Science:Math Exam Resources/Courses/MATH152/April 2010/Question A 02

MATH152 April 2010

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Question A 02
For questions A1-A3 below, $\mathbf {x}$ and $\mathbf {y}$ are the given vectors below. $\mathbf {x} =[1,-1,-1],\quad \mathbf {y} =[1,1,2],$ A2: Compute $\mathbf {x} \times \mathbf {y}$.

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.

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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, ${\begin{bmatrix}{\hat {\mathbf {i} }}&{\hat {\mathbf {j} }}&{\hat {\mathbf {k} }}\\1&-1&-1\\1&1&2\\\end{bmatrix}}.$ 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 ${\begin{aligned}{\begin{vmatrix}{\hat {\mathbf {i} }}&{\hat {\mathbf {j} }}&{\hat {\mathbf {k} }}\\1&-1&-1\\1&1&2\\\end{vmatrix}}=-1{\hat {\mathbf {i} }}-3{\hat {\mathbf {j} }}+2{\hat {\mathbf {k} }}=[-1,-3,2]\end{aligned}}$ which is precisely the definition of the cross product. Therefore $\mathbf {x} \times \mathbf {y} =[-1,-3,2]$.

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