MATH 152 Chapter 11.3

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Cross Product

Finding the direction of the cross product by the right-hand rule

Given $\mathbf {a} =\left\langle a,\,b,\,c\right\rangle$ and $\mathbf {b} =\left\langle x,\,y,\,z\right\rangle$,

${\begin{aligned}\mathbf {a} \times \mathbf {b} &=\left(\left|\mathbf {a} \right|\left|\mathbf {b} \right|\sin {\theta }\right)\mathbf {\hat {n}} \\&={\cancel {\left\langle bz-cy,\,cx-az,\,ay-bx\right\rangle }}\end{aligned}}$

• $\mathbf {\hat {n}}$ is unit vector perpendicular to both a and b (see right hand rule figure at right)
• θ is the angle between a and b
• Important: DO NOT MEMORIZE SECOND EQUATION

Matrix Definition

Review Determinants

${\begin{vmatrix}a&b\\c&d\end{vmatrix}}=ad-bc$

${\begin{aligned}{\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}}&=a{\begin{vmatrix}e&f\\h&i\end{vmatrix}}-b{\begin{vmatrix}d&f\\g&i\end{vmatrix}}+c{\begin{vmatrix}d&e\\g&h\end{vmatrix}}\\&=a\left(ei-fh\right)-b\left(di-fg\right)+c\left(dh-eg\right)\end{aligned}}$

Simpler Method

1. rewrite first two columns
2. add all the ↘ diagonals
3. subtract all the ↙ diagonals

${\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}}{\begin{matrix}a&b\\d&e\\g&h\end{matrix}}=aei+bfg+cdh-bdi-afh-ceg$

Vector Application

Given $\mathbf {a} =\left\langle a,\,b,\,c\right\rangle$ and $\mathbf {b} =\left\langle x,\,y,\,z\right\rangle$,

${\begin{aligned}\mathbf {a} \times \mathbf {b} &=\left(bz-cy\right)\mathbf {i} -\left(az-cx\right)\mathbf {j} +\left(ay-bx\right)\mathbf {k} \\&={\begin{vmatrix}b&c\\y&z\end{vmatrix}}\mathbf {i} -{\begin{vmatrix}a&c\\x&z\end{vmatrix}}\mathbf {j} +{\begin{vmatrix}a&b\\x&y\end{vmatrix}}\mathbf {k} \\&={\begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\a&b&c\\x&y&z\end{vmatrix}}\end{aligned}}$

Properties

Geometric

• a × b produces a vector perpendicular to both a and b (easier to understand visually)
• |a × b| = area of parallelogram formed by a and b
• Complete the parallelepiped (3D parallelogram) formed by 3 vectors a, b, and c:

  |(a × b) • c| = volume of parallelepiped.

Algebraic

• $\mathbf {a} \cdot \left(\mathbf {a} \times \mathbf {b} \right)=0$

  a × b is orthogonal to a, and $\mathbf {a} \cdot \mathbf {a} ^{\perp }=0$

• a × b ≠ b × a

Application: Torque

Vectors in play with torque

Torque $\left|\mathbf {\tau } \right|$ depends on length of wrench $\left|\mathbf {r} \right|$, force applied $\left|\mathbf {F} \right|$, and the angle between the force and the wrench θ (optimally 90° or ${\tfrac {\pi }{2}}$)

${\begin{aligned}\left|\mathbf {\tau } \right|&=\left|\mathbf {r} \right|\left|\mathbf {F} \right|\sin {\theta }\\&=\left|\mathbf {r} \times \mathbf {F} \right|\end{aligned}}$

Wednesday, December 1, 2010

Example 1

Given the points $P\left(-1,\,2,\,0\right)$, $Q\left(2,\,0,\,1\right)$, $R\left(-5,\,3,\,1\right)$, find a vector orthogonal to the plane containing these points.

${\begin{aligned}\mathbf {a} ={\overrightarrow {PQ}}&=\left\langle 3,\,-2,\,1\right\rangle \\\mathbf {b} ={\overrightarrow {PR}}&=\left\langle -4,\,1,\,1\right\rangle \\\mathbf {a} \times \mathbf {b} &={\begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\3&-2&1\\-4&1&1\end{vmatrix}}{\begin{matrix}\mathbf {i} &\mathbf {j} \\3&-2\\-4&1\end{matrix}}\\&=-2\mathbf {i} -4\mathbf {j} +3\mathbf {k} -3\mathbf {j} -\mathbf {i} -8\mathbf {k} \\&=-3\mathbf {i} -7\mathbf {j} -5\mathbf {k} =\left\langle -3,\,-7,\,-5\right\rangle \end{aligned}}$

Checking

$\left(\mathbf {a} \times \mathbf {b} \right)\cdot \mathbf {a} =0$

$\left(\mathbf {a} \times \mathbf {b} \right)\cdot \mathbf {b} =0$

Example 2

Area of $\triangle PQR$ = ${\tfrac {1}{2}}{\mbox{area of parallelogram}}$

$={\frac {1}{2}}\left({\overrightarrow {PQ}}\times {\overrightarrow {QR}}\right)$

Example 3

Volume of Parallelepiped defined by A(1,0,1), B(2,3,0), C(-1,1,4), and D(0,3,2)

${\begin{aligned}V_{\mbox{parallelepiped}}&=\left|\left(\mathbf {a} \times \mathbf {b} \right)\cdot \mathbf {c} \right|\\&={\begin{vmatrix}a_{1}&a_{2}&a_{3}\\b_{1}&b_{2}&b_{3}\\c_{1}&c_{2}&c_{3}\end{vmatrix}}\end{aligned}}$

1. a = <1,3,-1>
   b = <-1,3,1>
   c = <-2,1,3>
2. a × b = 6i - 0j + 6k
3. | (a × b) • c | = 6