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- Thread starter Ali Asadullah
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tiny-tim

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Can someone please give me an example.

Hi Ali!

A typical bound vector is a

the

(Actually, I notice that http://en.wikipedia.org/wiki/Free_vector" [Broken] uses a different definition of bound vector … it requires the

Also if we are given two bound vectors, is it possible to find the dot and cross product of two vectors.

Dot product and cross product of bound vectors work in exactly the same way as for free vectors (in the cross product case, one assumes that all the bound vectors have the same initial point).

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Hello Ali,

You have posted this in the physics, not maths sections so I will try to give you a physics answer.

Before we can have any vectors we must have a coordinate system and an origin.

You have probably been told that vectors are 'something with magnitude and direction'.

This is true of all types of vectors, but some types require further information.

Vectors that are not localised in space are called free vectors. This type only requires a magnitude and direction to fully specify them. Velocity is a good example of a free vector. A translation is another. A mechanical couple is a third.

Some vector types are localised in space. These are called localised or bound vectors. These vectors start or finish at a particular point in our coordinate system. To fully specify a bound vector we need to specify this point as well as a magnitude and direction. (We only need specify one point the other is given by the magnitude and direction.) Good examples of bound vectors are the displacement vector, the position vector and the moment of a force.

My three examples for free and bound vectors roughly correspond.

Notice I have not mentioned the Force vector. This is because it can take on either guise depending upon the application - you have to recognise which is involved.

As regards to calculations, you can freely mix n match localised and non localised vectors, adding, subtracting, forming dot and cross products. This is because the free vector will always include the point of application of the bound vector in its domain.

To illustrate consider the ship in the sketches.

Fig 1

Shows a ship drifting freely in the tide. It picks up the velocity of the tide, whcih is a free vector.

Fig2

The ship is now under power and has a heading velocity. This is also a free vector, as is the resultant velocity shown by the vector triangle.

Fig3

Shows what happens as the ship drops anchor at A. It no longer possesses any velocity, but it is subject to a force from the tide. This force has a moment about the anchor point and cause the ship to swing round to Fig 4.

Fig4

The ship is now at equilibrium in between the bound force of the tension in the anchor chain and the tidal force as before.

and yes, Wiki is correct.

You have posted this in the physics, not maths sections so I will try to give you a physics answer.

Before we can have any vectors we must have a coordinate system and an origin.

You have probably been told that vectors are 'something with magnitude and direction'.

This is true of all types of vectors, but some types require further information.

Vectors that are not localised in space are called free vectors. This type only requires a magnitude and direction to fully specify them. Velocity is a good example of a free vector. A translation is another. A mechanical couple is a third.

Some vector types are localised in space. These are called localised or bound vectors. These vectors start or finish at a particular point in our coordinate system. To fully specify a bound vector we need to specify this point as well as a magnitude and direction. (We only need specify one point the other is given by the magnitude and direction.) Good examples of bound vectors are the displacement vector, the position vector and the moment of a force.

My three examples for free and bound vectors roughly correspond.

Notice I have not mentioned the Force vector. This is because it can take on either guise depending upon the application - you have to recognise which is involved.

As regards to calculations, you can freely mix n match localised and non localised vectors, adding, subtracting, forming dot and cross products. This is because the free vector will always include the point of application of the bound vector in its domain.

To illustrate consider the ship in the sketches.

Fig 1

Shows a ship drifting freely in the tide. It picks up the velocity of the tide, whcih is a free vector.

Fig2

The ship is now under power and has a heading velocity. This is also a free vector, as is the resultant velocity shown by the vector triangle.

Fig3

Shows what happens as the ship drops anchor at A. It no longer possesses any velocity, but it is subject to a force from the tide. This force has a moment about the anchor point and cause the ship to swing round to Fig 4.

Fig4

The ship is now at equilibrium in between the bound force of the tension in the anchor chain and the tidal force as before.

and yes, Wiki is correct.

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This product yields the vector parallel to the line which is perpendicular to both the original skew lines.

I don't know if you have done any vector geometry, this is more advanced.

The vector equation of a line through a point a [tex]({a_1},{a_2},{a_3})[/tex] in a specific direction, ie parallel to a specific vector

or

(x,y,z) = [tex]({a_1},{a_2},{a_3})[/tex] + t[tex]({b_1},{b_2},{b_3})[/tex]

Let the two lines be

The vector perpendicular to both lines is

[tex]n\quad = \quad b \otimes b'[/tex]

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Hijacking someone else's thread with your own question is not good etiquette. You also get better results starting your own thread.

However your question does progress this thread so here is an answer.

You can only add (calculate the combined action of) bound vectors with different points of application by their action on something. When you do this you will not only obtain the vector sum by sliding one to form a triangle as you described, you will also find their vector product produces an effect.

For example consider two bound forces acting at different points on a body (rigid or deformable) . The body will translate according to the vector resultant of these forces. It will also rotate according to the resultant moment from these same forces.

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Thanks for your advice and help.

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