this with the cross product which is *not* commutative*)
u ⋅ (v+w) = u ⋅ v + u ⋅ w (distributive)
0 ⋅ u = 0 (zero multiplicative property)
(cu) ⋅ v = u⋅ (cv) = c(u⋅ v) scalars are also distributive
u ⋅ u = |u|^2

Answer 2

\({\bf{Alternative~Definition~of~Dot~Product}}\) u ⋅ v = |u||v|cos(theta)

\({\bf{Projecting~one~vector~onto~another:}}\) the resulting vector if the vector u is taken and "projected" to have the same direction as v while keeping the magnitude of u

\[proj _{v}u = \frac{ |u|(\cos \theta) v }{ |v| }=\frac{ (u⋅v)v }{ |v||v| } = \frac{ (u⋅v)v }{ |v|^2 }\]
w/ ucos(theta) being the scalar component, and v/|v| being the direction of v

ex: projecting a non-horizontal force onto a horizontal axis to calculate the horizontal component of force, and subsequently, work



Work of force F onto distance d = Fcos(theta) (d) = F⋅d

Answer 3

\({\bf{Cross-Products}}\) u × v = |u||v|sin(theta) * n where n is a unit vector perpendicular to both u and v

RHR


You might remember this from physics when you were working with electric/magnetic fields.

\({\bf{Properties}}\)
vectors u and v parallel iff u × v = 0
likewise a vector crossed w/ itself will also = 0

for scalars r and s, (ru) × (sv) = (rs)(u×v), can factor scalars out
v × u = -(u×v) *** important to remember, cross product, unlike dot product, is ***not*** commutative
0 × u = 0 (zero multiplicative property)
u × (v + w) = u × v + u × w, so a vector can be "distributed" across the sum of two vectors
(v + w) × u = v × u + w × u

triple cross-product:
u × (v × w) = (u ⋅ w)v - (u ⋅ v)w

*** | u × v | is the area of a parallelogram made by placing vectors u and v tail to tail ***

Answer 4

\({\bf{Determinant~Method~for~3D~vectors}}\) (if you are only given vectors in 2 dimensions you can simply set the third components as zero)

for u = u1 i + u2 j + u3 k
and v = v1 i + v2 j + v3 k
the cross product is
\[determinant~of \left[\begin{matrix}i & j & k \\ u1 & u2 & u3 \\ v1 & v2 & v3\end{matrix}\right] \]
which you can calculate with row expansion/column expansion, or the cross method, or whatever

Answer 5

\({\bf{Volume~of~Triangle~Given~Points}}\) draw a quick sketch of the points, define 2 vectors that define two sides of the triangle, make sure their tails are touching, and calculate the appropriate vectors, being mindful about direction. after that it's just 1/2 the cross product of the 2 vectors.



\({\bf{Volume~of~Paralellepiped}}\) pick a vertex of the shape, write vectors w, v, and u w/ their tails at one vertex

volume becomes |(u × v) ⋅ w|
honestly this took me a while to wrap my head around but as long as you u, v, and w share the same tail location just let u and v be the vectors defining one side of the parallelepiped and w be the vector remaining and you should be fine

Answer 6

Source material is section 12.3-12.4 of Thomas' Calculus, Early Transcendentals, 14th edition by Hass, Heil, Weir, et. al.

also I really don't care much for LaTeX sorrynotsorry