The "scalar product of vectors" is a very important
There are many ways of defining the product of two vectors.
e.g. $(3i + 2j) \times (4i-6j)= (3 \times 4)i+(2\times-6)j$
Turns out only ome definitions to be meaning or fuseful, and this one isnt' not at all conventional
there are two methods of multiply vectors, this or the cross product, which is next year
It follows the scalar product of two vectors $a$, $b$ are defined as: $a\cdot{b}=|a||b|\cos \theta$
Where $\theta$ is the angle between vectors arranged tail-to-tail or nose-to-nose.
Scalar product is always $\vec{a} \cdot \vec{b}$, NOT $\vec{a}\times \vec{b}$!
Why? $a\times b$ for vectors is actually the cross product, not the scalar product!{b}
Also, $ab$ is not a way of saying the dot product. It could be either dot or cross product.
When you take the dot product of two vectors, you get a real number.
Also, important!
The angle between the two vectors is always 0, acute, right, obtuse or straight.
We can identify these cases based on the dot product
0: $\cos\theta= 1$ $\vec{a}\cdot\vec{b}=|a||b|$
Acute: $\cos\theta > 0$
Right angle: $\cos\theta=0$
Obtuse: $\cos\theta<0$
Straight: $\cos\theta=-1$
For the two vectors $ai+bj,ci+dj$ $(a\vec{i}+b\vec{j})\cdot(c\vec{i}+d\vec{j})=(a\vec{i})\cdot(c\vec{i})+(a\vec{i})\cdot(d\vec{j})+(b\vec{j})\cdot(c\vec{i})+(b\vec{j})\cdot(d\vec{j})$ $=ac(\vec{i}\cdot\vec{i})+ad(\vec{i}\cdot\vec{j})+bc(\vec{j}\cdot\vec{i}) +bd(\vec{j}\cdot\vec{j})$ $= ac+bd$, because dot product of i and j will be 0 (i,j vectors are perpendicular), and dot product of itself is magnitude squared, and magnitude of i,j is 1!
