# Can anyone tell me that actually What vector multiplication is? [duplicate]

I am a high school student. Yesterday I attended a lecture on vector multiplication (first time). There was something I was not able to understand. The only concept of multiplication I had before is 2x3=6, means that if we write '3' 2 times we'll get 6.

But now there are two types of multiplications:

• Dot product
• Cross product

## Dot product

What I understood is that dot product of two vectors is a scalar quantity, But how?

Here if we multiply projection of A on B (A cos0) with B, finally that product would be in the direction of B?, am I right or wrong?

Secondly please specify the difference between this dot product and ordinary multiplication which I stated above (2x3=6)?

## Cross product:

By Cross product I understand that it's the product of the 'length of perpendicular produced for projecting A on B' and 'B'?

What is the purpose and meaning of this product?

And After multiplication product is multiplied by unit vector n, which perpendicular to the A and B.

Why the cross product is always perpendicular to the given vectors?

And if I am right that Cross product is the product of the 'length of perpendicular produced for projecting A on B' and 'B' then why that projection is not shown in this diagram of cross product (which is taken from wikipedia)?

## marked as duplicate by sammy gerbil, heather, Qmechanic♦May 18 '17 at 1:08

• This is a question about mathematics, not physics. – sammy gerbil May 17 '17 at 22:26
• – caverac May 17 '17 at 22:26
• Geometric Algebra. It actually is a pretty good physics question, because Geometric Algebra maps really well to space. – Rob Jun 6 '18 at 23:38
• Notice that the cross product and dot product are related: $a b = cos(\theta) + i sin(\theta) = (a \cdot b) + (a \wedge b)$. That wedge product is a (simpler) generalization of cross product. $(a \wedge b) = -I (a \times b)$ – Rob Jun 7 '18 at 0:01

It's really nice to see someone asking things rather than accepting them without a deeper understanding!

Dot Product:

For me, when you are doing a dot product, you are thinking on how a vector influences in another vector. For example... When you are playing Mario Kart, you have some boosts in the track:

$$\vec{a} \text{ } \cdot \text{ } \vec{b} = |a||b|\cos(\theta)$$ Consider the $\vec{a}$ the red vector and $\vec{b}$ the blue vector. If the angle between them is $0$, you'll see that you are going to get the maximum boost possible, because $\cos(0)=1$. When you start increasing the angle between them, you can see that the dot product is going to approach zero (can you tell why?).

Thinking that way, you can see that the the scalar, that is the result of a dot product, is telling you how much one vector is influencing the other one.

One other thing to build your intuition is to think about how to the computation of a dot product is done:

All images I've got from betterexplained.com, I recommend this site for building up your intuition.

Here if we multiply projection of A on B (A cos0) with B, finally that product would be in the direction of B?, am I right or wrong?

You are almost right, if you multiply the MAGNITUDE of the projection, with $\vec{b}$ you are going to do a normal scalar-vector multiplication, hence, the result is going to be a multiple of $\vec{b}$ in direction of $\vec{b}$

Secondly please specify the difference between this dot product and ordinary multiplication which I stated above (2x3=6)?

Just check how we compute a dot product and you'll see that there is some kind of relation with the algebraic multiplication you've stated!

Cross Product:

The result of a cross product: $$\vec{a} \text{ } \times \text{ } \vec{b} = \vec{c}$$ is a vector $\vec{c}$ where this vector is perpendicular to $\vec{a}$ and $\vec{b}$.

When we are looking for computing the MAGNITUDE of the vector in a cross product, you are going to get: $$|\vec{a} \text{ } \times \text{ } \vec{b}| = |a||b|\sin(\theta)$$ and by analysing that equation you can see that the area of the paralellogram made by $\vec{a}$ and $\vec{b}$ is going to be the magnitude of $\vec{c}$. (can you tell why?)

What is the purpose and meaning of this product?

Everytime in a problem when you just want the perpendicular part of a vector in a calculation you'll deal with the cross product. A lot of definitions in physics use cross product... Here's an example: $$\vec{f} = q(\vec{v} \text{ } \times \text{ }\vec{b})$$ That's an equation in electromagnetism that defines what is going to be the force $\vec{f}$ acting on a charge $q$ that is with the velocity $\vec{v}$ in a magnetic field $\vec{b}$.

Why the cross product is always perpendicular to the given vectors?

The reason why the cross product is a vector perpendicular has to do with its computation, that deals with determinant.

I don't know if I should get into in explaining why its vector result is perpendicular because it's not that trivial since its computation deals with determinant, a thing in math that most people use but don't know what it is! If you really want to, comment this post that I'll try to explain!

When I said "can you tell why", I want to see if you are understanding what I'm trying to explain... If you can't figure it out, comment and I'll help you!

Hope it helps!

Keep studying and being curious, but some things are hard to prove without a deeper mathematical approach, and you'll need to handle them by intuition until you get some maturity in math (I'm not that mature in a lot of the situations! hahaha)

• Good answer, but I think torque would be a better example for the cross product, since it's a more everyday occurrence and the OP has probably not studied electromagnetism yet. You can then explain that rotations vectors are defined as parallel to the axis of rotation and thus perpendicular to the torquing force and radius, necessitating a cross product. – Mark H May 17 '17 at 23:21
• @MarkH You're right mate! Good point about using torques! Thank you for liking it, It's good to see someone with a deeper understanding linking what I've tried to explain! – Bruno Reis May 17 '17 at 23:27
• @BrunoReis >determinant, a thing in math that most people use but don't know what it is! I'll be very thankful if you can explain me What It Is? I've also posted a question related to this math.stackexchange.com/questions/160328/… – M.Ahmad Aug 27 '17 at 15:25

A Geometric View of Vector Products

Vectors have geometric properties independent of coordinate systems. The vector dot product $\mathbf A \cdot \mathbf B = AB \cos(\theta)$, projects the length of vector $\mathbf A$ onto the direction of vector $\mathbf B$; it is the shadow projection of the one vector onto the other. This operation is linear over vector addition, and the definition is symmetric, so the dot product is commutative, $\mathbf A \cdot \mathbf B =\mathbf B \cdot \mathbf A$. It is used to determine angles between vectors as well as lengths and distances. The condition $\mathbf A \bullet \mathbf B = 0$ is a test for orthogonality.

The geometrical meaning of the vector cross product $\mathbf A \times \mathbf B = AB \sin(\theta) \mathbf n$ is obtained by sliding vector $\mathbf A$ along the length of vector $\mathbf B$, always remaining in their joint plane,and with $\mathbf A$ remaining parallel to itself. This is done by hooking the right hand thumb about vector $\mathbf B$ as a guide, and then pushing with that hand to mark out the area of a parallelogram. The “right hand rule” orients $\mathbf n$ with your right thumb, making it normal to the plane of the parallelogram. So in addition to the determination of areas and angles, the creation of the unit vector ˆn determines the orientation of the plane formed by the two vectors. This operator is linear over vector addition, but note that the "right hand rule" reverses the orientation if the order of the vectors is reversed: $\mathbf B \times \mathbf A = -\mathbf A \times \mathbf B$. The test $\mathbf A \times \mathbf B = 0$ is a test for parallelism.

Note that ordinary multiplication also has a geometric meaning: 2 x 3 is the area of a rectangle 2 wide by 3 long; rotation of the rectangle by 90 degrees shows that the same area is obtained with 3 x 2.