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I do not understand the properties of these cross product problems.

For #1: $(\vec{i} \times \vec{j} ) \times \vec{k} = \vec{k} \times \vec{k} = 0$. So I assume that any vector, like "$\vec{k}$" crossed with itself will equal 0?

Then I have no idea how to solve these problems:

Number 2: $\text{ }$$\vec{k}\times (\vec{i} - 2\vec{j})$

3: $\text{ }$$(\vec{j}-\vec{k}) \times (\vec{k} - \vec{j})$

4: $\text{ }$$(\vec{i} + \vec{j} ) \times (\vec{i} - \vec{j})$

Note: all the "$\times$"s are cross products.

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I know links aren't always the most helpful, but I would refer you to: en.wikipedia.org/wiki/Cross_product#Algebraic_properties I often find it helpful to have the list of properties as reference when learning how to deal with a new operation (like the cross product). –  Drew Christianson Sep 8 '12 at 20:42
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3 Answers

up vote 5 down vote accepted

If you have a vector $\vec{a}$, the cross product $\vec{a}\times\vec{a}$ is always zero. Actually, if $\vec{a}\times\vec{b}=0$, $\vec{a}$ and $\vec{b}$ are multiples of each other.

Note that for $k\times(i - 2j)$, we have $k\times i-2k\times j=j+2i$. This is one property: $\vec{a}\times(\vec{b}+\vec{c})=\vec{a}\times\vec{b}+\vec{a}\times\vec{c}$.

For the third, we have $(j-k)\times(k - j)=j\times(k - j)-k\times(k - j)$ and finally $j\times k - j\times j-k\times k +k\times j=i-0-0-i=0$

The fourth you can do the same way.

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Note that for the third, $(k-j)$ could also be seen as a multiple (-1) of $(j-k)$ giving the same answer. –  Mark Bennet Sep 8 '12 at 20:43
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True, much easier to think as this. –  Ataias Reis Sep 8 '12 at 20:44
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$\times:\mathbb{R}^3\times\mathbb{R}^3\mapsto\mathbb{R}^3$ is an antisymmetric bilinear form. This means that $$ \left(a\vec{x}+b\vec{y}\right)\times\vec{z}=a\left(\vec{x}\times\vec{z}\right)+b\left(\vec{y}\times\vec{z}\right)\tag{1} $$ and $$ \vec{y}\times\vec{x}=-\vec{x}\times\vec{y}\tag{2} $$ It is not associative; that is, one cannot claim that $\left(\vec{x}\times\vec{y}\right)\times\vec{z}=\vec{x}\times\left(\vec{y}\times\vec{z}\right)$.

An immediate consequence of $(2)$ is that $$ x\times x=0\tag{3} $$ because $x\times x=-x\times x$.

A consequence of $(1)$ and $(2)$ is that $$ \begin{align} \vec{z}\times\left(a\vec{x}+b\vec{y}\right) &=-\left(a\vec{x}+b\vec{y}\right)\times\vec{z}\\ &=-a\left(\vec{x}\times\vec{z}\right)-b\left(\vec{y}\times\vec{z}\right)\\ &=a\left(\vec{z}\times\vec{x}\right)+b\left(\vec{z}\times\vec{y}\right)\tag{4} \end{align} $$ On the canonical basis for $\mathbb{R}^3$, $\{i,j,k\}$, $\times$ is defined by $(1)$, $(2)$, $(3)$, and $$ \begin{align} i\times j&=k\\ j\times k&=i\\ k\times i&=j \end{align}\tag{5} $$ Hopefully, applying these properties to your problems should help.

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Hint: 1) use the fact that $$i\times j = k = - j \times i,$$ $$j\times k = i = - k \times j,$$ $$k\times i = j = - i \times k.$$ 2) use distributive law.

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