12.4 Cross Product

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Determinants

A determinant of order 2 is defined by ∣abcd∣=ad−bc\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc A determinant of order 3 is defined by ∣a1a2a3b1b2b3c1c2c3∣=a1∣b2b3c2c3∣−a2∣b1b3c1c3∣+a3∣b1b2c1c2∣\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}

Cross Product

The cross product of a⃗\vec{a} and b⃗\vec{b} is a vector, sometimes called the "vector product". The product is a vector that is perpendicular to a⃗\vec{a} and b⃗\vec{b}. It is calculated with a determinant 3x3 matrix that includes the standard basis vector and the components of the two vectors. a⃗×b⃗=∣i^j^k^a1a2a3b1b2b3∣\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_{1}&a_{2}&a_{3}\\b_{1}&b_{2}&b_{3}\end{vmatrix}

  • To visualize which direction the cross product is in, use the right hand rule:
  • a⃗×b⃗\vec{a}\times\vec{b} is orthogonal to a⃗\vec{a} and b⃗\vec{b}.
  • If θ\theta is the angle between a⃗\vec{a} and b⃗\vec{b} (0≤θ≤π0\leq\theta\leq\pi), then the magnitude of the cross product, ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin{\theta}
    • Corollary: nonzero vector a⃗\vec{a} and b⃗\vec{b} are parallel if ∣a⃗×b⃗∣=0|\vec{a}\times\vec{b}|=0 because the angle between the vectors is 0∘0^{\circ} or 180∘180^{\circ}.

Area of Traingle

Calculate the area of a trangle with points P(1,4,6)P(1,4,6), Q(−2,5,1)Q(-2,5,1), and R(1,−1,1)R(1,-1,1). PQ⃗=<−2−1,5−4,1−6>=<−3,1,7>\vec{PQ}=<-2-1,5-4,1-6>=<-3,1,7> and PR⃗=<0,−5,−5>\vec{PR}=<0,-5,-5> A=12bh=12absin⁡θA=\frac{1}{2}bh=\frac{1}{2}ab\sin{\theta} A=12∣PQ⃗∣∣PR⃗∣sin⁡PA=\frac{1}{2}|\vec{PQ}||\vec{PR}|\sin{P} A=12∣PQ⃗×PR⃗∣A=\frac{1}{2}|\vec{PQ}\times\vec{PR}| A=12∣i^j^k^−31−70−55∣A=\frac{1}{2}\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-3&1&-7\\0&-5&5\end{vmatrix} A=12∣<−40,−15,15>∣=12(−40)2+(−15)2+(152)A=\frac{1}{2}|<-40,-15,15>|=\frac{1}{2}\sqrt{(-40)^2+(-15)^2+(15^2)} A=5282A=\frac{5}{2}\sqrt{82}

Properties of Cross Product

If a⃗\vec{a}, b⃗\vec{b}, and c⃗\vec{c} are vectors and cc is a scalar:

  1. a⃗×b⃗\vec{a}\times\vec{b}=−b⃗×a⃗-\vec{b} \times \vec{a}
  2. ca⃗×b⃗=c(a⃗×b⃗)=a⃗×cb⃗c\vec{a}\times\vec{b}=c(\vec{a}\times\vec{b})=\vec{a}\times c\vec{b}
  3. a⃗×(b⃗+c⃗)=a⃗×b⃗+a⃗×c⃗\vec{a}\times (\vec{b}+\vec{c}) = \vec{a}\times\vec{b}+\vec{a}\times \vec{c}
  4. (a⃗+b⃗)×c=a⃗×c+b⃗×c(\vec{a}+\vec{b}) \times c = \vec{a} \times c + \vec{b} \times c
  5. a⃗⋅(b⃗×c⃗)=(a⃗×b⃗)⋅c⃗\vec{a}\cdot(\vec{b} \times \vec{c}) = ({\vec{a}\times\vec{b}}) \cdot \vec{c}
  6. a⃗×(b⃗×c⃗)=(a⃗⋅c⃗)b⃗−(a⃗⋅b⃗)c⃗\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}
  7. Cross product is not commutative.

Triple Products

a⃗⋅(b⃗×c⃗)\vec{a} \cdot (\vec{b} \times \vec{c}) is called the "scalar triple product" a⃗⋅(b⃗×c⃗)=∣a1a2a3b1b2b3c1c2c3∣  ⟹  R\vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix}a_{1}&a_{2}&a_{3}\\b_{1}&b_{2}&b_{3}\\c_{1}&c_{2}&c_{3}\end{vmatrix} \implies \mathbb{R}

The volume of a parallelpiped determined by the vectors a⃗\vec{a}, b⃗\vec{b}, and c⃗\vec{c} is the magnitude (always positive like volume) of their scalar triple product. V=∣a⃗⋅(b⃗×c⃗)∣V= |\vec{a} \cdot (\vec{b} \times \vec{c})| If the volume is zero, that means that the vectors are coplanar, or on the same plane.

Torque

The torque vector (τ⃗\vec{\tau}) is the coss product of position (r⃗\vec{r}) and force (F⃗\vec{F}) vector. τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F} Torque measures the tendency of the body to rotate about the origin. The direction of τ⃗\vec{\tau} is the axis of rotation. ∣τ⃗∣=∣r⃗×F⃗∣=∣r⃗∣∣F⃗∣sin⁡θ|\vec{\tau}|=|\vec{r} \times \vec{F}| = |\vec{r}||\vec{F}|\sin{\theta} The magnitude of torque is equal to the area of the parallelogram determined by r⃗\vec{r} and F⃗\vec{F}.