## Force Vector Component?

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Boris Lagutin

### Force Vector Component?

In Supplementary Problems (MIT 18.02 online course: http://ocw.mit.edu/courses/mathematics/ ... upp_notes/) I faced problem like: Find the component of the force F = 3i-3j+2k in the direction of the vector 5i+4j-7k. Could someone explain what does "component of Force vector in the direction of another vector" mean?

So how I understood and solved it. I understood it as a projection of the Force vector (F) <3i-3j+2k> on another vector (V) <5i+4j-7k> and solved it so:

1) Find cosine of the angle between F and V by using dot product:

cos(?)=(<3,-3,2>*<5,4,-7>)/(|F| |V|) = (15-12-14)/(sqrt{22}*sqrt{90})~-0.24721 so angle = arccos(-0.24721) ~ 104.31

2) Find the projection (component) of the Force vector on the Vector: the component = |cos(104.31)*sqrt{22}| ~ |-1.16|

I guess my solution is wrong! The right solution would be: 1) unit vector = <5i, 4j, -7k>/|V| then 2) dot product F*u BUT WHY???
Boris Lagutin

### Re: Force Vector Component?

Ok. I understood the steps of how to find a scalar value of Force vector projection in the direction of another vector. However, I have not understood how the steps lead to calculating the projection scalar value. I know what dot product and unit vector are. I know that dot product of any two vectors equals "0" it means these vector are perpendicular to each other. But how dot product of a unit vector and Force vector may give a scalar value of Force vector projection, I don't understand!?

Could someone explain or give a reliable resource where there is a proof of this conception?

Thanks a lot.
joe
Posts: 755
Joined: Fri Aug 29, 2003 11:57 am
Location: Sydney

### Re: Force Vector Component?

Have a look at Physclips' introduction to vectors:
http://www.animations.physics.unsw.edu. ... ectors.htm
and especially the sections on unit vectors and scalar products.

The component of a in the direction of b is just a times the cos of the angle between a and b.
You know from the definition that a.b = (ab times the cos of that angle).
It follows that you can calculate the component of a in the direction of b as a.b^, where b^ is a unit vector in the direction of b. Now
b^ = b/b, where b = |b| so
The component of a in the direction of b = a.b/b

Joe

PS Good to hear from you again, I hope all is going well. Let me know how the MIT course goes.
Boris Lagutin

### Re: Force Vector Component?

Thank you very much, Joe.
Boris Lagutin

### Re: Force Vector Component?

After short preview of linear algebra coming back to multi-variable calculus. So there is a plane equation, for example, 3i+2j+5k = 8 and points A(-2,1,-3) and B(1,5,-4). We need to identify where the points are located on the plane, upper or lower with respect to the plane. We just put the points' coordinates into 3i+2j+5k=8 appropriately and check which values the equation gives. If the resulting value is more than 8 it means a point is above the plane otherwise a point is under. But my question is: as I know we use vectors components (not points' coordinates) to find whether a vector is on a plane so why it is available to put points' coordinates into a plane equation? As known a point is different from a vector.

Thanks.
Last edited by Boris Lagutin on Sun Jul 05, 2015 5:30 am, edited 2 times in total.
a.s.h.
Posts: 110
Joined: Tue Jan 28, 2014 3:50 pm

### Re: Force Vector Component?

The point (x,y,z) has position vector xi + yj + zk (so the coordinates of the point are essentially the same as the components of the position vector for that point, e.g. the vector 1i + 3j – 2k corresponds to the point (1,3,-2), where by 'corresponds', I mean that it is the vector that 'starts' at the origin (0,0,0) and 'ends' at the point (1,3,-2)), so points and vectors are quite 'similar' in this sense. Also, technically that plane equation should be 3x + 2y + 5z = 8, because we are saying the plane is the set of all points (x,y,z) (or vectors xi +yj +zk) such that 3x + 2y + 5z = 8. The 'equation' 3i+2j+5k = 8 is not mathematically correct because the LHS is a vector whereas the RHS is a scalar.

So substituting the coordinates (i.e. the components of the position vector of that point) of the point into the LHS of the equation is the right thing to do.
a.s.h.
Posts: 110
Joined: Tue Jan 28, 2014 3:50 pm

### Re: Force Vector Component?

Boris Lagutin wrote:We need to identify where the points are located on the plane, upper or lower with respect to the plane. We just put the points' coordinates into 3i+2j+5k=8 appropriately and check which values the equation gives. If the resulting value is more than 8 it means a point is above the plane otherwise a point is under.
In general, we should not do this as it will not always work. For example, consider the plane 2x + 4y – z = 6, and the point P(1,1,-1). If we substitute this point's coordinates into the LHS, we get 2 + 4 + 1 = 7 > RHS. However, the point (1,1,-1) is NOT above the plane! It as actually below, because the point on the plane with x = 1 and y = 1 is (1,1,0), and this point has a greater z-value than that of (1,1,-1), so is actually above the point P. (cont. in next post due to character limit)
a.s.h.
Posts: 110
Joined: Tue Jan 28, 2014 3:50 pm

### Re: Force Vector Component?

So in general, to determine whether a given point A is above or below a plane Π, we should find the z-value of the point on Π that has the same x- and y-values as that of A (by substituting these into the equation for Π and solving for z), and then compare this z-value to that of A.
Boris Lagutin

### Re: Force Vector Component?

Great Thanks a.s.h.. I meant vectors' components, of course. It's difficult to study by oneself without real-time instructors. Many small details questions appear while learning a new stuff and answers to them are critical often to understand it properly. In my case I need to think more by myself. Sometimes I wastes time.
Yes. I thought about a point is a vector from the origin but even after your nice explanation I have some obscuration. By the way, I understood your comment clearly but having something worries me. I think that it is because I yet don't feel a space coordinate system and its elements positioning well. I somehow cannot grasp all possible ways of how it may be. So my obscuration comes from here, I suppose.
Boris Lagutin

### Re: Force Vector Component?

In lecture 2 "Determinants" of MIT 18.02 course: http://ocw.mit.edu/courses/mathematics/ ... -lectures/ there are diagrams which I don't understand which vector the instructor means. In the second diagram vector A is located on a place in which a vector B is located in the first diagram. Moreover, he calculates a dot product A and B, however, there is no any vector B on the second diagram. May I mix up?
Attachments
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a.s.h.
Posts: 110
Joined: Tue Jan 28, 2014 3:50 pm

### Re: Force Vector Component?

I took a look at the lecture now, and I think what the lecturer did in the diagrams is as follows.

The aim was to compute the area of the parallelogram spanned by two arbitrary vectors A = <a1, a2> and B = <b1, b2>. To do this, we needed to find |A||B|sin(θ), as the lecture explained.The purpose of the first diagram is simply to show that we will find this by considering the dot product of A' and B, where A' is a 90° counter-clockwise rotation of A (in other words, a diagram of the problem, always a good idea for geometry problems).

The second diagram's purpose is to illustrate how to find the components of A' in terms of the components of A (which we need to do to compute A'B). The dot product being taken is with the B from the first diagram, since the two diagrams are used for illustrating parts of the same problem.
Boris Lagutin

### Re: Force Vector Component?

Thank you very much. Have a nice weekend
Boris Lagutin

### Re: Force Vector Component?

Another interesting point. In the lecture 7 "Exam Review" in MIT 18.02 course since 46 min. 00 sec.: http://ocw.mit.edu/courses/mathematics/ ... am-review/ the instructor tells that we just put a point's coordinates to the plane equation to find an equation of a plane for this point (2, 1, 0) and a given normal vector (as I understand it, hopefully right). However, you, a.s.h., wrote that in this case we assume that the point is a vector that comes from the origin if I understood right. Therefore, the plane equation must equal "0" if the plane comes through the origin. But x + y + 2z = 3? It seems that the vector comes through the origin but the plane which the vector is lying in does not come through the origin. How may it happen?

thanks.
Last edited by Boris Lagutin on Tue Jul 07, 2015 1:58 am, edited 1 time in total.
a.s.h.
Posts: 110
Joined: Tue Jan 28, 2014 3:50 pm

### Re: Force Vector Component?

I did not say (or at least mean) that a point IS a vector. What I meant is that the relationship between the POINT P(x,y,z) and the VECTOR p = <x,y,z> = xi + yj + zk is that the vector p can be obtained by joining considering the vector connecting the origin (the point O(0,0,0)) to the point P.

Also, I didn't emphasise this before, but a vector is actually independent of where it is in space, all the matters is its length and direction. So say in 2D (i.e. ℝ^2), the vector connecting the point (0,1) to (3,0) (which is found by subtracting coordinates of (0,1) from (3,0), and is hence <3,-1>) is the same as the vector connecting the point (0,0) to (3,-1). This is because both of these vectors actually have the same length and same direction (they are parallel and have same orientation – you can check this by drawing a diagram). And in terms of components, both vectors are <3,-1>, i.e. same vector.

(cont. in next post due to character limit.)
Last edited by a.s.h. on Tue Jul 07, 2015 1:43 am, edited 1 time in total.
a.s.h.
Posts: 110
Joined: Tue Jan 28, 2014 3:50 pm

### Re: Force Vector Component?

So when we are given a normal vector to a plane, we can actually think of moving it around anywhere in space, as long as we keep its direction pointing the same way, as this will mean the vector is still the same. So think of being given the normal vector n as meaning that if we moved it from the origin and instead placed it on the plane (keeping the same direction), then this vector would be perpendicular to plane.