Cross Product

Cross Product #

\( \boxed{ \begin{aligned} \mathbf{a\times b} &=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} =\begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix} \mathbf{i} -\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix} \mathbf{j} +\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix} \mathbf{k} \\ &=(a_2b_3-a_3b_2)\mathbf{i}-(a_1b_3-a_3b_1)\mathbf{j}+(a_1b_2-a_2b_1)\mathbf{k} \\ &=\langle a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1\rangle \end{aligned} } \)

Revised

09 Apr 2023

Programming Environment #

import numpy as np

Determinants #

Definition of Determinant of Order 2 #

\( \begin{aligned} \begin{vmatrix} a & b \\ c & d \end{vmatrix} =ad-bc \end{aligned} \)

Definition of Determinant of Order 3 #

\( \begin{aligned} \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} &=a_1\begin{vmatrix} b_2 & b_3 \\ c_2 & c_3 \end{vmatrix} -a_2\begin{vmatrix} b_1 & b_3 \\ c_1 & c_3 \end{vmatrix} +a_3\begin{vmatrix} b_1 & b_2 \\ c_1 & c_2 \end{vmatrix} =a_1(b_2c_3-b_3c_2)-a_2(b_1c_3-b_3c_1)+a_3(b_1c_2-b_2c_1) \end{aligned} \)

Construction of the cross product #

Let \(\mathbf{a}\) and \(\mathbf{b}\) be two nonzero vectors and let \(\mathbf{c}\) be a vector that is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\).

\( \begin{aligned} \mathbf{a\perp c} \implies \mathbf{a\cdot c} &=0 =\mathbf{b\cdot c} \impliedby \mathbf{b\perp c} \\ b_3(a_1c_1+a_2c_2+a_3c_3) &=0 =a_3(b_1c_1+b_2c_2+b_3c_3) \\ a_1b_3c_1+a_2b_3c_2+a_3b_3c_3 &=0 =a_3b_1c_1+a_3b_2c_2+a_3b_3c_3 \\ a_1b_3c_1+a_2b_3c_2+\cancel{a_3b_3c_3}-a_3b_1c_1-a_3b_2c_2-\cancel{a_3b_3c_3} &=0 \\ (a_1b_3-a_3b_1)c_1+(a_2b_3-a_3b_2)c_2 &=0 \\ pc_1+qc_2 &=0 \,\,\,\,\,\text{where}\,\,\,\,\, c_1=q=a_2b_3-a_3b_2 \,\,\,\,\, c_2=-p=a_3b_1-a_1b_3 \\ a_1(a_2b_3-a_3b_2)+a_2(a_3b_1-a_1b_3)+a_3c_3 &=b_1(a_2b_3-a_3b_2)+b_2(a_3b_1-a_1b_3)+b_3c_3 \\ \cancel{a_1a_2b_3}-a_1a_3b_2+a_2a_3b_1-\cancel{a_1a_2b_3}+a_3c_3 &=a_2b_1b_3-\cancel{a_3b_1b_2}+\cancel{a_3b_1b_2}-a_1b_2b_3+b_3c_3 \\ a_3c_3-b_3c_3 &=a_1a_3b_2-a_2a_3b_1+a_2b_1b_3-a_1b_2b_3 \\ c_3(a_3-b_3) &=a_3(a_1b_2-a_2b_1)-b_3(a_1b_2-a_2b_1) \\ c_3 &=\frac{\cancel{(a_3-b_3)}(a_1b_2-a_2b_1)}{\cancel{(a_3-b_3)}} \\ c_3 &=a_1b_2-a_2b_1 \end{aligned} \)

A vector \(\mathbf{c}\) that is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\) is thus

\( \mathbf{c} =\langle c_1,c_2,c_3\rangle =\langle a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1\rangle \)

Definition of the cross product #

Let \(\mathbf{a}=\langle a_1,a_2,a_3\rangle\) and \(\mathbf{b}=\langle b_1,b_2,b_3\rangle\).

The cross product of \(\mathbf{a}\) and \(\mathbf{b}\) is

\( \begin{aligned} \mathbf{a\times b} &=\langle a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1\rangle \\ &=(a_2b_3-a_3b_2)\mathbf{i}-(a_1b_3-a_3b_1)\mathbf{j}+(a_1b_2-a_2b_1)\mathbf{k} \\ &=\begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix} \mathbf{i} -\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix} \mathbf{j} +\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix} \mathbf{k} \\ &=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} \end{aligned} \)

Properties of the cross product #

\(\mathbf{a\times a=0}\)#

The cross product of a vector with itself is the zero vector.

PROOF

\( \begin{aligned} \mathbf{a\times a} &=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ a_1 & a_2 & a_3 \end{vmatrix} \\ &=\begin{vmatrix} a_2 & a_3 \\ a_2 & a_3 \end{vmatrix} \mathbf{i} -\begin{vmatrix} a_1 & a_3 \\ a_1 & a_3 \end{vmatrix} \mathbf{j} +\begin{vmatrix} a_1 & a_2 \\ a_1 & a_2 \end{vmatrix} \mathbf{k} \\ &=(a_2a_3-a_3a_2)\mathbf{i}-(a_1a_3-a_3a_1)\mathbf{j}+(a_1a_2-a_2a_1)\mathbf{k} \\ &=\langle a_2a_3-a_3a_2,a_3a_1-a_1a_3,a_1a_2-a_2a_1\rangle \\ &=\mathbf{0} \end{aligned} \)

\(\blacksquare\)

\(\mathbf{a\perp(a\times b)\perp b}\)#

The cross product of two vectors is orthogonal to both original vectors.

PROOF

\( \begin{aligned} \mathbf{a}\cdot(\mathbf{a\times b}) &=\langle a_1,a_2,a_3\rangle\cdot\langle a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1\rangle \\ &=a_1(a_2b_3-a_3b_2)-a_2(a_1b_3-a_3b_1)+a_3(a_1b_2-a_2b_1) \\ &=a_1a_2b_3-a_1a_3b_2-a_1a_2b_3+a_2a_3b_1+a_1a_3b_2-a_2a_3b_1 \\ &=0 \end{aligned} \)

Alternatively

\( \begin{aligned} \mathbf{a}\cdot(\mathbf{a\times b}) &=\begin{vmatrix} a_1 & a_2 & a_3 \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} &&\text{by the definition of the scalar triple product} \\ &=a_1\begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix} -a_2\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix} +a_3\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix} \\ &=a_1(a_2b_3-a_3b_2)-a_2(a_1b_3-a_3b_1)+a_3(a_1b_2-a_2b_1) \\ &=a_1a_2b_3-a_1a_3b_2-a_1a_2b_3+a_2a_3b_1+a_1a_3b_2-a_2a_3b_1 \\ &=0 \end{aligned} \)

\( \begin{aligned} \mathbf{b}\cdot(\mathbf{a\times b}) &=\langle b_1,b_2,b_3\rangle\cdot\langle a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1\rangle \\ &=b_1(a_2b_3-a_3b_2)-b_2(a_1b_3-a_3b_1)+b_3(a_1b_2-a_2b_1) \\ &=a_2b_1b_3-a_3b_1b_2-a_1b_2b_3+a_3b_1b_2+a_1b_2b_3-a_2b_1b_3 \\ &=0 \end{aligned} \)

Alternatively

\( \begin{aligned} \mathbf{b}\cdot(\mathbf{a\times b}) &=\begin{vmatrix} b_1 & b_2 & b_3 \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} &&\text{by the definition of the scalar triple product} \\ &=b_1\begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix} -b_2\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix} +b_3\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix} \\ &=b_1(a_2b_3-a_3b_2)-b_2(a_1b_3-a_3b_1)+b_3(a_1b_2-a_2b_1) \\ &=a_2b_1b_3-a_3b_1b_2-a_1b_2b_3+a_3b_1b_2+a_1b_2b_3-a_2b_1b_3 \\ &=0 \end{aligned} \)

\(\blacksquare\)

\(\mathbf{\|a\times b\|}=\mathbf{\|a\|\|b\|}\sin\theta\)#

Let \(0\le\theta\le\pi\) be the angle between \(\mathbf{a}\) and \(\mathbf{b}\).

\(\mathbf{\|a\times b\|}=\mathbf{\|a\|\|b\|}\sin\theta\)

PROOF

\( \begin{aligned} \|\mathbf{a\times b}\|^2 &=(a_2b_3-a_3b_2)^2+(a_3b_1-a_1b_3)^2+(a_1b_2-a_2b_1)^2 \\ &=a_2^2b_3^2-2a_2a_3b_2b_3+a_3^2b_2^2+a_3^2b_1^2-2a_1a_3b_1b_3+a_1^2b_3^2+a_1^2b_2^2-2a_1a_2b_1b_2+a_2^2b_1^2 \\ &=a_1^2b_2^2+a_1^2b_3^2+a_2^2b_1^2+a_2^2b_3^2+a_3^2b_1^2+a_3^2b_2^2-2a_1a_2b_1b_2-2a_1a_3b_1b_3-2a_2a_3b_2b_3 \\ &=(a_1^2b_2^2+a_1^2b_3^2+a_2^2b_1^2+a_2^2b_3^2+a_3^2b_1^2+a_3^2b_2^2)-(2a_1a_2b_1b_2+2a_1a_3b_1b_3+2a_2a_3b_2b_3) \\ &=(a_1^2b_2^2+a_1^2b_3^2+a_2^2b_1^2+a_2^2b_3^2+a_3^2b_1^2+a_3^2b_2^2)+[a_1^2b_1^2+a_2^2b_2^2+a_3^2b_3^2] \\ &-(2a_1a_2b_1b_2+2a_1a_3b_1b_3+2a_2a_3b_2b_3)-[a_1^2b_1^2+a_2^2b_2^2+a_3^2b_3^2] \\ &=a_1^2b_1^2+a_1^2b_2^2+a_1^2b_3^2+a_2^2b_1^2+a_2^2b_2^2+a_2^2b_3^2+a_3^2b_1^2+a_3^2b_2^2+a_3^2b_3^2 \\ &-(a_1^2b_1^2+a_1a_2b_1b_2+a_1a_3b_1b_3+a_1a_2b_1b_2+a_2^2b_2^2+a_2a_3b_2b_3+a_1a_3b_1b_3+a_2a_3b_2b_3+a_3^2b_3^2) \\ &=(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)-(a_1b_1+a_2b_2+a_3b_3)^2 \\ &=\mathbf{\|a\|}^2\mathbf{\|b\|}^2-(\mathbf{a\cdot b})^2 \\ &=\mathbf{\|a\|}^2\mathbf{\|b\|}^2-(\mathbf{\|a\|\|b\|}\cos\theta)^2 \\ &=\mathbf{\|a\|}^2\mathbf{\|b\|}^2-\mathbf{\|a\|}^2\mathbf{\|b\|}^2\cos^2\theta \\ &=\mathbf{\|a\|}^2\mathbf{\|b\|}^2(1-\cos^2\theta) \\ &=\mathbf{\|a\|}^2\mathbf{\|b\|}^2\sin^2\theta \\ \|\mathbf{a\times b}\| &=\mathbf{\|a\|\|b\|}\sin\theta \end{aligned} \)

\(\blacksquare\)

\(\mathbf{a\parallel b}\iff\mathbf{a\times b=0}\)#

Two nonzero vectors \(\mathbf{a}\) and \(\mathbf{b}\) are parallel iff their cross product \(\mathbf{a\times b}\) is the zero vector \(\mathbf{0}\).

PROOF

1

\(\mathbf{a\parallel b}\implies\mathbf{a\times b=0}\)

If two nonzero vectors \(\mathbf{a}\) and \(\mathbf{b}\) are parallel then their cross product \(\mathbf{a\times b}\) is the zero vector \(\mathbf{0}\).

\( \mathbf{a\parallel b} \implies \theta=0\lor\theta=\pi \implies \sin\theta=0 \implies \|\mathbf{a\times b}\|=0 \implies \mathbf{a\times b}=\mathbf{0} \)

2

\(\mathbf{a\times b=0}\implies\mathbf{a\parallel b}\)

If the cross product of two nonzero vectors \(\mathbf{a}\) and \(\mathbf{b}\) is the zero vector \(\mathbf{0=a\times b}\) then they are parallel.

\(\blacksquare\)

\(A_\text{parallelogram}=\|\mathbf{a\times b}\|\)#

The magnitude \(\mathbf{\|a\times b\|}\) of the cross product is equal to the area of the parallelogram determined by \(\mathbf{a}\) and \(\mathbf{b}\).

If \(\mathbf{a}\) and \(\mathbf{b}\) are represented as directed line segments with the same initial point, then they determine a parallelogram with

base \(\mathbf{\|a\|}\)

height \(\mathbf{\|b\|}\sin\theta\)

area \(A_\text{parallelogram}=\text{base}\times\text{height}=(\mathbf{\|a\|})(\mathbf{\|b\|}\sin\theta)=\mathbf{\|a\times b\|}\)

\(V_\text{parallelepiped}=\mathbf{\|a\cdot(b\times c)\|}\)#

Noncommutativity of cross product #

\( \boxed{ \mathbf{a\times b\ne b\times a} } \)

Standard basis vectors

\( \begin{aligned} \mathbf{k} =\mathbf{i\times j} &\ne\mathbf{j\times i} =\mathbf{-k} \\ \mathbf{i} =\mathbf{j\times k} &\ne\mathbf{k\times j} =\mathbf{-i} \\ \mathbf{j} =\mathbf{k\times i} &\ne\mathbf{i\times k} =\mathbf{-j} \end{aligned} \)

Nonassociativity of cross product #

\( \boxed{ \mathbf{a\times(b\times c)} \ne\mathbf{(a\times b)\times c} } \)

Standard basis vectors

\( \begin{aligned} \mathbf{0} =\mathbf{i\times 0} =\mathbf{i\times(i\times i)} &=\mathbf{(i\times i)\times i} =\mathbf{0\times i} =\mathbf{0} \\ \mathbf{-j} =\mathbf{i\times k} =\mathbf{i\times(i\times j)} &\ne\mathbf{(i\times i)\times j} =\mathbf{0\times j} =\mathbf{0} \\ \mathbf{-k} =\mathbf{i\times -j} =\mathbf{i\times(i\times k)} &\ne\mathbf{(i\times i)\times k} =\mathbf{0\times k} =\mathbf{0} \\ &... \\ \mathbf{0} =\mathbf{i\times i} =\mathbf{i\times(j\times k)} &=\mathbf{(i\times j)\times k} =\mathbf{k\times k} =\mathbf{0} \\ &... \end{aligned} \)

\(\mathbf{a\times b=-b\times a}\)#

PROOF

\( \begin{aligned} \mathbf{a\times b} &=(a_2b_3-a_3b_2)\mathbf{i}-(a_1b_3-a_3b_1)\mathbf{j}+(a_1b_2-a_2b_1)\mathbf{k} \\ &=-[(b_2a_3-b_3a_2)\mathbf{i}-(b_1a_3-b_3a_1)\mathbf{j}+(b_1a_2-b_2a_1)\mathbf{k}] \\ &=-\mathbf{b\times a} \end{aligned} \)

\(\blacksquare\)

\((c\mathbf{a})\mathbf{\times b}=c(\mathbf{a\times b})=\mathbf{a\times}(c\mathbf{b})\)#

PROOF

\( \begin{aligned} (c\mathbf{a})\times\mathbf{b} &=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ ca_1 & ca_2 & ca_3 \\ b_1 & b_2 & b_3 \end{vmatrix} \\ &=\begin{vmatrix} ca_2 & ca_3 \\ b_2 & b_3 \end{vmatrix} \mathbf{i} -\begin{vmatrix} ca_1 & ca_3 \\ b_1 & b_3 \end{vmatrix} \mathbf{j} +\begin{vmatrix} ca_1 & ca_2 \\ b_1 & b_2 \end{vmatrix} \mathbf{k} \\ &=(ca_2b_3-ca_3b_2)\mathbf{i}-(ca_1b_3-ca_3b_1)\mathbf{j}+(ca_1b_2-ca_2b_1)\mathbf{k} \\ &=(a_2cb_3-a_3cb_2)\mathbf{i}-(a_1cb_3-a_3cb_1)\mathbf{j}+(a_1cb_2-a_2cb_1)\mathbf{k} \\ &=\begin{vmatrix} a_2 & a_3 \\ cb_2 & cb_3 \end{vmatrix} \mathbf{i} -\begin{vmatrix} a_1 & a_3 \\ cb_1 & cb_3 \end{vmatrix} \mathbf{j} +\begin{vmatrix} a_1 & a_2 \\ cb_1 & cb_2 \end{vmatrix} \mathbf{k} \\ &=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ cb_1 & cb_2 & cb_3 \end{vmatrix} \\ &=\mathbf{a}\times(c\mathbf{b}) \\ &=c(a_2b_3-a_3b_2)\mathbf{i}-c(a_1b_3-a_3b_1)\mathbf{j}+c(a_1b_2-a_2b_1)\mathbf{k} \\ &=c[(a_2b_3-a_3b_2)\mathbf{i}-(a_1b_3-a_3b_1)\mathbf{j}+(a_1b_2-a_2b_1)\mathbf{k}] \\ &=c\left[ \begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix} \mathbf{i} -\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix} \mathbf{j} +\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix} \mathbf{k} \right] \\ &=c\left[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} \right] \\ &=c(\mathbf{a\times b}) \end{aligned} \)

\(\blacksquare\)

\(\mathbf{a\times(b+c)=a\times b+a\times c}\)#

PROOF

\( \begin{aligned} \mathbf{a\times(b+c)} &=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1+c_1 & b_2+c_2 & b_3+c_3 \end{vmatrix} \\ &=\begin{vmatrix} a_2 & a_3 \\ b_2+c_2 & b_3+c_3 \end{vmatrix} \mathbf{i} -\begin{vmatrix} a_1 & a_3 \\ b_1+c_1 & b_3+c_3 \end{vmatrix} \mathbf{j} +\begin{vmatrix} a_1 & a_2 \\ b_1+c_1 & b_2+c_2 \end{vmatrix} \mathbf{k} \\ &=[a_2(b_3+c_3)-a_3(b_2+c_2)]\mathbf{i}-[a_1(b_3+c_3)-a_3(b_1+c_1)]\mathbf{j}+[a_1(b_2+c_2)-a_2(b_1+c_1)]\mathbf{k} \\ &=(a_2b_3+a_2c_3-a_3b_2-a_3c_2)\mathbf{i}-(a_1b_3+a_1c_3-a_3b_1-a_3c_1)\mathbf{j}+(a_1b_2+a_1c_2-a_2b_1-a_2c_1)\mathbf{k} \\ &=(a_2b_3-a_3b_2+a_2c_3-a_3c_2)\mathbf{i}-(a_1b_3-a_3b_1+a_1c_3-a_3c_1)\mathbf{j}+(a_1b_2-a_2b_1+a_1c_2-a_2c_1)\mathbf{k} \\ &=(a_2b_3-a_3b_2)\mathbf{i}+(a_2c_3-a_3c_2)\mathbf{i}-(a_1b_3-a_3b_1)\mathbf{j}-(a_1c_3-a_3c_1)\mathbf{j}+(a_1b_2-a_2b_1)\mathbf{k}+(a_1c_2-a_2c_1) \\ &=(a_2b_3-a_3b_2)\mathbf{i}-(a_1b_3-a_3b_1)\mathbf{j}+(a_1b_2-a_2b_1)\mathbf{k}+(a_2c_3-a_3c_2)\mathbf{i}-(a_1c_3-a_3c_1)\mathbf{j}+(a_1c_2-a_2c_1)\mathbf{k} \\ &=\begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix} \mathbf{i} -\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix} \mathbf{j} +\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix} \mathbf{k} +\begin{vmatrix} a_2 & a_3 \\ c_2 & c_3 \end{vmatrix} \mathbf{i} -\begin{vmatrix} a_1 & a_3 \\ c_1 & c_3 \end{vmatrix} \mathbf{j} +\begin{vmatrix} a_1 & a_2 \\ c_1 & c_2 \end{vmatrix} \mathbf{k} \\ &=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} +\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ c_1 & c_2 & c_3 \end{vmatrix} \\ &=\mathbf{a\times b+a\times c} \end{aligned} \)

\(\blacksquare\)

\(\mathbf{(a+b)\times c=a\times c+b\times c}\)#

Scalar Triple Product \(\mathbf{a\cdot(b\times c)=(a\times b)\cdot c}\)#

\( \mathbf{a\cdot(b\times c)=(a\times b)\cdot c} \)

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

PROOF

\( \begin{aligned} \mathbf{a\cdot(b\times c)} &=a_1(b_2c_3-b_3c_2)+a_2(b_3c_1-b_1c_3)+a_3(b_1c_2-b_2c_1) \\ &=a_1b_2c_3-a_1b_3c_2+a_2b_3c_1-a_2b_1c_3+a_3b_1c_2-a_3b_2c_1 \\ &=a_2b_3c_1-a_3b_2c_1+a_3b_1c_2-a_1b_3c_2+a_1b_2c_3-a_2b_1c_3 \\ &=(a_2b_3-a_3b_2)c_1+(a_3b_1-a_1b_3)c_2+(a_1b_2-a_2b_1)c_3 \\ &=\mathbf{(a\times b)\cdot c} \end{aligned} \)

\(\blacksquare\)

Vector Triple Product \(\mathbf{a\times(b\times c)=(a\cdot c)b-(a\cdot b)c}\)#

PROOF

\( \begin{aligned} \mathbf{a\times(b\times c)} &=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_2c_3-b_3c_2 & b_3c_1-b_1c_3 & b_1c_2-b_2c_1 \end{vmatrix} \\ &=\begin{vmatrix} a_2 & a_3 \\ b_3c_1-b_1c_3 & b_1c_2-b_2c_1 \end{vmatrix} \mathbf{i} -\begin{vmatrix} a_1 & a_3 \\ b_2c_3-b_3c_2 & b_1c_2-b_2c_1 \end{vmatrix} \mathbf{j} +\begin{vmatrix} a_1 & a_2 \\ b_2c_3-b_3c_2 & b_3c_1-b_1c_3 \end{vmatrix} \mathbf{k} \\ &=[a_2(b_1c_2-b_2c_1)-a_3(b_3c_1-b_1c_3)]\mathbf{i}-[a_1(b_1c_2-b_2c_1)-a_3(b_2c_3-b_3c_2)]\mathbf{j}+[a_1(b_3c_1-b_1c_3)-a_2(b_2c_3-b_3c_2)]\mathbf{k} \\ &=(a_2b_1c_2-a_2b_2c_1-a_3b_3c_1+a_3b_1c_3)\mathbf{i}-(a_1b_1c_2-a_1b_2c_1-a_3b_2c_3+a_3b_3c_2)\mathbf{j}+(a_1b_3c_1-a_1b_1c_3-a_2b_2c_3+a_2b_3c_2)\mathbf{k} \\ &=[(a_2b_1c_2-a_2b_2c_1-a_3b_3c_1+a_3b_1c_3)+(a_1b_1c_1-a_1b_1c_1)]\mathbf{i} \\ &+[(-a_1b_1c_2+a_1b_2c_1+a_3b_2c_3-a_3b_3c_2)+(a_2b_2c_2-a_2b_2c_2)]\mathbf{j} \\ &+[(a_1b_3c_1-a_1b_1c_3-a_2b_2c_3+a_2b_3c_2)+(a_3b_3c_3-a_3b_3c_3)]\mathbf{k} \\ &=a_1b_1c_1\mathbf{i}+a_2b_1c_2\mathbf{i}+a_3b_1c_3\mathbf{i}-(a_1b_1c_1\mathbf{i}+a_2b_2c_1\mathbf{i}+a_3b_3c_1\mathbf{i}) \\ &+a_1b_2c_1\mathbf{j}+a_2b_2c_2\mathbf{j}+a_3b_2c_3\mathbf{j}-(a_1b_1c_2\mathbf{j}+a_2b_2c_2\mathbf{j}+a_3b_3c_2\mathbf{j}) \\ &+a_1b_3c_1\mathbf{k}+a_2b_3c_2\mathbf{k}+a_3b_3c_3\mathbf{k}-(a_1b_1c_3\mathbf{k}+a_2b_2c_3\mathbf{k}+a_3b_3c_3\mathbf{k}) \\ &=(a_1c_1+a_2c_2+a_3c_3)(b_1\mathbf{i}+b_2\mathbf{j}+b_3\mathbf{k})-(a_1b_1+a_2b_2+a_3b_3)(c_1\mathbf{i}+c_2\mathbf{j}+c_3\mathbf{k}) \\ &=\mathbf{(a\cdot c)b-(a\cdot b)c} \end{aligned} \)

\(\blacksquare\)

Exercises #

[EXERCISE]

Determine a vector perpendicular to the plane that passes through the points \(P(1,4,6),Q(-2,5,-1),R(1,-1,1)\).

The plane is determined by the vectors \(\vec{PQ},\vec{PR}\).

\( \begin{aligned} \vec{PQ} &=(-2-1)\mathbf{i}+(5-4)\mathbf{j}+(-1-6)\mathbf{k} =-3\mathbf{i}+\mathbf{j}-7\mathbf{k} \\ &=\langle-3,1,-7\rangle \\ \vec{PR} &=(1-1)\mathbf{i}+(-1-4)\mathbf{j}+(1-6)\mathbf{k} =-5\mathbf{j}-5\mathbf{k} \\ &=\langle0,-5,-5\rangle \\ \vec{PQ}\times\vec{PR} &=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & 1 & -7 \\ 0 & -5 & -5 \end{vmatrix} =\begin{vmatrix} 1 & -7 \\ -5 & -5 \end{vmatrix} \mathbf{i} -\begin{vmatrix} -3 & -7 \\ 0 & -5 \end{vmatrix} \mathbf{j} +\begin{vmatrix} -3 & 1 \\ 0 & -5 \end{vmatrix} \mathbf{k} =-40\mathbf{i}-15\mathbf{j}+15\mathbf{k} \\ &=\langle-40,-15,15\rangle \end{aligned} \)

Any nonzero scalar multiple

\(t\langle-40,-15,15\rangle\,\,\,t\in\mathbb{R}\)

of \(\vec{PQ}\times\vec{PR}\) is perpendicular to the plane

\(a\langle-3,1,-7\rangle+b\langle0,-5,-5\rangle\,\,\,a,b\in\mathbb{R}\)

!!!

\( \begin{aligned} t(\vec{PQ}\times\vec{PR})\cdot(a\vec{PQ}+b\vec{PR}) &=\langle-40t,-15t,15t\rangle\cdot\langle-3a,a,-7a\rangle+\langle0,-5b,-5b\rangle \\ &=\langle-40t,-15t,15t\rangle\cdot\langle-3a,a-5b,-7a-5b\rangle \\ &=(-40t)(-3a)+(-15t)(a-5b)+(15t)(-7a-5b) \\ &=120at-15at+75bt-105at-75bt \\ &=0 \end{aligned} \)

[EXERCISE]

Find the area of the triangle with vertices \(P(1,4,6),Q(-2,5,-1),R(1,-1,1)\).

\(\vec{PQ}\) and \(\vec{PR}\) determine a parallelogram with area

\( \begin{aligned} A_\text{parallelogram} &=\|\vec{PQ}\times\vec{PR}\| =\sqrt{(-40)^2+(-15)^2+(15)^2} =\sqrt{2050} =\sqrt{25\cdot82} \\ &=5\sqrt{82} \\ A_{\Delta PQR} =\frac{1}{2}A_\text{parallelogram} &=\frac{5}{2}\sqrt{82} \end{aligned} \)

[EXERCISE]

Show that the vectors

\( \begin{aligned} \mathbf{a} &=\langle1,4,-7\rangle \\ \mathbf{b} &=\langle2,-1,4\rangle \\ \mathbf{c} &=\langle0,-9,18\rangle \end{aligned} \)

are coplanar via the scalar triple product.

\( \begin{aligned} \mathbf{a\cdot(b\times c)} &=\begin{vmatrix} 1 & 4 & -7 \\ 2 & -1 & 4 \\ 0 & -9 & 18 \\ \end{vmatrix} \\ &=\begin{vmatrix} -1 & 4 \\ -9 & 18 \\ \end{vmatrix} -4\begin{vmatrix} 2 & 4 \\ 0 & 18 \\ \end{vmatrix} -7\begin{vmatrix} 2 & -1 \\ 0 & -9 \\ \end{vmatrix} \\ &=(-1)(18)-(4)(-9)-4[(2)(18)-(4)(0)]-7[(2)(-9)-(-1)(0)] \\ &=-18+36-144+126 \\ &=0 \end{aligned} \)

Figures #

[W] Hamilton, William (1805-1865)

Terms #

[W] Cross Product

Cross Product

Contents