# Math Help - Cross Product in Linear Algebra!

1. ## Cross Product in Linear Algebra!

Find the solutions to v in:

(-i + 2j + 3k) X v = i + 5j - 3k

I have tried this question multiple times using the method of solving three equations in 3 unknowns but keep going around in circles!

Any help would be very much appreciated!

2. Originally Posted by caelestis
Find the solutions to v in:

(-i + 2j + 3k) X v = i + 5j - 3k

I have tried this question multiple times using the method of solving three equations in 3 unknowns but keep going around in circles!

Any help would be very much appreciated!
Let $v = v_1 i + v_2 j + v_3 k$.

Do you get the following system of linear equations:

$-3v_2 + 2 v_3 = 1$ .... (1)

$3v_1 + v_3 = 5$ .... (2)

$-2v_1 - v_2 = -3$ .... (3)

Note: There are an infinite number of solutions since (1) - 2 (2) = (3).

3. Originally Posted by mr fantastic
Let $v = v_1 i + v_2 j + v_3 k$.

Do you get the following system of equations:

$-3v_2 + 2 v_3 = 1$ .... (1)

$3v_1 + v_3 = 5$ .... (2)

$-2v_1 - v_2 = -3$ .... (3)

Yes I have the exact same system of equations as you but after a few steps... two of the equations cancel each other out!

Do you know why this happens or what it means??

4. Originally Posted by caelestis
Yes I have the exact same system of equations as you but after a few steps... two of the equations cancel each other out!

Do you know why this happens or what it means??
See the note at the end of my edited first post.

The general solution is found by letting one of the unknowns equal a parameter. Eg. Let $v_3 = t$ where t is any real number. Then:

$v_1 = \frac{5 - t}{3}$

$v_2 = \frac{2t-1}{3}$

$v_3 = t$.

If all you want is just one concrete solution, substitute a convenient value for t. t = 0 is pretty convenient ......

5. Hello
Originally Posted by caelestis
Do you know why this happens or what it means??
Given two vectors $\vec{a}$ and $\vec{c}$, if a vector $\vec{b}$ is a solution of $\vec{a}\times\vec{v}=\vec{c}\,\,\,\,\,\,(1)$ then for all $t\in\mathbb{R}$ one has $\vec{a}\times(\vec{b}+t\vec{a})=\vec{a}\times \vec{b}+\vec{0}=\vec{c}$ : every vector $\vec{b}+t\vec{a}$ is also a solution of (1). That's why if there is one solution then there is an infinite number of solutions.

The initial question can also be answered using another method : Given two vectors $\vec{a}$ and $\vec{c}$, (both different from $\vec{0}$) we want to solve $\vec{a}\times \vec{v}=\vec{c}\,\,\,\,\,(1)$ for $\vec{v}$.

_ If $\vec{a}$ and $\vec{c}$ aren't orthogonal, this equation has no solution.

_ If $\vec{a}\perp \vec{c}$ then $\vec{v}$ has to be orthogonal to $\vec{c}$ too. As $\vec{a}$ and $\vec{a}\times \vec{c}$ aren't collinear and are both orthogonal to $\vec{c}$, the solutions of (1) can be written $\vec{v}=\alpha \vec{a}+\beta \vec{a}\times \vec{c}$ for $\alpha,\beta \in\mathbb{R}$. If we put this in (1), we get

\begin{aligned}\vec{a}\times(\alpha \vec{a}+\beta \vec{a}\times \vec{c})=\vec{c}& \Longleftrightarrow \beta \vec{a}\times (\vec{a}\times \vec{c})=\vec{c}\\
& \Longleftrightarrow -\beta \vec{a}^2\vec{c}=\vec{c}\,\,\,\,\,(\text{triple product expansion})\\
& \Longleftrightarrow \beta=-\frac{1}{\vec{a}^2}\end{aligned}

There is no condition on $\alpha$ and the value of $\beta$ is $-\frac{1}{a^2}$. In your case, $\vec{a}=(-1,2,3)$ and $\vec{c}=1,5,-3)$ which gives $\vec{a}^2=1+2^2+3^2=14$ hence the solutions are

$
\vec{v}=\alpha\begin{pmatrix}-1\\2\\3\end{pmatrix}+\frac{1}{14}\begin{pmatrix}-1\\2\\3\end{pmatrix}\times\begin{pmatrix}1\\5\\-3\end{pmatrix}=\alpha\begin{pmatrix}-1\\2\\3\end{pmatrix}+\begin{pmatrix}3/2\\0\\1/2\end{pmatrix}$

which happens to be Mr. Fantastic's solution. $\left(\alpha=\frac{t}{3}-\frac{1}{6}\right)$