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

17Calculus - Vector Cross Product Application - Triple Scalar Product

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The triple scalar product is a result of combining the dot product with the cross product. Some other names for the triple scalar product are scalar triple product, mixed product and box product.
First, let's define what it is and then discuss a couple of properties.

Definition and Notation

If we have three vectors in space, \(\vec{u} = u_x\hat{i}+u_y\hat{j}+u_z\hat{k}\), \(\vec{v} = v_x\hat{i}+v_y\hat{j}+v_z\hat{k}\) and \(\vec{w} = w_x\hat{i}+w_y\hat{j}+w_z\hat{k}\), then the triple scalar product is defined to be \( \vec{u} \cdot (\vec{v} \times \vec{w}) \)
The calculation of this can be done as follows
\( \vec{u} \cdot (\vec{v} \times \vec{w}) = \begin{vmatrix} u_x & u_y & u_z \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{vmatrix} \)
Let's look at where this comes from.

Theorem: Triple Scalar Product

If we have three vectors in space,
\(\vec{u} = u_x\hat{i}+u_y\hat{j}+u_z\hat{k}\), \(\vec{v} = v_x\hat{i}+v_y\hat{j}+v_z\hat{k}\) and \(\vec{w} = w_x\hat{i}+w_y\hat{j}+w_z\hat{k}\),
then the triple scalar product is \( \vec{u} \cdot (\vec{v} \times \vec{w}) = \begin{vmatrix} u_x & u_y & u_z \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{vmatrix}\)

Proof - - To prove this, we will calculate \( \vec{u} \cdot (\vec{v} \times \vec{w}) \) and then calculate the determinant to show that we get the same result. Let's start with the cross product.

\( \vec{v} \times \vec{w} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{vmatrix} = \) \( (v_y w_z - v_z w_y)\hat{i} - \) \( (v_x w_z - w_x v_z)\hat{j} + \) \( (v_x w_y - w_x v_y)\hat{k} \)

Taking the dot product of \(\vec{u}\) and the last equation gives us

\( (u_x\hat{i}+u_y\hat{j}+u_z\hat{k}) \cdot [ (v_y w_z - v_z w_y)\hat{i} - \) \( (v_x w_z - w_x v_z)\hat{j} + \) \( (v_x w_y - w_x v_y)\hat{k} ] = \) \( u_x(v_y w_z - v_z w_y) - \) \( u_y(v_x w_z - w_x v_z) + \) \( u_z(v_x w_y - w_x v_y) \)

We could certainly multiply the components of vector \(\vec{u}\) through each factor, but for reasons you will see later, we will leave the equation as it is. To sum up, we have calculated

\( \vec{u} \cdot (\vec{v} \times \vec{w}) = \) \( u_x(v_y w_z - v_z w_y) - \) \( u_y(v_x w_z - w_x v_z) + \) \( u_z(v_x w_y - w_x v_y) ~~~~~ (1) \)

Okay, now let's calculate the determinant
\( \begin{vmatrix} u_x & u_y & u_z \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{vmatrix} \)

We will go across the top row (although going down the first column will give us the same result, going across the top row makes the algebra come out the way want it to).

\( \begin{vmatrix} u_x & u_y & u_z \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{vmatrix} = u_x(v_y w_z - v_z w_y) - u_y(v_x w_z - w_x v_z) + u_z(v_x w_y - w_x v_y) \)

Notice this last equation is the same as equation (1) above. This is easier to see since we did not multiply out all factors in Equation (1). So, we have shown

\( \vec{u} \cdot (\vec{v} \times \vec{w}) = \begin{vmatrix} u_x & u_y & u_z \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{vmatrix}\)       [qed]

The triple scalar product is so named because the result is a scalar.

Properties

The triple scalar product can also be evaluated in one of the following forms.
\( \vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{b} \cdot (\vec{c} \times \vec{a}) = \vec{c} \cdot (\vec{a} \times \vec{b}) \)
The parentheses may be omitted since evaluating the dot product first yields a scalar and it doesn't make sense to take the cross product of a scalar with a vector.
This property also holds \( [\vec{a} \cdot (\vec{b} \times \vec{c})]\vec{a} = (\vec{a} \times \vec{b}) \times (\vec{a} \times \vec{c}) \)
When the triple scalar product is zero, the 3 vectors are coplanar.

Applications

Geometrically, the triple scalar product \( \vec{a} \cdot (\vec{b} \times \vec{c} ) \) is the (signed) volume of the parallelepiped defined by the three vectors given (see figure on the right). The word 'signed' means that the result can be positive or negative depending on the orientation of the vectors.
You can probably now see that when the 3 vectors are coplanar, the parallelepiped is flat and has no volume, so the triple scalar product is zero.

Here are several videos that explains this in more detail. The first two are especially good and the third contains a proof.

Dr Chris Tisdell - Scalar triple product [9mins-30secs]

video by Dr Chris Tisdell

Dr Chris Tisdell - Scalar triple product and volume [11mins-49secs]

video by Dr Chris Tisdell

Larson Calculus - Proof - Geometric Property of the Triple Scalar Product [1mins-40secs]

video by Larson Calculus

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 scalar product \(\vec{A}\cdot\vec{B}\times\vec{C}\).

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 scalar product \(\vec{A}\cdot\vec{B}\times\vec{C}\).

Final Answer

\(\vec{A}\cdot\vec{B}\times\vec{C}=-9\)

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 scalar product \(\vec{A}\cdot\vec{B}\times\vec{C}\).

Solution

First, we need to find the cross product.

\(\vec{B}\times\vec{C}\) = \( (3\hat{i}-\hat{j}-\hat{k}) \times (-\hat{i}-\hat{k}) = \) \(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\\\ 3 & -1 & -1 \\\\ -1 & 0 & -1 \end{vmatrix} \) \(= (1-0)\hat{i} - (-3-1)\hat{j} + (0-1)\hat{k} \) \(= \hat{i} +4\hat{j} -\hat{k}\)

Now we do the dot product.
\( (\hat{i}-2\hat{j}+2\hat{k}) \cdot (\hat{i} +4\hat{j} -\hat{k}) = \) \( (1)(1) + (-2)(4) + (2)(-1) = \) \( 1-8-2 = -9\)

1266 video

Final Answer

\(\vec{A}\cdot\vec{B}\times\vec{C}=-9\)

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Show that the vectors \(\vec{A} = 2\hat{i}-3\hat{j}+4\hat{k}\), \(\vec{B} = 6\hat{i}+2\hat{j}+\hat{k}\) and \(\vec{C} = 6\hat{i}+10\hat{j}-7\hat{k}\) are coplanar.

Problem Statement

Show that the vectors \(\vec{A} = 2\hat{i}-3\hat{j}+4\hat{k}\), \(\vec{B} = 6\hat{i}+2\hat{j}+\hat{k}\) and \(\vec{C} = 6\hat{i}+10\hat{j}-7\hat{k}\) are coplanar.

Final Answer

Since the triple scalar product is zero, the vectors are coplanar.

Problem Statement

Show that the vectors \(\vec{A} = 2\hat{i}-3\hat{j}+4\hat{k}\), \(\vec{B} = 6\hat{i}+2\hat{j}+\hat{k}\) and \(\vec{C} = 6\hat{i}+10\hat{j}-7\hat{k}\) are coplanar.

Solution

We know that if the 3 vectors are coplanar, then the triple scalar product is zero. Although not explicitly stated here, we can assume the reverse, i.e. if the triple scalar product is zero, then the vectors are coplanar. [Note: The assumption of the reverse being true cannot be always be made. Be careful when making this assumption.]
So let's calculate the triple scalar product to see if it is zero.
\( \vec{A}\cdot \vec{B} \times \vec{C} = 2\hat{i}-3\hat{j}+4\hat{k} \cdot 6\hat{i}+2\hat{j}+\hat{k} \times 6\hat{i}+10\hat{j}-7\hat{k} = \) \(\begin{vmatrix} 2 & -3 & 4 \\ 6 & 2 & 1 \\ 6 & 10 & -7 \end{vmatrix} = \) \( 2(-14-10) - (-3)(-42 - 6) + 4(60-12) = 0 \)

Final Answer

Since the triple scalar product is zero, the vectors are coplanar.

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

Set 2 - squared identities

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

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

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

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

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

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

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

 

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

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

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

 

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

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

 

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

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

 

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