Question

How do I find a vector cross product on a $\text{TI-89}$ ?

Answer 1

In this question we will see the steps on how to do a vector cross product generally and then see the steps on how to do the product on a $\text{TI-89}$, which is a scientific calculator which has inbuilt function for doing vector cross products.

Complete step-by-step solution:
A vector is a quantity which has both a magnitude and a direction, they can be added by joining both the tails of the vector and to subtract vectors the direction of the vector we want to subtract should be reversed and then added as usual for subtraction.
Now when multiplication is considered in vectors, there can be two types of vector products which are dot product and cross product. The result of the dot product is a scalar quantity and the result of cross product is a vector.
The general way of denoting a vector in two dimensions is in the form of $x=ai+bj$, where $x$ is the two-dimensional vector and the general way of denoting a vector in three dimensions can be done as $x=ai+bj+ck$.
The determinant notation is used to find the vector cross product.
Let’s consider two vectors:
$x=3i+2j+1k$ and $y=i+4j+1k$
The steps to do the cross product on a $\text{TI-89}$ are:
$1)$ press the $\text{ }\\!\\![\\!\\!\text{ 2nd }\\!\\!]\\!\\!\text{ }$ and then $5$ to enter into the $\text{ }\\!\\![\\!\\!\text{ math }\\!\\!]\\!\\!\text{ }$ menu.
$2)$ scroll down to the fourth option named $\text{ }\\!\\![\\!\\!\text{ matrix }\\!\\!]\\!\\!\text{ }$ and press $\text{ }\\!\\![\\!\\!\text{ enter }\\!\\!]\\!\\!\text{ }$
$3)$ scroll down to the third option named $[\det ]$ and press $\text{ }\\!\\![\\!\\!\text{ enter }\\!\\!]\\!\\!\text{ }$
$4)$enter the vectors $i,j,k$ and coefficients of each vector separated by a comma, and all the three rows separated by a semi-colon and enclose the matrix in parenthesis.
$5)$press $\text{ }\\!\\![\\!\\!\text{ enter }\\!\\!]\\!\\!\text{ }$and the answer would be displayed.
Generally, the determinant notation is used to compute the cross product.
It can be done as:
$\Rightarrow \left( \begin{matrix} 3 \\\ 2 \\\ 1 \\\ \end{matrix} \right)\times \left( \begin{matrix} 1 \\\ 4 \\\ 1 \\\ \end{matrix} \right)=\left| \begin{matrix} i & j & k \\\ 3 & 2 & 1 \\\ 1 & 4 & 1 \\\ \end{matrix} \right|$
On simplifying the determinant, we get:
$\Rightarrow \left( \begin{matrix} 3 \\\ 2 \\\ 1 \\\ \end{matrix} \right)\times \left( \begin{matrix} 1 \\\ 4 \\\ 0 \\\ \end{matrix} \right)=\left| \begin{matrix} 2 & 1 \\\ 4 & 1 \\\ \end{matrix} \right|i-\left| \begin{matrix} 3 & 1 \\\ 1 & 1 \\\ \end{matrix} \right|j+\left| \begin{matrix} 3 & 2 \\\ 1 & 4 \\\ \end{matrix} \right|k$
On taking the determinant, we get:
$\Rightarrow \left( \begin{matrix} 3 \\\ 2 \\\ 1 \\\ \end{matrix} \right)\times \left( \begin{matrix} 1 \\\ 4 \\\ 1 \\\ \end{matrix} \right)=-2i-1j+10k$
Which can be written as:
$\Rightarrow \left( \begin{matrix} 3 \\\ 2 \\\ 1 \\\ \end{matrix} \right)\times \left( \begin{matrix} 1 \\\ 4 \\\ 1 \\\ \end{matrix} \right)=\left( \begin{matrix} -2 \\\ -2 \\\ 10 \\\ \end{matrix} \right)$

Note: It is to be remembered that a $\text{TI-89}$ can also do vector product by the $\text{CrossP()}$ function with the arguments as $\text{CrossP( }\\!\\![\\!\\!\text{ 3,2,1 }\\!\\!]\\!\\!\text{ , }\\!\\![\\!\\!\text{ 1,4,0 }\\!\\!]\\!\\!\text{ )}$
Various other vector operations such as the dot product using the $\text{dot P()}$ function and it can also do multiplication of scalars with vectors.