Question

Can someone please explain cross product and dot product rule to me?
@Michele_Liano

Answer 1

@Michele_Laino

Answer 2

dot product, can also called the \(external\) product, since to each pair of vectors, corresponds a real number, namely an object of different type

whereas, the cross product can be called an \(internal\) product, since to each pair of vectors corresponds another vector, namely an object of the same type

Answer 3

for example, if we consider the space of geometric vectors, namely each vector is defined by its component:
\[{\mathbf{v}} = \left( {x,y,z} \right)\]
we can write these formulas

Answer 4

Isnt this related to "determinant"?

Answer 5

yes! of course:
here is the scalar product:
\[{{\mathbf{v}}_1} \cdot {{\mathbf{v}}_2} = {x_1}{x_2} + {y_1}{y_2} + {z_1}{z_2}\]

Answer 6

wherein:
\[\begin{gathered}
{{\mathbf{v}}_1} = \left( {{x_1},{y_1},{z_1}} \right) \hfill \\
{{\mathbf{v}}_2} = \left( {{x_2},{y_2},{z_2}} \right) \hfill \\
\end{gathered} \]

Answer 7

vector of A dot vector of be is abcostheta,right?

Answer 8

b* not be.

Answer 9

yes! provided that \(\theta\) is the angle between A and B

Answer 10

then
two perpendicular lines will be zero,won't they?

Answer 11

\[î + j = 0\]
If there perpendicular?

Answer 12

Because cos 90 is 0.

Answer 13

yes! since \(\cos 90=0\)

Answer 14

or \(\cos (\pi/2)=0\)

Answer 15

now I write the cross product formula, please wait a moment

Answer 16

okay!

Answer 17

here is the cross product, please note that the result is another vector:
\[\begin{gathered}
{{\mathbf{v}}_1} \times {{\mathbf{v}}_2} = \det \left( {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
{{x_1}}&{{y_1}}&{{z_1}} \\
{{x_2}}&{{y_2}}&{{z_2}}
\end{array}} \right) = \hfill \\
\hfill \\
= {\mathbf{i}}\left( {{y_1}{z_2} - {y_2}{z_1}} \right) + {\mathbf{j}}\left( {{z_1}{x_2} - {z_2}{x_1}} \right) + {\mathbf{k}}\left( {{x_1}{y_2} - {x_2}{y_1}} \right) \hfill \\
\end{gathered} \]

Answer 18

wherein, the vectors \[{\mathbf{i}},{\mathbf{j}},{\mathbf{k}}\]
are unit vectors along the axes x, y, and z respectively

Answer 19

shouldn't it be -j(z_1x_2 -z_2x_1)

Answer 20

the vector \(v_1 \times v_2\) is a vector orthogonal to the plane defined by the vectors \(v_1\) and \(v_2\)

Answer 21

it should be:
\[ - {\mathbf{j}}\left( { - {z_1}{x_2} + {z_2}{x_1}} \right)\]

Answer 22

Then are my notes wrong? :/

Answer 23

Ops.sorry I misread the post.
There're correct.

Answer 24

furthermore, we can write:

\[\left| {{{\mathbf{v}}_1} \times {{\mathbf{v}}_2}} \right| = \left| {{{\mathbf{v}}_1}} \right|\;\left| {{{\mathbf{v}}_2}} \right|\;\sin \alpha ,\quad 0 \leqslant \alpha \leqslant 180\]