We are trying to find the cross product between \(\displaystyle u\) and \(\displaystyle v\).

Recall the formula for cross product.

If  \(\displaystyle u=(u_x,u_y,u_z)\), and \(\displaystyle v=(v_x,v_y,v_z)\), then

\(\displaystyle u\times v=(u_yv_z-u_zv_y, -u_xv_z+u_zv_x, u_xv_y-u_yv_x)\).

Now apply this to our situation.

\(\displaystyle u\times v=(4(10)-5(6),-2(10)+5(-2), 2(6)-4(-2))\)

\(\displaystyle u\times v=(40-30,-20-10, 12+8)\)

\(\displaystyle u\times v=(10,-30,20)\)

We are trying to find the cross product between \(\displaystyle u\) and \(\displaystyle v\).

Recall the formula for cross product.

If  \(\displaystyle u=(u_x,u_y,u_z)\), and \(\displaystyle v=(v_x,v_y,v_z)\), then

\(\displaystyle u\times v=(u_yv_z-u_zv_y, -u_xv_z+u_zv_x, u_xv_y-u_yv_x)\).

Now apply this to our situation.

\(\displaystyle u\times v=(0(1)-1(0),-1(1)+1(0), 1(0)-0(0))\)

\(\displaystyle u\times v=(0-0,-1+0, 0-0)\)

\(\displaystyle u\times v=(0,-1,0)\)

This is true. The cross product is defined this way. The dot product however can be taken for two vectors of dimension n (provided that both vectors are the same dimension).

By definition, the order of the dot product of two vectors does not matter, as the final output is a scalar.  However, the cross product of two vectors will change signs depending on the order that they are crossed.  Therefore 

\(\displaystyle \begin{matrix}\vec{A}\cdot \vec{B}=\vec{B}\cdot \vec{A}, & & \vec{A} \times \vec{B}= - \vec{B}\times \vec{A} \end{matrix}\).

We have the following equation that relates the cross product of two vectors \(\displaystyle \vec{A} \times \vec{B}\) to the relative angle between them \(\displaystyle \theta\), written as

\(\displaystyle \frac{\vec{A} \times \vec{B}}{\left| \vec{A} \right| \left| \vec{B} \right|}= \sin \theta\).

From this, we can see that the numerator, or cross product, will be \(\displaystyle 0\) whenever \(\displaystyle \sin \theta=0\).  This will be true for all even multiples of \(\displaystyle \frac{\pi}{2}\).  Therefore, we find that the cross product of two vectors will be \(\displaystyle 0\) for \(\displaystyle \theta = 0(0^{\circ}), \pi (180^{\circ}), 2\pi (360^{\circ})\).

It is not possible to take the cross product of \(\displaystyle 2\)-component vectors. The definition of the cross product states that the two vectors must each have \(\displaystyle 3\) components. So the above problem is impossible.

To evaluate the cross product, we use the determinant formula

\(\displaystyle \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1& a_2 & a_3 \\ b_1& b_2 & b_3 \end{vmatrix}\)

So we have

\(\displaystyle < 1,0,4> \times < 1,1,1>\)

\(\displaystyle = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1& 0 & 4 \\ 1& 1 & 1 \end{vmatrix}\)

\(\displaystyle = \begin{vmatrix} 0 & 4 \\ 1 & 1 \end{vmatrix} \mathbf{i} -\begin{vmatrix} 1 & 4 \\ 1& 1 \end{vmatrix} \mathbf{j} + \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} \mathbf{k}\). (Use cofactor expansion along the top row. This is typically done when taking any cross products)

\(\displaystyle = -4\mathbf{i}-(-3)\mathbf{j}+1\mathbf{k}\)

\(\displaystyle = < -4,3,1>\)

To evaluate the cross product, we use the determinant formula

\(\displaystyle \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1& a_2 & a_3 \\ b_1& b_2 & b_3 \end{vmatrix}\)

So we have

\(\displaystyle < 1,1,0> \times < 8,0,8>\)

\(\displaystyle = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1& 1 & 0 \\ 8& 0 & 8 \end{vmatrix}\)

\(\displaystyle = \begin{vmatrix} 1 & 0 \\ 0 & 8 \end{vmatrix} \mathbf{i} -\begin{vmatrix} 1 & 0 \\ 8& 8 \end{vmatrix} \mathbf{j} + \begin{vmatrix} 1 & 1 \\ 8 & 0 \end{vmatrix} \mathbf{k}\). (Use cofactor expansion along the top row. This is typically done when taking any cross products)

\(\displaystyle = 8\mathbf{i}-(8)\mathbf{j}+(-8)\mathbf{k}\)

\(\displaystyle = < 8,-8,-8>\)

To find the cross product, we solve for the determinant of the matrix

\(\displaystyle \begin{vmatrix} i&j &k \\ 4&3 &7 \\ 5&7 &10 \end{vmatrix}\)

The determinant equals

\(\displaystyle =det(\begin{vmatrix} 3&7 \\ 7&10 \end{vmatrix})i-det(\begin{vmatrix} 4&7 \\ 5&10 \end{vmatrix})j+det(\begin{vmatrix} 4&3 \\ 5&7 \end{vmatrix})k\)

\(\displaystyle =(3*10-7*7)i-(4*10-5*7)j+(4*7-5*3)k\)

\(\displaystyle =-19i-5j+13k\)

\(\displaystyle =< -19,-5,13>\)

As the cross-product.

To find the cross product, we solve for the determinant of the matrix

\(\displaystyle \begin{vmatrix} i&j &k \\ 1&0 &0 \\ 0&1 &0 \end{vmatrix}\)

The determinant equals

\(\displaystyle =det(\begin{vmatrix} 0&0 \\ 1&0 \end{vmatrix})i-det(\begin{vmatrix} 1&0 \\ 0&0 \end{vmatrix})j+det(\begin{vmatrix} 1&0 \\ 0&1 \end{vmatrix})k\)

\(\displaystyle =(0*0-1*0)i-(1*0-0*0)j+(1*1-0*0)k\)

\(\displaystyle =0i-0j+1k\)

\(\displaystyle =< 0,0,1>\)

As the cross-product.