Question

why there is cosine angle in dot product and sine angle in cross product

Answer 1

The primary purpose of "cross product" is to calculate areas. If you have two vectors, u⃗ and v⃗ , the area of the parallelogram having those two vectors as two sides is, of course, "base times height". The "base" is the length of one of the vector, |u⃗ |, say. To Find the "height", draw a line from the tip of v⃗ perpendicular to u⃗ . That gives a right triangle having |v⃗ | as hypotenuse. The "height" is the length of the "opposite side" which is given by |v⃗ |sin(θ) where θ is the angle between the vectors. That is, the area of the parallelogram formed by vectors u⃗ and v⃗ is |u⃗ ||v⃗ |sin(θ).

A primary use of the dot product, on the other hand, is to find the projection of one vector on the other. Again, draw a line from the tip of u⃗ perpedicular to v⃗ . The projection of u⃗ onto v⃗ is now the near side of the right triangle produced. Its length is |u|cos(θ). To find the actual vector projection, multiply that length by a unit vector in the direction of v⃗ which is v⃗ /|v⃗ . That is, the vector projection of u⃗ on v⃗ is given by |u⃗ |cos(θ)[v⃗ /|v⃗ |] which we can simplify by multiplying both numerator and denominator by |v⃗ |:
|u⃗ ||u⃗ |cos(θ)|v⃗ |2v⃗

The numerator of that fraction is u⃗ ⋅v⃗ .

Answer 2

This is for dot product
\[v_1=(x_1,y_1)\]\[v_2=(x_2,y_2)\]\[v_1*v_2=x_1x_2+y_1y_2\rightarrow v_1*v_2/(\left|v_1 \right| \left|v_2\right| )=\]\[(x_1/\left| v_1 \right| )(x_2/\left| v_2 \right| )+(y_1/\left| v_1 \right| )(y_2/\left| v_2 \right| )=\cos(\alpha_1)\cos(\alpha_2)+\sin(\alpha_1)\sin(\alpha_2)=\cos(\alpha_2-\alpha_1)\]
Then:

\[v_1*v_2=\left| v_1\right| \left| v_2 \right| \cos(\alpha_2-\alpha_1)\]

Same reasoning for v1 x v2

Answer 3

\[v_1 \times v_2=x_1y_2-y_1x_2\]\[(v_1 \times v_2)/(\left| v_1\right|\left| v_2\right|)=\cos(\alpha_1)\sin(\alpha_2)-\sin(\alpha_1)\cos(\alpha_2)=\sin(\alpha_2-\alpha_1)\]
then\[v_1 \times v_2 =\left| v_1\right| \left| v_2\right| \sin(\alpha_2-\alpha_1)\]