\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus - Vector Cross Product Application - Triple Vector Product

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The triple vector product (or vector triple product, as it is sometimes called) is so named because the result is a vector. [For comparison, see the triple scalar product.]

When you have three vectors, \( \vec{u}\), \(\vec{v}\) and \(\vec{w}\), the triple vector product is defined as \( \vec{u} \times \vec{v} \times \vec{w} \).

Practice

Given \(\vec{A}=\hat{i}-2\hat{j}+2\hat{k}\), \(\vec{B} = 3\hat{i}-\hat{j}-\hat{k}\) and \(\vec{C}=-\hat{i}-\hat{k}\), calculate the triple vector product \(\vec{A}\times\vec{B}\times\vec{C}\).

Problem Statement

Given \(\vec{A}=\hat{i}-2\hat{j}+2\hat{k}\), \(\vec{B} = 3\hat{i}-\hat{j}-\hat{k}\) and \(\vec{C}=-\hat{i}-\hat{k}\), calculate the triple vector product \(\vec{A}\times\vec{B}\times\vec{C}\).

Final Answer

\(\vec{A}\times\vec{B}\times\vec{C}=-7\hat{i}-\hat{j}+7\hat{k}\)

Problem Statement

Given \(\vec{A}=\hat{i}-2\hat{j}+2\hat{k}\), \(\vec{B} = 3\hat{i}-\hat{j}-\hat{k}\) and \(\vec{C}=-\hat{i}-\hat{k}\), calculate the triple vector product \(\vec{A}\times\vec{B}\times\vec{C}\).

Solution

Let's work left to right.
\(\vec{A} \times \vec{B} = (\hat{i}-2\hat{j}+2\hat{k}) \times (3\hat{i}-\hat{j}-\hat{k}) = \) \(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 3 & -1 & -1 \end{vmatrix} = \) \((2-(-2))\hat{i} - (-1-6)\hat{j} + (-1-(-6))\hat{k} = \) \(4\hat{i} + 7\hat{j} +5\hat{k} = \vec{v}\)
Now we need to calculate \(\vec{v} \times \vec{C}\).
\( (4\hat{i} + 7\hat{j} +5\hat{k}) \times (-\hat{i}-\hat{k}) = \) \(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 7 & 5 \\ -1 & 0 & -1 \end{vmatrix} = \) \((-7-0)\hat{i} - (-4-(-5))\hat{j} +(0-(-7))\hat{k} = \) \( -7\hat{i} -\hat{j} +7\hat{k}\)

Final Answer

\(\vec{A}\times\vec{B}\times\vec{C}=-7\hat{i}-\hat{j}+7\hat{k}\)

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Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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