I have read in a book about bound vectors that we can not move them i mean that they can not move parallel to any location. Can someone please give me an example. Also if we are given two bound vectors, is it possible to find the dot and cross product of two vectors.
Hi Ali! A typical bound vector is a force vector … the line of application of the force is an essential part of the force, since two equal forces along different parallel lines will have a different effect (when applied to a body which is allowed to rotate). (Actually, I notice that http://en.wikipedia.org/wiki/Free_vector uses a different definition of bound vector … it requires the initial point of a bound vector to be fixed in space, while so far as I know, in physics, it is only necessary for the line of a bound vector to be fixed in space.) 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).
Sir i am asking about a vector which can not be moved from its position i.e what is given on Wikipedia. Sir i think you have told me about a sliding vector which can be moved any where on its line of action. Thanks
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.
Thank you Studiot, one last question can we find the product of two bound vectors which don't have same initial point. What i think that for cross-product, vectors should have same initial point.(If we have free vectors, we can get same initial point by sliding a vector to a position parallel to itself) but now we have bound vectors which we can not move or slide. So we can not find the cross product of two bound vectors with different initial points. Am i right sir and this was the case of cross product, can we find the dot product of two bound vectors with different initial points.
Vectors directed along two skew lines can still form a vector cross product. Don't forget we have to work in 3D for this stuff. 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 b [tex]({b_1},{b_2},{b_3})[/tex] is r = a +tb where r is the position vector and t is a parameter (scalar) 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 r = a +tb and r = a'+sb' The vector perpendicular to both lines is [tex]n\quad = \quad b \otimes b'[/tex]
i recently watched these posts, but may i ask can we find the sum of two bound vectors which don't have same initial point. can we just treat them like free vectors, and move them freely paralell to find the sum of vector (to determine the direction and magnitude), while keeping the real position(ie, initial point of the resultant vector) unknown?
Hello hoshii and welcome, 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.