Constructing Orthogonal Vectors, Geometric Implications

1. Last time we ended with discussing how to project a vector onto another. So if we consider projecting {\vec{u}} onto {\vec{v}}, we can write this as
(1)\displaystyle  \mathop{\mathrm{proj}}\nolimits_{\vec{v}}\vec{u} = \vec{u}_{\|}
Observe, we have another vector constructed
(2)\displaystyle  \vec{u}_{\bot} = \vec{u}-\vec{u}_{\|}
which is orthogonal to {\vec{v}}. We have a closed form expression for projection, namely
(3)\displaystyle  \mathop{\mathrm{proj}}\nolimits_{\vec{v}}\vec{u} = (\vec{u}\cdot\widehat{v})\widehat{v}
where {\widehat{v}=\vec{v}/\|\vec{v}\|} is a unit vector. But do we have a closed form expression for {\vec{u}_{\bot}}?

2. Cross-Product. The cross product of {\vec{u}} and {\vec{v}} is
(4)\displaystyle  \vec{u}\times\vec{v} = \begin{vmatrix} \widehat{\textbf{\i}} & \widehat{\textbf{\j}} & \widehat{\mathbf{k}}\\ u_{1} & u_{2} & u_{3}\\ v_{1} & v_{2} & v_{3} \end{vmatrix} = \bigl(\|\vec{u}\|\|\vec{v}\|\sin(\theta)\bigr)\widehat{u}
Note we are using notation from linear algebra writing, recursively,
(5)\displaystyle  \begin{aligned} \det(A) &= \begin{vmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{vmatrix} \\ &= a_{11}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{12} \begin{vmatrix} a_{21} & a_{23}\\ a_{31} & a_{33} \end{vmatrix} +a_{13}\begin{vmatrix} a_{21} & a_{22}\\ a_{31} & a{32} \end{vmatrix} \end{aligned}
(6)\displaystyle  \begin{vmatrix} a & b\\ c & d \end{vmatrix} = ad-bc.

Remark 1. Observe this implies {\widehat{\textbf{\i}}\times\widehat{\textbf{\j}}=\widehat{\textbf{k}}}, {\widehat{\textbf{\j}}\times\widehat{\textbf{k}}=\widehat{\textbf{\i}}}, and {\widehat{\textbf{k}}\times\widehat{\textbf{\i}}=\widehat{\textbf{\j}}}.

Remark 2. The cross-product takes two vectors, and produces a third vector. It does not produce a scalar (a number, unlike the dot product).

Pop quiz: let {\vec{u}} and {\vec{v}} be vectors. Is {\vec{u}\times\vec{v}=\vec{v}\times\vec{u}}?

Example 1. Consider {\vec{u}=\langle2,1,-3\rangle} and {\vec{v}=\langle1,-2,1\rangle}. What is {\vec{u}\times\vec{v}}?

Solution: we find
(7)\displaystyle  \begin{aligned} \vec{u}\times\vec{v} &= \begin{vmatrix} \widehat{\textbf{\i}} & \widehat{\textbf{\j}} &\widehat{\textbf{k}}\\ 2 & 1 & -3\\ 1 & -2 & 1 \end{vmatrix}\\ &= (1\cdot1-(-3)\cdot(-2))\widehat{\textbf{\i}} - (2\cdot1-(-3)\cdot1)\widehat{\textbf{\j}} +(2\cdot(-2)-1\cdot1)\widehat{\textbf{k}}\\ &=(1-6)\widehat{\textbf{\i}} - (2+3)\widehat{\textbf{\j}} +(-4-1)\widehat{\textbf{k}}\\ &=\langle-5,5,-5\rangle \end{aligned}
Another approach would have been to write
(8)\displaystyle  \vec{u}\times\vec{v} = (2\widehat{\textbf{\i}} + \widehat{\textbf{\j}} -3\widehat{\textbf{k}}) \times(\widehat{\textbf{\i}} -2 \widehat{\textbf{\j}} +\widehat{\textbf{k}})
and used the cross-product’s anticommutativity to do the calculations.

3. Parallelogram Area. Consider three distinct points {P}, {Q}, and {R}. We can construct a parallelogram, as in the following diagram:

We see that {\overrightarrow{PQ}\times\overrightarrow{PR}=\vec{N}}, then the area of the parallelogram is {\|\vec{N}\|}.

4. Parallelepiped Volume. If we work in 3-space, and we have a six-sided region whose sides are each parallelograms, we call this region a parallelepiped. Observe that we only need 3 vectors to specify the vertices: {\vec{u}}, {\vec{v}}, and {\vec{w}}. Then we consider {\vec{u}+\vec{v}}, {\vec{u}+\vec{w}}, {\vec{v}+\vec{w}}, and {\vec{u}+\vec{v}+\vec{w}} for the remaining vertices. What is the volume of this region?

Lets draw a diagram:

Lets first consider the face described by {\overrightarrow{PQ}=\vec{u}} and {\overrightarrow{PR}=\vec{v}}. We see the parallelepiped may be considered as a “stack” of such faces, whose height is given by the third vector {\vec{w}}. Then we see the area of the face is shaded in the diagram, and algebraically it’s given by {\vec{u}\times\vec{v}}, and this produces a vector whose magnitude is the area of the face. When we “dot” this with {\mathop{\mathrm{proj}}\nolimits_{\vec{u}\times\vec{v}}\vec{w}}, it’s intuitively taking the product of the “area of a face” ({\vec{u}\times\vec{v}}) and the “height of the parallelepiped” ({\mathop{\mathrm{proj}}\nolimits_{\vec{u}\times\vec{v}}\vec{w}}) producing the volume
(9)\displaystyle  \mbox{volume } = (\vec{u}\times\vec{v})\cdot\vec{w}.
Note this can be negative, and this just tells us information regarding the parallelepiped’s orientation.


About Alex Nelson

I like math. I like programming. Most of all, I love puzzles.
This entry was posted in Calculus, Cross Product, Geometry, Vector Calculus and tagged . Bookmark the permalink.

One Response to Constructing Orthogonal Vectors, Geometric Implications

  1. Pingback: Constructing Planes | My Math Blog

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s