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

17Calculus - Vector Cross Product Application - Triple Vector Product

17Calculus

Coordinate Systems

Vectors

Using Vectors

Applications

Vector Functions

Partial Derivatives

Partial Integrals

Double Integrals - 2Int

Triple Integrals - 3Int

Practice

Vector Fields

Calculus Tools

Learning Tools

Articles

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

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

Log in to rate this practice problem and to see it's current rating.

Really UNDERSTAND Calculus

Topics You Need To Understand For This Page

Related Topics and Links

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

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

 

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

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

 

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

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

 

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

To bookmark this page and practice problems, log in to your account or set up a free account.

Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

how to take good notes

Practice

The Humongous Book of Calculus Problems

Save Up To 50% Off SwissGear Backpacks Plus Free Shipping Over $49 at eBags.com!

Shop Amazon - Rent Textbooks - Save up to 80%

As an Amazon Associate I earn from qualifying purchases.

Page Sections

Math Word Problems Demystified

Save Up To 50% Off SwissGear Backpacks Plus Free Shipping Over $49 at eBags.com!

Shop Amazon - Used Textbooks - Save up to 90%

As an Amazon Associate I earn from qualifying purchases.

Do NOT follow this link or you will be banned from the site!

When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.

Links and banners on this page are affiliate links. We carefully choose only the affiliates that we think will help you learn. Clicking on them and making purchases help you support 17Calculus at no extra charge to you. However, only you can decide what will actually help you learn. So think carefully about what you need and purchase only what you think will help you.

We use cookies on this site to enhance your learning experience.

17calculus

Copyright © 2010-2021 17Calculus, All Rights Reserved     [Privacy Policy]     [Support]     [About]

mathjax.org
Real Time Web Analytics