TC3 → Stan Brown → TI-83/84/89 → Multiplying Vectors
revised 31 Dec 2013 (What’s New?)

Multiplying Vectors on TI-83/84

Copyright © 2005–2014 by Stan Brown, Oak Road Systems

Summary: You can easily compute scalar products (dot products) on your TI-83 or TI-84 with built-in functions. A simple-follow-on lets you compute the length or magnitude of a vector.

This page also gives a program, ready for downloading or keying in, that computes those quantities plus the vector product (cross product) of two vectors, the angle between them, and the area of the parallelogram that they form.

Contents:

Built-in Functions for Handling Vectors

Your TI-83 or TI-84 can do operations on lists of numbers. Though it doesn’t know about vectors as such, you can consider a list of two or three numbers to be the components of a vector in the plane or in space.

Use curly braces { } around the list of components. To get a left or right curly brace, press the [2nd] key and then the left or right parenthesis.

Multiplication by a Scalar

Example: Suppose vector a is [2,−3] and you want to display the vector −7a. Here’s how.

Enter the scalar. [(-)] [7]
Enter the multiplication sign. [×] (displays as *)
Enter the vector in curly braces, with commas separating the components. [2nd ( makes {] [2] [,] [(-)] [3] [2nd ) makes }]
Display the result. screen display for this example [ENTER]

Dot Product

The dot product of two vectors u and v is formed by multiplying their components and adding. In the plane, u·v = u1v1+u2v2; in space it’s u1v1+u2v2+u3v3.

If you tell the TI-83/84 to multiply two lists, it multiplies the elements of the two lists to make a third list. The sum of the elements of that third list is the dot product of the vectors.

Example: If u = [2,3,1] and v = [4,−3,2], find u·v.

First invoke the summation function. Press [2nd STAT makes LIST] [] [5] to make “sum(” appear.
Enter the first vector, using curly braces as before. [2nd ( makes {] [2] [,] [3] [,] [1] [2nd ) makes }]
Multiply. Press [×] and * appears on the screen.
Enter the second vector. [2nd ( makes {] [4] [,] [(-)] [3] [,] [2] [2nd ) makes }]
Close the parenthesis from “sum(” and do the calculation. (u·v = 2×4 + 3×(−3) + 1×2 = 8−9+2 = 1.) screen display for this example [)] [ENTER]

Length of a Vector

You know that the length or magnitude of vector v is found by

||v|| = √(v1²+v2²)

You know also that the dot product of a vector with itself is

v·v = v1v1+v2v2 = v1²+v2²

which is the square of the length of v. Therefore

||v|| = √(v·v)

With the TI-83/84, there’s no need to enter the vector twice to find its length. Instead, just square the list of components — the calculator interprets this as squaring every component. Then take the square root of the sum.

Example: Find the length of vector a = [2,−5,−3].

Set up the square root. Press [2nd  makes ]. Notice that the calculator supplies a left parenthesis for you.
Set up the sum. [2nd STAT makes LIST] [] [5]
Enter the vector. [2nd ( makes {] [2] [,] [(-)] [5] [,] [(-)] [3] [2nd ) makes }]
Square all the components and close both parentheses.
 
You can check the calculator’s result by computing the length manually: ||v|| = √2²+(−5)²+(−3)²) = 4+25+9 = √38, which is about 6.16.
square root of sum of list 2, −5, −3, squared, equals 6.1644 [] [)] [)] [ENTER]

A Program for Harder Operations

I wrote a little TI-83/84 program to compute interesting results from two vectors, up to and including the cross product or vector product, and I offer it on this Web page.

Getting the Program

There are three methods to get the program into your calculator:

Running the Program

However you get it into your calculator, run the program VECPRODS. TI-83/84 screen with prompts for three components of each vector

Example (with my thanks to Jason Duguay): Find the dot product and cross product of

u = [0.894, 0.447, 0]   and   v = [−600, 200, −300]

As soon as you run the program, it prompts you for the three components of each vector. (If your vectors have only x and y components, enter 0 for the z components.) The program then displays two screens of information about the two vectors and their products.

 
program output screen 1: U={.894, .447, 0}, len U=.99952239, V={minus 600, 200, minus 300}, len V=700
program output screen 2: U dot V= minus 447, theta=129.7081, U cross V = {minus 134.1, 268.2, 447}, area=538.25928

You’ll see the following information, in order:

Program Variables

The program stores its results in several variables, which are left afterward for your use:

To access any of them, press the [ALPHA] key and then the key for the letter such as [ALPHA 6 makes V], or press [2nd STAT makes LIST] and then croll up to find the desired list name.

If you want to delete a list, press [2nd + makes MEM] [2] [4], cursor to each one, and press [DEL]. To delete an ordinary variable, press [2nd + makes MEM] [2] [2], cursor to each one, and press [DEL].

The program stores your graphics settings in GDB0 and then automatically deletes that variable after restoring your settings. You don’t care about this unless you're using GDB0 for your own purposes, which would be quite unusual.

More on the Cross Product

definition of cross product as a determinant and as length of normal vector Any two nonzero vectors u and v determine a unique plane, assuming they’re not parallel.

The cross product u×v is a third vector, which is defined in two ways as shown at right. Its magnitude is ||u|| ||v|| sin θ, the magnitude of the first times the magnitude of the second times the sine of the angle between them – this is also the area of a parallelogram whose sides are u and v. The cross product vector is normal (perpendicular) to the plane containing the two vectors, indicated by the unit normal vector n.

But which unit normal vector, since there are two? (Think of one pointing above the plane and one pointing below.) The answer is the infamous right-hand rule: if you hold your right hand so that the fingers curl from the first vector toward the second, then your thumb will point in the direction of the cross product vector u×v.

How do you evaluate the cross product, in component form? The answer is the determinant you see above. Doesn’t help? There are two main ways to evaluate a 3-by-3 determinant:

  1. The way I like best is to rewrite the first two columns to the right of the determinant and then take the six products shown:

    determinant evaluated by method of six products

    The products going down to the right have a plus sign, and those going up to the right have a minus sign. This is Sarrus’s rule, due to the Frenchman J.P. Sarrus (1789–1861).

  2. Some people prefer to evaluate the determinant by minors, the method of Pierre Simon, Marquis de Laplace (1749–1827). Remember that the second minor has its columns in reverse order from the original determinant!

    determinant evaluated by method of minors

Naturally, the two methods always give the same result (barring computational errors). Here’s a manual computation for the same cross product that the TI-83 or TI-84 calculated earlier:

[3, 5, -7] cross [-6, 4, -1] as a determinant = [5(−1)−4(−7)]i + [(−7)((−6)−3(−1)]j + [3(4)−(−6)5]k = 23i + 45j + 42k

What’s New


This page is used in instruction at Tompkins Cortland Community College in Dryden, New York; it’s not an official statement of the College. Please visit www.tc3.edu/instruct/sbrown/ to report errors or ask to copy it.

For updates and new info, go to http://www.tc3.edu/instruct/sbrown/ti83/