## 17Calculus - Vector Cross Product Application - Triple Vector Product

Using Vectors

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Double Integrals - 2Int

Triple Integrals - 3Int

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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 $$\vec{A}\times\vec{B}\times\vec{C}$$.

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

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 $$\vec{v} \times \vec{C}$$.
$$(4\hat{i} + 7\hat{j} +5\hat{k}) \times (-\hat{i}-\hat{k}) =$$ $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 7 & 5 \\ -1 & 0 & -1 \end{vmatrix} =$$ $$(-7-0)\hat{i} - (-4-(-5))\hat{j} +(0-(-7))\hat{k} =$$ $$-7\hat{i} -\hat{j} +7\hat{k}$$

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

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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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