Cross with the scalar product

Steve writes:

The scalar product between two vectors is the sum of the product of the components. Let

$$\mathbf{u}=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$$

Since the components of $\bf v$ are all non-zero I can choose anything for $y$ and $z$ so long as I choose $x= -(2y+3z)$. This is the equation of a plane $P: x+2y+3z=0$, which is perpendicular to the vector ${\bf v}$. This makes sense: the scalar product of two vectors is zero if and only if they are perpendicular to each other.

Next, taking a cross product between two vectors always gives a vector which is perpendicular to the original two vectors. So, I should be able to make any vector perpendicular to ${\bf v}$ by taking cross products. These are all vectors lying in the plane $P$.

To decide which vectors ${\bf u}$ solve ${\bf u}\cdot {\bf v}=0$ for a given vector ${\bf v}$ I need to decide which vectors are perpendicular to ${\bf v}$.

To do this, I could take any linear combination of ${\bf i \times v}, {\bf j \times v}$ and ${\bf j \times v}$

$$\mathbf{w}\times \mathbf{i}=\left(\begin{array}{c}4\\ 5\\ 6\end{array}\right)\times \left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 4& 5& 6\\ 1& 0& 0\end{array}\right|=\left(\begin{array}{c}0\\ 6\\ -5\end{array}\right)$$

etc.

From this it seems that I just need to switch 2 non-zero components and change the sign of one of them to make a vector perpendicular to a given vector. In hindsight, this is obvious!

Doug writes:

Solution 2:

Dot $$$\left(\begin{array}{c}1\\ 2\\ 3\end{array}\right)$$$ with a general vector $$$\left(\begin{array}{c}a\\ b\\ c\end{array}\right)$$$ and you want to find a,b,c such that a + 2b + 3c = 0, so for example a = -2b - 3c, so there are an infinite number of solutions. Pick any 2 and find the other one, for example c=1,b=1,a=-5.

For the 2nd part, again cross $\mathbf{v}$, using a general vector, and find with the "determinant method":

$$\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 1& 2& 3\\ a& b& c\end{array}\right|=\left(\begin{array}{c}2c-3b\\ 3a-c\\ b-2a\end{array}\right),$$ so again you can choose a,b,c such that this is satisfied, and there are an infinite number of possibilities.

To quickly create a vector $\mathbf{u}$ such that $\mathbf{u}\cdot\mathbf{v} = \mathbf{0}$, I first looked at the general case with vectors $$$\left(\begin{array}{c}a\\ b\\ c\end{array}\right)$$$ and $$$\left(\begin{array}{c}d\\ e\\ f\end{array}\right)$$$ so ad+be+cf = 0, so I could choose any 2 numbers e & f, then calculate $d = \frac{-(be+cf)}{a}$. But we want to do this "quickly", so we might simplify $e=f=-1$, which gives $$\mathbf{u}=\left(\begin{array}{c}x\\ -1\\ -1\end{array}\right),$$ where $x = \frac{b+c}{a}$.

However, when I tried this solution on the examples given, I found that with the final example, there was a divide by zero error.

To get round this, we instead add a caveat to the algorithm to swap a,b,c to ensure that there is a non-zero denominator. So in the final case, we set $y = \frac{a+c}{b} = 0 \Rightarrow $$\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)$$\cdot$$\left(\begin{array}{c}-1\\ 0\\ -1\end{array}\right)$$ = 0$.

In general, the cross product of two vectors is: $$\left(\begin{array}{c}a\\ b\\ c\end{array}\right)\times \left(\begin{array}{c}d\\ e\\ f\end{array}\right)=\left(\begin{array}{c}bf-ce\\ cd-af\\ ae-bd\end{array}\right)$$ so if we let f=e=0, and d=1, then $\mathbf{w} = $$\left(\begin{array}{c}0\\ c\\ -b\end{array}\right)$$$ is always the result of crossing $\mathbf{v}$ with $$$\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)$$$, for example, which is probably the most efficient method of calculating this. 

 

Solution 1:

Under development

${\bf u}\cdot {\bf v}=\pmatrix {u_1\cr u_2\cr u_3}\cdot\pmatrix{1\cr 2\cr 3}$= $u_1 + 2u_2 + 3u_3$ =0

We therefore have 1 independent equation and 3 unknows which results in an infinite number of solutions.

One vector which satisifies this relationship is ${\bf u} = \pmatrix {0\cr0\cr0}$

If we take a general vector ${\bf z} = \pmatrix {z_1\cr z_2\cr z_3}$ then:

$$\mathbf{v}\times \mathbf{z}=\left(\begin{array}{c}1\\ 2\\ 3\end{array}\right)\times \left(\begin{array}{c}{z}_1\\ z_2\\ z_3\end{array}\right)=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 1& 2& 3\\ z_1& z_2& z_3\end{array}\right|=\left(\begin{array}{c}2z_3-3z_2\\ 3z_1-z_3\\ z_2-2z_1\end{array}\right)=\left(\begin{array}{c}{w}_1\\ w_2\\ w_3\end{array}\right)$$

This gives us a relationship between the components of ${\bf z}$ and ${\bf w}$ in the form of 3 simultaneous equations. 

Solving equation 1 and equation 3 simultaneously:

from equation3 we find that $z_2 = w_3 +2z_2$

if we now sustitute $z_2$ into equation1 we find that $-6z_1 + 2z_3 = 3w_3 + w_1$

This result is actually equation2 but scaled which implies that there are actually only two independent equations but still 3 unknows $( z_1, z_2, z_3)$. We are therefore unable to find a unique solution, there are once again an infinite number of solutions.

An alternative method of deducing this result would be to rewrite the equations in matrix form, a 0 determinant to this matrix would indicate that there are an infinite number of solutions.

In matrix form:

$$\left[\begin{array}{ccc}0& -3& 2\\ 3& 0& -1\\ -2& 1& 0\end{array}\right]$$

If we now take the determinant by expanding by the top row.

Determinant = 0 - -3( 0 - (-1)(-2) ) + 2( (3)(1) - 0) = -6 +6 = 0

Note: The above matrix is anti-symmetric, an anti-symmetric matrix of this form always results from taking the cross product of two vectors, the determinant of an odd order (in this case 3 x 3) anti-symmetric matrx will always give a zero determinant.

Define any vector ${\bf V}$= $\pmatrix {v_1\cr v_2\cr v_3}$

${\bf v}\cdot {\bf u}=\pmatrix{u_1\cr u_2\cr u_3}\cdot\pmatrix{v_1\cr v_2\cr v_3}$= $u_1v_1 + u_2v_2 + u_3v_3$ =0

If ${\bf V}\cdot {\bf U}$ = 0 then the two vectors are orthogonal, to determine ${\bf u}$ we therefore need to find a vector which is perpendicular to ${\bf v}$

Define vector ${\bf z}$= $\pmatrix {z_1\cr z_2\cr z_3}$

The cross product of any 2 vectors will yield a vector perpendicular to both of the original vectors, the cross product of ${\bf v}$ and ${\bf z}$ will therefore give a vector which is perpendicular to ${\bf v}$ (and ${\bf z}$), hence the cross product gives a vector which satisfies the condition for ${\bf u}$.

${\bf V }x{\bf Z}$=$\pmatrix {v_1\cr v_2\cr v_3} x \pmatrix{z_1\cr z_2\cr z_3}$=$$\left[\begin{array}{ccc}i& j& k\\ v_1& v_2& v_3\\ z_1& z_2& z_3\end{array}\right]$$ = $\pmatrix {u_1\cr u_2\cr u_3}$

then

$ v_2z_3 - v_3z_2 = u_1$

$v_3z_1 - v_1z_3 = u_2$

$v_1z_2 - v_2z_1 =u_3$

${\bf u}= \pmatrix { v_2z_3 - v_3z_2\cr v_3z_1 - v_1z_3 \cr v_1z_2 - v_2z_1 }$

Note: ${\bf z}$ can be selected as any vector.

Example: Let${\bf z }= \pmatrix { 1\cr 5 \cr 11 }$ and ${\bf v}=\pmatrix { 1\cr 2 \cr 3 }$

Then from the above formula ${\bf u} = \pmatrix { 7\cr -8 \cr 3 }$

And ${\bf U}\cdot {\bf V} = 7 -16 + 9 = 0$

if we now need

${\bf V }x{\bf U }={\bf W}$

$ v_2u_3 - v_3u_2 = w_1$

$v_3u_1 - v_1u_3 = w_2$

$v_1u_2 - v_2u_1 =w_3$

if we let $u_1 = k$, then:

$u_2 = \frac{w_3 - Kv_2}{v_1}$

$u_3 = \frac{-w_2 + Kv_3}{v_1}$

so ${\bf U} =\pmatrix {K\cr \frac{w_3 - Kv_2}{v_1}\cr \frac{-w_2 + Kv_3}{v_1}}$ (Valid when $v_1$ is non zero)

similarly by letting $u_2$ and $u_3$ = K in turn we find 2 further possibilities

${\bf U} =\pmatrix {\frac{-w_3 +Kv_1}{v_2} \cr K\cr \frac{w_1 + Kv_3}{v_2}}$ (Valid when $v_2$ is non zero)

${\bf U} =\pmatrix {\frac{w_2 +Kv_1}{v_3} \cr \frac{-w_1 + Kv_2}{v_3}\cr K}$ (Valid when $v_3$ is non zero)