Science:Math Exam Resources/Courses/MATH152/April 2011/Question A 02

MATH152 April 2011

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Question A 02
Given the vectors ${\begin{aligned}\mathbf {x} &=[0,-3,2]\\\mathbf {y} &=[1,1,1]\\\end{aligned}}$ Compute the cross product $\mathbf {x} \times \mathbf {y}$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Think about how to write the cross product in terms of a determinant to be computed.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. If we think of denoting the unit vectors in the $x,y,z$ directions as ${\hat {\mathbf {i} }},{\hat {\mathbf {j} }},{\hat {\mathbf {k} }}$ respectively then we can turn our vectors $\mathbf {x}$ and $\mathbf {y}$ into the following matrix, ${\begin{bmatrix}{\hat {\mathbf {i} }}&{\hat {\mathbf {j} }}&{\hat {\mathbf {k} }}\\0&-3&2\\1&1&1\\\end{bmatrix}}.$ Firstly notice that we align the components of the vectors such that the $x$ components are in the column with the heading ${\hat {\mathbf {i} }}$, the $y$ components are in the column with the heading ${\hat {\mathbf {j} }}$, and the $z$ components are in the column with the heading ${\hat {\mathbf {k} }}$. Secondly, this may look odd as a matrix since one of the rows is a vector. This matrix is artificial but if we treat it as if it were a standard matrix and computed its determinant, we would get ${\begin{aligned}{\begin{vmatrix}{\hat {\mathbf {i} }}&{\hat {\mathbf {j} }}&{\hat {\mathbf {k} }}\\0&-3&2\\1&1&1\\\end{vmatrix}}=-5{\hat {\mathbf {i} }}+2{\hat {\mathbf {j} }}+3{\hat {\mathbf {k} }}=[-5,2,3]\end{aligned}}$ which is precisely the definition of the cross product. Therefore $\mathbf {x} \times \mathbf {y} =[-5,2,3]$.