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Example 1 of Triple Product (Dot Product and Cross Product)
Example problem: Given three vectors: \(\mathop u\limits_{}^ \to = \left\langle {\begin{array}{*{20}{c}}1\\0\\1\end{array}} \right\rangle \), \(\mathop v\limits^ \to = \left\langle {\begin{array}{*{20}{c}}2\\1\\0\end{array}} \right\rangle \), and \(\mathop w\limits^ \to = \left\langle {\begin{array}{*{20}{c}}1\\1\\{ – 1}\end{array}} \right\rangle \) find the triple product, \(\mathop u\limits_{}^ \to \cdot \left( {\mathop v\limits^ \to \times \mathop w\limits^ \to } \right)\)
Find the cross product first.
To do this easily, write both vectors horizontally, and find the 2×2 determinant while covering your hand over the column you want to find.
Basically, you have two 1×3 vectors (row x column), the result will also be 1×3.
We’ll show this below. Note that “i”, “j” and “k” just represent the first, second, and third dimensions/components, i is NOT the imaginary number here!
\[\begin{array}{*{20}{c}}i&j&k\\2&1&0\\1&1&{ – 1}\end{array}\]
We want v x w, so we write v first and then w.
To get the first component, put your hand over the first column, then find the 2×2 determinant, which is just a product of a diagonal minus the product of the other diagonal.
The first component is (2)(0) – (1)(1) = -1
Now put your hand over the second column, and find the 2×2 determinant, BUT, you need to add a negative sign to your answer.
This is because of the cofactor matrix which you can write out, but basically the middle determinant needs to be negative. The second component is -[(2)(0) – (1)(1)] = 1
And the final component, put your hand to cover the third column and find the determinant, which is (2)(1) – (1)(2) = 0
Now just put the three components into a vector.
\[\left( {\mathop v\limits^ \to \times \mathop w\limits^ \to } \right) = \left\langle {\begin{array}{*{20}{c}}{ – 1}\\2\\1\end{array}} \right\rangle \]
Then we find the dot product. To find the dot product, multiply the components by each other and sum them up.
\[\mathop u\limits_{}^ \to \cdot \left( {\mathop v\limits^ \to \times \mathop w\limits^ \to } \right) = (1)( – 1) + (0)(2) + (1)(1)\]
\[\mathop u\limits_{}^ \to \cdot \left( {\mathop v\limits^ \to \times \mathop w\limits^ \to } \right) = 0\]
So the triple product is 0.
Double checking the triple product (cross product and dot product) using MATLAB
You can double check your results by using a program called MATLAB, which specializes in using arrays (vectors and matrices), use the following code script m-file and feel free to replace the numbers with the numbers for your vectors to find your triple product:
%bai-gaming.com/math-guides
u = [1;0;1]
v = [2;1;0]
w = [1;1;-1]
dot(u,(cross(v,w)))And you will get a result of 0 in MATLAB as well! Nice!
Click here to see another harder problem where we find the triple product!