commit bdfd5a870e8f304ee180497b07de1b747994b8eb
parent c67760bfbc09fb4623a462b345a1927d7c2e3c68
Author: Bradley Gannon <bradley@bradleygannon.com>
Date: Thu, 19 Mar 2026 21:12:48 -0400
Solve 1.{2..5}
Diffstat:
| M | main.typ | | | 193 | +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ |
1 file changed, 193 insertions(+), 0 deletions(-)
diff --git a/main.typ b/main.typ
@@ -1 +1,194 @@
+#let tag(equation, tag) = math.equation(numbering: _ => tag, block: true, equation)
+#let vec(v) = math.bold(math.upright(v))
+#let unit(u) = math.hat(math.bold(math.upright(u)))
+#let cross = math.times
+#let definedAs = math.eq.triple
+#let implies = math.arrow.r.double
+#set heading(numbering: (..nums) => {
+ let n = nums.pos()
+ if n.len() <= 2 {
+ numbering("1.1", ..n)
+ }
+})
+
+= Vector Analysis
+
+== Vector Algebra
+
+=== Problem 1.1
+
+=== Problem 1.2
+
+The cross product is not associative. We will show this by providing a counterexample.
+
+Let $vec(A)$, $vec(B)$, and $vec(C)$ be vectors, with $vec(C)$ perpendicular to both $vec(A)$ and
+$vec(B)$, and $vec(A)$ and $vec(B)$ neither perpendicular nor parallel.
+
+$vec(A) cross vec(B)$ is equal to $A B sin theta_(vec(A) angle vec(B)) unit(n)_(vec(A) cross
+vec(B))$ by definition. Since $vec(C)$ is perpendicular to both $vec(A)$ and $vec(B)$, their cross
+product is parallel to $vec(C)$. The cross product of any two parallel vectors is $vec(0)$ because
+the angle between them is $0$, and $sin (0) = 0$. So, $(vec(A) cross vec(B)) cross vec(C) = vec(0)$.
+
+$vec(B) cross vec(C)$ is equal to $B C sin theta_(vec(B) angle vec(C)) unit(n)_(vec(B) cross
+vec(C))$ by definition. The result is perpendicular to $vec(C)$, which means that it must be in the
+plane of $vec(A)$ and $vec(B)$. But, the result is not parallel to $vec(A)$ because $vec(B)$ is not
+perpendicular to $vec(A)$. So, $vec(A) cross (vec(B) cross vec(C)) eq.not vec(0)$.
+
+The left and right sides of the definition of associativity are not equal for this set of vectors,
+which means they are a counterexample. Therefore the cross product is not associative. $qed$
+
+=== Problem 1.3
+
+We will use the same cube arrangement as Example 1.2. Let $vec(A)$ be the body diagonal extending
+from the origin to $(1, 1, 1)$, and let $vec(B)$ be the body diagonal extending from $(1, 0, 0)$ to
+$(0, 1, 1)$. In component form, these are
+
+$
+vec(A) = (1 - 0) unit(x) + (1 - 0) unit(y) + (1 - 0) unit(z) = unit(x) + unit(y) + unit(z) \
+vec(B) = (0 - 1) unit(x) + (1 - 0) unit(y) + (1 - 0) unit(z) = -unit(x) + unit(y) + unit(z) "."
+$
+
+The magnitudes of these vectors are both $sqrt(3)$.
+
+$
+A = sqrt(vec(A) dot vec(A)) = sqrt(1 dot 1 + 1 dot 1 + 1 dot 1) = sqrt(3) \
+B = sqrt(vec(B) dot vec(B)) = sqrt((-1) dot (-1) + 1 dot 1 + 1 dot 1) = sqrt(3)
+$
+
+We can find the angle between these vectors by first computing the dot product in component form and
+then isolating the angle $theta$ in abstract form.
+
+$
+vec(A) dot vec(B) = 1 dot (-1) + 1 dot 1 + 1 dot 1 = 1 \
+vec(A) dot vec(B) definedAs A B cos theta_(vec(A) vec(B)) implies theta_(vec(A) vec(B)) = cos^(-1) ((vec(A) dot vec(B)) / (A B)) \
+theta_(vec(A) vec(B)) = cos^(-1) (1 / (sqrt(3) sqrt(3))) = cos^(-1) (1 / 3) approx #rect($1.231 "radians"$)
+$
+
+Note that the dot product implicitly considers vectors that all extend from the same origin, so by
+applying it to $vec(B)$ we have "moved" it so that its tail is at the same origin as $vec(A)$. This
+is acceptable because it does not change the magnitude or direction of the vector.
+
+=== Problem 1.4
+
+The cross product of any two non-parallel vectors points normal to the plane formed by the two
+inputs. For convenience, we select two of the edges of the shaded triangle to be our input vectors
+such that their cross product will have the same direction as $unit(n)$.
+
+Let $vec(A)$ be the edge that lies in the $x y$ plane, and let $vec(B)$ be the edge extending from
+$(1, 0, 0)$ to $(0, 0, 3)$. In component form, these are
+
+$
+vec(A) = (0 - 1) unit(x) + (2 - 0) unit(y) + (0 - 0) unit(z) = -unit(x) + 2 unit(y) \
+vec(B) = (0 - 1) unit(x) + (0 - 0) unit(y) + (3 - 0) unit(z) = -unit(x) + 3 unit(z) "."
+$
+
+Their cross product is
+
+$
+vec(A) cross vec(B) &= (A_y B_z - A_z B_y) unit(x) + (A_z B_x - A_x B_z) unit(y) + (A_x B_y - A_y B_x) unit(z) \
+&= (2 dot 3 - 0 dot 0) unit(x) + (0 dot (-1) - (-1) dot 3) unit(y) + ((-1) dot 0 - 2 dot (-1)) unit(z) \
+&= 6 unit(x) + 3 unit(y) + 2 unit(z) "."
+$
+
+The result of $vec(A) cross vec(B)$ has the same direction as $unit(n)$, but it is not a unit
+vector. We must divide by its magnitude to compute $unit(n)$.
+
+$
+unit(n) = (vec(A) cross vec(B)) / (|vec(A) cross vec(B)|)
+= (6 unit(x) + 3 unit(y) + 2 unit(z)) / (sqrt(6^2 + 3^2 + 2^2))
+= 1/7 (6 unit(x) + 3 unit(y) + 2 unit(z))
+$
+
+The components of $unit(n)$ are $#rect($unit(n)_x = 6/7, unit(n)_y = 3/7, "and" unit(n)_z = 2/7
+"."$)$
+
+=== Problem 1.5
+
+The *BAC-CAB* rule is
+
+#tag($ vec(A) cross (vec(B) cross vec(C)) = vec(B) (vec(A) dot vec(C)) - vec(C) (vec(A) dot vec(B)) $, "(Eq. 1.17)")
+
+which we can prove by expanding both sides in component form. We will evaluate each side in
+isolation and then compare them.
+
+First, we evaluate the left side.
+
+$
+vec(A) cross (vec(B) cross vec(C)) =&
+vec(A) cross ((B_y C_z - B_z C_y) unit(x)
++ (B_z C_x - B_x C_z) unit(y)
++ (B_x C_y - B_y C_x) unit(z)) \
+=& (A_y (B_x C_y - B_y C_x) - A_z (B_z C_x - B_x C_z)) unit(x) \
+&+ (A_z (B_y C_z - B_z C_y) - A_x (B_x C_y - B_y C_x)) unit(y) \
+&+ (A_x (B_z C_x - B_x C_z) - A_y (B_y C_z - B_z C_y)) unit(z) \
+=& (A_y (B_x C_y - B_y C_x) - A_z (B_z C_x - B_x C_z)) unit(x) \
+&+ (A_z (B_y C_z - B_z C_y) - A_x (B_x C_y - B_y C_x)) unit(y) \
+&+ (A_x (B_z C_x - B_x C_z) - A_y (B_y C_z - B_z C_y)) unit(z) \
+=& (A_y B_x C_y + A_z B_x C_z - A_y B_y C_x - A_z B_z C_x) unit(x) \
+&+ (A_x B_y C_x + A_z B_y C_z - A_x B_x C_y - A_z B_z C_y) unit(y) \
+&+ (A_x B_z C_x + A_y B_z C_y - A_x B_x C_z - A_y B_y C_z) unit(z)
+$
+
+And now the right side.
+
+$
+vec(B) (vec(A) dot vec(C)) - vec(C) (vec(A) dot vec(B)) =&
+vec(B) (A_x C_x + A_y C_y + A_z C_z) - vec(C) (A_x B_x + A_y B_y + A_z B_z) \
+=& B_x (A_x C_x + A_y C_y + A_z C_z) unit(x) \
+&+ B_y (A_x C_x + A_y C_y + A_z C_z) unit(y) \
+&+ B_z (A_x C_x + A_y C_y + A_z C_z) unit(z) \
+&- C_x (A_x B_x + A_y B_y + A_z B_z) unit(x) \
+&- C_y (A_x B_x + A_y B_y + A_z B_z) unit(y) \
+&- C_z (A_x B_x + A_y B_y + A_z B_z) unit(z) \
+=& (A_x B_x C_x + A_y B_x C_y + A_z B_x C_z - A_x B_x C_x - A_y B_y C_x - A_z B_z C_x) unit(x) \
+&+ (A_x B_y C_x + A_y B_y C_y + A_z B_y C_z - A_x B_x C_y - A_y B_y C_y - A_z B_z C_y) unit(y) \
+&+ (A_x B_z C_x + A_y B_z C_y + A_z B_z C_z - A_x B_x C_z - A_y B_y C_z - A_z B_z C_z) unit(z) \
+=& (A_y B_x C_y + A_z B_x C_z - A_y B_y C_x - A_z B_z C_x) unit(x) \
+&+ (A_x B_y C_x + A_z B_y C_z - A_x B_x C_y - A_z B_z C_y) unit(y) \
+&+ (A_x B_z C_x + A_y B_z C_y - A_x B_x C_z - A_y B_y C_z) unit(z)
+$
+
+Both sides are equal, which proves the *BAC-CAB* rule. $qed$
+
+=== Problem 1.6
+
+=== Problem 1.7
+
+=== Problem 1.8
+
+=== Problem 1.9
+
+=== Problem 1.10
+
+== Differential Calculus
+
+== Integral Calculus
+
+== Curvilinear Coordinates
+
+== The Dirac Delta Function
+
+== The Theory of Vector Fields
+
+= Electrostatics
+
+= Potentials
+
+= Electric Fields in Matter
+
+= Magnetostatics
+
+= Magnetic Fields in Matter
+
+= Electrodynamics
+
+= Conservation Laws
+
+= Electromagnetic Waves
+
+= Potentials and Fields
+
+= Radiation
+
+= Electrodynamics and Relativity
