§AERE 351 Homework 1

Sep 25, 2025

§1.

Given:

\begin{align*} A &= A_x i + A_y j + A_z k \\ B &= B_x i + B_y j + B_z k \\ C &= C_x i + C_y j + C_z k \end{align*}

§a.

Given:

A \cdot A = |A|^2

Left hand side:

\begin{align*} A \cdot A &= A_x \cdot A_x + A_y \cdot A_y + A_z \cdot A_z \\ &= \boxed{A_x^2 + A_y^2 + A_z^2} \end{align*}

Right hand side:

\begin{align*} |A|^2 &= \left[ \sqrt{A_x^2 + A_y^2 + A_z^2} \right]^2 \\ &= \boxed{A_x^2 + A_y^2 + A_z^2} \end{align*}

§b.

Given:

A \cdot (B \times C) = (A \times B) \cdot C

Left hand side:

\begin{align*} A \cdot (B \times C) &= A \cdot \begin{vmatrix} i & j & k \\ B_x & B_y & B_z \\ C_x & C_y & C_z \end{vmatrix} \\ &= A \cdot \left[ (B_y C_z - B_z C_y) i + (B_z C_x - B_x C_z) j + (B_x C_y - B_y C_x) k \right] \\ &= \boxed{A_x B_y C_z - A_x B_z C_y + A_y B_z C_x - A_y B_x C_z + A_z B_x C_y - A_z B_y C_x} \end{align*}

Right hand side:

\begin{align*} (A \times B) \cdot C &= \begin{vmatrix} i & j & k \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix} \cdot C \\ &= \left[ (A_y B_z - A_z B_y) i + (A_z B_x - A_x B_z) j + (A_x B_y - A_y B_x) k \right] \cdot C \\ &= A_y B_z C_x - A_z B_y C_x + A_z B_x C_y - A_x B_z C_y + A_x B_y C_z - A_y B_x C_z \\ &= \boxed{A_x B_y C_z - A_x B_z C_y + A_y B_z C_x - A_y B_x C_z + A_z B_x C_y - A_z B_y C_x} \end{align*}

§c.

Given:

A \times (B \times C) = B (A \cdot C) - C (A \cdot B)

Left hand side:

\begin{align*} A \times (B \times C) &= A \times \begin{vmatrix} i & j & k \\ B_x & B_y & B_z \\ C_x & C_y & C_z \end{vmatrix} \\ &= A \times \left[ (B_y C_z - B_z C_y) i + (B_z C_x - B_x C_z) j + (B_x C_y - B_y C_x) k \right] \\ &= \begin{vmatrix} i & j & k \\ A_x & A_y & A_z \\ B_y C_z - B_z C_y & B_z C_x - B_x C_z & B_x C_y - B_y C_x \end{vmatrix} \\ &= \boxed{ \begin{bmatrix} A_y B_x C_y - A_y B_y C_x - A_z B_z C_x + A_z B_x C_z \\ A_z B_y C_z - A_z B_z C_y - A_x B_x C_y + A_x B_y C_x \\ A_x B_z C_x - A_x B_x C_z - A_y B_y C_z + A_y B_z C_y \end{bmatrix} } \end{align*}

Right hand side:

\begin{align*} B (A \cdot C) - C (A \cdot B) &= B (A_x C_x + A_y C_y + A_z C_z) - C (A_x B_x + A_y B_y + A_z B_z) \\ &= \begin{bmatrix} B_x A_x C_x + B_x A_y C_y + B_x A_z C_z \\ B_y A_x C_x + B_y A_y C_y + B_y A_z C_z \\ B_z A_x C_x + B_z A_y C_y + B_z A_z C_z \end{bmatrix} - \\ &\quad~ \begin{bmatrix} C_x A_x B_x + C_x A_y B_y + C_x A_z B_z \\ C_y A_x B_x + C_y A_y B_y + C_y A_z B_z \\ C_z A_x B_x + C_z A_y B_y + C_z A_z B_z \end{bmatrix} \\ &= \begin{bmatrix} A_x B_x C_x + A_y B_x C_y + A_z B_x C_z \\ A_x B_y C_x + A_y B_y C_y + A_z B_y C_z \\ A_x B_z C_x + A_y B_z C_y + A_z B_z C_z \end{bmatrix} - \\ &\quad~ \begin{bmatrix} A_x B_x C_x + A_y B_y C_x + A_z B_z C_x \\ A_x B_x C_y + A_y B_y C_y + A_z B_z C_y \\ A_x B_x C_z + A_y B_y C_z + A_z B_z C_z \end{bmatrix} \\ &= \boxed{ \begin{bmatrix} A_y B_x C_y - A_y B_y C_x - A_z B_z C_x + A_z B_x C_z \\ A_z B_y C_z - A_z B_z C_y - A_x B_x C_y + A_x B_y C_x \\ A_x B_z C_x - A_x B_x C_z - A_y B_y C_z + A_y B_z C_y \end{bmatrix} } \end{align*}

§2.

Given:

(A \times B) \cdot (C \times D) = \boxed{(A \cdot C) (B \cdot D) - (A \cdot D) (B \cdot C)}

Using identity (b):

(A \times B) \cdot (C \times D) = ((A \times B) \times C ) \cdot D

Flipping order of cross project:

((A \times B) \times C ) \cdot D = -(C \times (A \times B)) \cdot D

Using identity (c):

C \times (A \times B) = A (C \cdot B) - B (C \cdot A)

Plugging it back in:

-(C \times (A \times B)) \cdot D = -(A (C \cdot B) - B (C \cdot A)) \cdot D

Distribute sign:

-(A (C \cdot B) - B (C \cdot A)) \cdot D = (B (C \cdot A) - A (C \cdot B)) \cdot D

Distribute dot product:

(B \cdot D) (C \cdot A) - (A \cdot D) (C \cdot B)

Rearrange:

(B \cdot D) (C \cdot A) - (A \cdot D) (C \cdot B) = \boxed{(A \cdot C) (B \cdot D) - (A \cdot D) (B \cdot C)}

§3.

Given:

\begin{align*} A &= 8 i + 9 j + 12 k \\ B &= 9 i + 6 j + 1 k \\ C &= 3 i + 5 j + 10 k \end{align*}

The normal of the plane that

A

and

B

lie on is similar to their cross product:

\begin{align*} C_n &= A \times B \\ &= \begin{vmatrix} i & j & k \\ 8 & 9 & 12 \\ 9 & 6 & 1 \end{vmatrix} \\ &= (9 - 72) i + (108 - 8) j + (48 - 81) k \\ &= -63 i + 100 j -33 k \end{align*}

The unit vector normal of the plane:

\begin{align*} \hat u_n &= \frac{C_n}{|C_n|} \\ &= \frac{-63 i + 100 j -33 k}{\sqrt{3969 + 10000 + 1089}} \\ &= -0.513401 i + 0.814923 j -0.268924 k \end{align*}

The projection of a vector onto a plane using the normal vector:

\begin{align*} C_{AB} &= C - (C \cdot \hat u_n) \hat u_n \\ &= C - (3 \cdot -0.513401 + 5 \cdot 0.814923 + 10 \cdot -0.268924) \hat u_n \\ &= C + 0.154828 \hat u_n \\ &= C + (-0.079489 i + 0.126173 j - 0.041637 k) \\ &= \boxed{2.920511 i + 5.126173 j + 9.958363 k} \end{align*}

§4.

The position (in meters) as a function of time (in seconds):

\begin{align*} x &= \sin 3t \\ y &= \cos t \\ z &= \sin 2t \end{align*}

Time of interest:

t = 3 s

§a.

\begin{align*} \dot x &= 3 \cos 3t \\ \dot y &= -\sin t \\ \dot z &= 2 \cos 2t \end{align*}

\begin{align*} v &= \dot x i + \dot y j + \dot z k \\ v(3) &= 3 \cos 9 ~ i - \sin 3 ~ j + 2 \cos 6 ~ k \\ &= \boxed{(-2.733391 i - 0.141120 j + 1.920341 k) \frac{m}{s}} \end{align*}

§b.

\begin{align*} |v| &= \sqrt{\dot x ^ 2 + \dot y ^ 2 + \dot z ^ 2} \\ |v(3)| &= \sqrt{7.471426 + 0.019915 + 3.687710} \\ &= \boxed{3.343509 \frac{m}{s}} \end{align*}

§c.

\begin{align*} \hat u_t &= \frac{v}{|v|} \\ \hat u_t(3) &= \frac{-2.733391 i - 0.141120 j + 1.920341 k}{3.343509} \\ &= \boxed{-0.817522 i - 0.042207 j + 0.574349 k} \end{align*}

§d.

\begin{align*} \theta_x &= \arccos u_{t,x} = \arccos - 0.817522 = \boxed{144.8 \deg} \\ \theta_y &= \arccos u_{t,y} = \arccos - 0.042207 = \boxed{92.42 \deg} \\ \theta_z &= \arccos u_{t,z} = \arccos \phantom{-} 0.574349 = \boxed{54.95 \deg} \end{align*}

§e.

\begin{align*} \ddot x &= -9 \sin 3t \\ \ddot y &= -\cos t \\ \ddot z &= -4 \sin 2t \end{align*}

\begin{align*} a &= \ddot x i + \ddot y j + \ddot z k \\ a(3) &= -9 \sin 9 ~ i - \cos 3 ~ j + 4 \sin 6 ~ k \\ &= \boxed{(-3.709066 i - 0.989992 j + 1.117662 k) \frac{m}{s^2}} \end{align*}

§f.

The binormal vector lies perpendicular to the plane in which the normal and tangential vectors lie. I don't have the normal vector yet but the the normalized acceleration and the tangential vector from before should also lie in the plane.

|a| = \sqrt{\ddot x ^ 2 + \ddot y ^ 2 + \ddot z ^ 2} = \boxed{3.9983 \frac{m}{s^2}}

\begin{align*} \frac{a}{|a|} &= \frac{-3.709066 i - 0.989992 j + 1.117662 k}{3.9983} \\ &= -0.927661 i - 0.247603 j + 0.279534 k \end{align*}

\begin{align*} \hat u_t \times \frac{a}{|a|} &= \begin{bmatrix} -0.817522 \\ -0.042207 \\ 0.574349 \end{bmatrix} \times \begin{bmatrix} -0.927661 \\ -0.247603 \\ 0.279534 \end{bmatrix} \\ &= 0.130412 i - 0.304276 j + 0.163267 k \end{align*}

\begin{align*} \hat u_b &= \frac{\hat u_t \times \frac{a}{|a|}}{\left| \hat u_t \times \frac{a}{|a|} \right|} \\ &= \boxed{0.353309 i - 0.824335 j + 0.442318 k} \end{align*}

§g.

\begin{align*} \hat u_n &= \hat u_b \times \hat u_t \\ &= \begin{bmatrix} 0.353309 \\ -0.824335 \\ 0.442318 \end{bmatrix} \times \begin{bmatrix} -0.817522 \\ -0.042207 \\ 0.574349 \end{bmatrix} \\ &= \boxed{-0.454787 i -0.564527 j -0.688824 k} \end{align*}

§h.

Thankfully, I can reuse the normalized acceleration vector from before:

\frac{a}{|a|} = -0.927661 i - 0.247603 j + 0.279534 k

The components:

\begin{align*} \phi_x &= \arccos - 0.927661 = 2.758896 \\ \phi_y &= \arccos - 0.247603 = 1.821002 \\ \phi_z &= \arccos \phantom{-} 0.279534 = 1.287488 \end{align*}

§i.

\begin{align*} a_t &= a \cdot \hat u_t \\ &= \begin{bmatrix} -3.709066 \\ -0.989992 \\ 1.117662 \end{bmatrix} \frac{m}{s^2} \cdot \begin{bmatrix} -0.817522 \\ -0.042207 \\ 0.574349 \end{bmatrix} \\ &= \boxed{3.71596 \frac{m}{s^2}} \end{align*}

§j.

\begin{align*} a_n &= a \cdot \hat u_n \\ &= \begin{bmatrix} -3.709066 \\ -0.989992 \\ 1.117662 \end{bmatrix} \frac{m}{s^2} \cdot \begin{bmatrix} -0.513401 \\ 0.814923 \\ -0.268924 \end{bmatrix} \\ &= \boxed{0.796905 \frac{m}{s^2}} \end{align*}

§k.

\begin{align*} \rho &= \frac{|v|^2}{|a_n|} \\ &= \frac{\left( 3.343509 m/s \right)^2}{0.796905 m/s^2} \\ &= \boxed{14.0281m} \end{align*}

§l.

The center should be a radius away from the position in the direction of the normal vector:

P_0 = r + \rho \hat u_n

Surprisingly, the question has not made me compute

r

yet:

\begin{align*} r &= \begin{bmatrix} \sin 3t \\ \cos t \\ \sin 2t \end{bmatrix} \\ r(3) &= \begin{bmatrix} \sin 9 \\ \cos 3 \\ \sin 6 \end{bmatrix} \\ &= 0.412118 i - 0.989992 j - 0.279415 k \end{align*}

The second term:

\rho \hat u_n = -7.20204 i + 11.4318 j - 3.77249 k

Summing:

P_0 = \boxed{(-6.78992i + 10.4418j -4.05191k)m}

§5.

Given:

\begin{align*} \omega &= 2k ~ rad/s \\ \dot \omega &= -5k ~ rad/s^2 \\ \ddot \omega &= 3k ~ rad/s^3 \\ F &= 15 i + 10 j ~ N \end{align*}

First derivative:

\begin{align*} \dot F &= \omega \times F \\ &= \left< 0, 0, 2 \right> \times \left< 15, 10, 0 \right> \\ &= \left< -20, 30, 0 \right> N/s \\ \end{align*}

Second derivative:

\begin{align*} \ddot F &= \frac{d \dot F}{dt} \\ &= \frac{d}{dt} \left( \omega \times F \right) \\ &= \dot \omega \times F + \omega \times \dot F \\ &= \left< 0, 0, -5 \right> \times \left< 15, 10, 0 \right> + \left< 0, 0, 2 \right> \times \left< -20, 30, 0 \right> \\ &= \left< -10, -115, 0 \right> N/s^2 \\ \end{align*}

Third derivative:

\begin{align*} \overset{...} F &= \frac{d \ddot F}{dt} \\ &= \frac{d}{dt} \left( \dot \omega \times F + \omega \times \dot F \right) \\ &= \ddot \omega \times F + 2 \dot \omega \times \dot F + \omega \times \ddot F \\ &= \left < 0, 0, 3 \right> \times \left< 15, 10, 0 \right> + 2 \left< 0, 0, -5 \right> \times \left< -20, 30, 0 \right> + \left < 0, 0, 2 \right> \times \left< -10, -115, 0 \right> \\ &= \left< 500, 225, 0 \right> N/s^3 \\ &= \boxed{500 i + 225 j ~ N/s^3} \end{align*}

§6.

Given:

\begin{align*} v &= [0, 1, 0]^T \\\\ v' &= R_1(v) \\ e_1 &= [0, 0, 1]^T \\ \theta_1 &= 30deg \\\\ v'' &= R_2(v') \\ e_2 &= [0, 1, 0]^T \\ \theta_2 &= 60deg \\\\ \end{align*}

R_1

followed by

R_2

\begin{align*} v' &= R_z v \\ &= \begin{bmatrix} \cos \theta_1 & -\sin \theta_1 & 0 \\ \sin \theta_1 & \cos \theta_1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \\ &= \begin{bmatrix} -\sin \theta_1 \\ \cos \theta_1 \\ 0 \end{bmatrix} \\\\ v'' &= R_y v' \\ &= \begin{bmatrix} \cos \theta_2 & 0 & \sin \theta_2 \\ 0 & 1 & 0 \\ -\sin \theta_2 & 0 & \cos \theta_2 \end{bmatrix} \begin{bmatrix} -\sin \theta_1 \\ \cos \theta_1 \\ 0 \end{bmatrix} \\ &= \begin{bmatrix} -\sin \theta_1 \cos \theta_2 \\ \cos \theta_1 \\ \sin \theta_1 \sin \theta_2 \end{bmatrix} \\ &= \boxed{\left< -\frac{1}{4}, \frac{\sqrt 3}{2}, \frac{\sqrt 3}{4} \right>} \end{align*}

R_2

followed by

R_1

\begin{align*} v' &= R_y v \\ &= \begin{bmatrix} \cos \theta_2 & 0 & \sin \theta_2 \\ 0 & 1 & 0 \\ -\sin \theta_2 & 0 & \cos \theta_2 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \\ &= \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \\\\ v'' &= R_z v' \\ &= \begin{bmatrix} \cos \theta_1 & -\sin \theta_1 & 0 \\ \sin \theta_1 & \cos \theta_1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \\ &= \boxed{ \begin{bmatrix} -\sin \theta_1 \\ \cos \theta_1 \\ 0 \end{bmatrix} \neq \begin{bmatrix} -\sin \theta_1 \cos \theta_2 \\ \cos \theta_1 \\ \sin \theta_1 \sin \theta_2 \end{bmatrix} } \end{align*}

R_2 R_1 v

is not the same as

R_1 R_2 v