Force Vector Component?
Moderator: msmod
Re: Force Vector Component?
I am sorry that I understood you wrong. Ok. We have just stepped closer to what is interesting. Well, the vector components may be positioned anywhere in space but the vector direction stay the same. Another question comes from here. Really we can take any plane formula and any vector (doesn't matter which direction) and find that plane equation. Please, notice that a normal vector in this plane equation remains the same. To give us a valid plane equation a vector that we take to insert in should be perpendicular to this normal vector. Also we know that RHS number shows us that how far the plane is from the origin (it doesn't tell us its exact direction). continue on the next post...
Last edited by Boris Lagutin on Tue Jul 07, 2015 10:18 am, edited 4 times in total.
Re: Force Vector Component?
In this way, by inserting any vector (its components) into the plane equation we will get some number on RHS BUT this vector may be not perpendicular the normal vector contained in this plane equation. In lecture 7 (above) the instructor just take a point (supposed any point) assuming it as a vector coming from the origin (if I understand right) and insert it into the equation then get RHS number which says that the plane is above the origin. BUT how we may be sure that this vector is perpendicular to the plane normal vector in this case? Any vector (its components) inserted in the plane equation gives some number and this number (except "0") says us that the plane is above or lower than the origin is!
thanks.
thanks.
Last edited by Boris Lagutin on Tue Jul 07, 2015 4:35 am, edited 4 times in total.
Re: Force Vector Component?
Let's see the following situation (drawing below) when vector OS is not parallel to jaxis but vector A is parallel to jaxis and is in the plane. So we have the plane equation ax+by+cz = d, where "d" is a number devoting a distance from the origin to the plane. Now we insert x, y and z components of vector OS into the plane equation and, of course, we get some number (except "0") but IT DOES NOT MEAN that A PLANE which VECTOR "OS" IS IN is THE SAME PLANE which VECTOR A IS IN just SHIFTED by some number on RHS! In this case I don't understand why the instructor easily insert a point (assuming it as a vector) in the plane equation in the lecture 7 so it may be valid only if the vector is perpendicular to a normal vector contained in this plane equation?!
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Last edited by Boris Lagutin on Tue Jul 07, 2015 5:02 am, edited 6 times in total.
Re: Force Vector Component?
The problem #5 which the instructor talks about is here: http://ocw.mit.edu/courses/mathematics/ ... prac1a.pdf
The point P_o(2,1,0) is in the plane formed by vectors P_oP_1 and P_oP_2 , however, it doesn't mean that a vector which comes from the origin to this point is parallel to the plane!?
P.S. if we insert components of some vector in a plane equation and get some number, except "0", on RHS we cannot be sure that this vector is perpendicular to a normal vector contained in this plane equation!
thanks a lot.
The point P_o(2,1,0) is in the plane formed by vectors P_oP_1 and P_oP_2 , however, it doesn't mean that a vector which comes from the origin to this point is parallel to the plane!?
P.S. if we insert components of some vector in a plane equation and get some number, except "0", on RHS we cannot be sure that this vector is perpendicular to a normal vector contained in this plane equation!
thanks a lot.
Re: Force Vector Component?
To sum up, I should say how I understand a plane equation: ax+by+cz = d. It means that any vector the components x, y, z of which we insert into the equation must be perpendicular to the normal vector <a,b,c> contained in this equation despite a value of "d". Of course, if we get "0" on RHS we are sure that a vector which we insert is perpendicular to the normal vector. But if we get another number we CANNOT say that we just have the same plane just shifted up or down with respect to the origin. Moreover, if the inserted vector is not perpendicular the plane normal vector the plane equation is violated and it is INVALID in this case!
Thanks a lot.
Thanks a lot.
Re: Force Vector Component?
Let P be a point in the plane in 3D space. Let p be the position vector of this point (i.e. the vector connecting the origin O to P).Boris Lagutin wrote:[...] inserting any vector [...] into the plane equation we will get some number on RHS BUT this vector may be not perpendicular the normal vector [...] In lecture 7 (above) the instructor just take a point (supposed any point) assuming it as a vector coming from the origin (if I understand right) and insert it into the equation then get RHS number [...]. BUT how we may be sure that this vector is perpendicular to the plane normal vector in this case?
Then when we say the vector n is normal to the plane, we don't mean that n will be perpendicular to the vector p. What we mean is that n will be perpendicular to any vector connecting two points in the plane.
e.g. consider the plane x + y + z = 1. Two points on this plane are A = (1,0,0) and B = (0,1,0).
(cont.)
Re: Force Vector Component?
These points A and B have position vectors a = <1,0,0> and b = <0,1,0>, and a normal to the plane is n = <1,1,1>. The points A and B lie on the plane, so the vector connecting A to B, AB (= b – a = <1,1,0>) is perpendicular to n (you can check this using dot products if you have learnt this). This is because the definition of a line ℓ being perpendicular to a plane is that any line segment that lies in the plane is perpendicular to ℓ. There is no requirement that a or b must be perpendicular to n. In fact, this would be the case if and only if the RHS of the plane equation were 0.
Re: Force Vector Component?
Correct; a vector from O to a point in the plane does not have to be parallel to the plane. In fact, this would be the case if and only if the plane passes through the origin. However, any vector connecting two points in the plane is going to be parallel to the plane.Boris Lagutin wrote:The point P_o(2,1,0) is in the plane formed by vectors P_oP_1 and P_oP_2 , however, it doesn't mean that a vector which comes from the origin to this point is parallel to the plane!?
Re: Force Vector Component?
That's right. This vector would be perpendicular to the normal vector if and only if the RHS is 0. This is because the equation of a plane in terms of dot products can be written in this form: n•x = n•a, where n is a normal vector, a is the position vector of a point given to be in the plane, and x = <x,y,z> is the position vector of an arbitrary point in ℝ^3 (so when we say n•x = n•a is the equation of the plane, we mean that the plane is the set of all points whose position vectors satisfy that equation).Boris Lagutin wrote:P.S. if we insert components of some vector in a plane equation and get some number, except "0", on RHS we cannot be sure that this vector is perpendicular to a normal vector contained in this plane equation!
(cont.)
Re: Force Vector Component?
Now if the RHS is 0, then we have n•x = 0, which means any point in the plane must satisfy n•x = 0, which is the same as saying that its vector must be perpendicular to n.
If the RHS is nonzero, we have n•x = nonzero number, which tells us any point with position vector x in the plane is not perpendicular to n, since their dot product is nonzero.
Also, note that the RHS d of the plane equation ax + by + cz = d is 0 if and only if the plane passes through the origin.
This means that the vector n is perpendicular to the position vector of any point in the plane if and only if the plane passes through the origin.
However, it is always true that the normal to a plane is perpendicular to a vector connecting any two points in the plane (because this is just the definition of a line being perpendicular to a plane).
If the RHS is nonzero, we have n•x = nonzero number, which tells us any point with position vector x in the plane is not perpendicular to n, since their dot product is nonzero.
Also, note that the RHS d of the plane equation ax + by + cz = d is 0 if and only if the plane passes through the origin.
This means that the vector n is perpendicular to the position vector of any point in the plane if and only if the plane passes through the origin.
However, it is always true that the normal to a plane is perpendicular to a vector connecting any two points in the plane (because this is just the definition of a line being perpendicular to a plane).
Re: Force Vector Component?
Thank you. Let me ask you a question. Why does the instructor take the position vector coming from the origin to the point P_o (2, 1,0) and insert its components in the plane equation whose normal vector is not perpendicular to this position vector in lecture 7 since 46 min. 00 sec.: http://ocw.mit.edu/courses/mathematics/ ... amreview/ ?? In my opinion it's evident that the normal vector <1,1,2> in the plane equation x+y+2z = (any number) is not perpendicular to the position vector <2,1,0> (doesn't matter which number is on RHS).
By doing this operation he gets a plane equation x+y+2z = 3 but I'm sorry the normal vector <1,1,2> is NOT perpendicular to the position vector <2,1,0>. Is such a plane equation is VALID (please, answer "yes" or "not" and if "yes" then why)?
Thanks a lot.
By doing this operation he gets a plane equation x+y+2z = 3 but I'm sorry the normal vector <1,1,2> is NOT perpendicular to the position vector <2,1,0>. Is such a plane equation is VALID (please, answer "yes" or "not" and if "yes" then why)?
Thanks a lot.
Re: Force Vector Component?
P.S. Where my doubts come from. They come from a statement that if a number on RHS is not "0" it means that a plane is shifted by this number with respect to the origin. I understand it in the following way. If we insert any vector components into a plane equation in which RHS is some number (nonzero) and we get a dotproduct equal to this number on RHS it means that this vector is in this plane and perpendicular to this plane equation normal vector. Maybe I mistake?
Re: Force Vector Component?
Answer: yes (explanation in next post)Boris Lagutin wrote:Thank you. Let me ask you a question. Why does the instructor take the position vector coming from the origin to the point P_o (2, 1,0) and insert its components in the plane equation whose normal vector is not perpendicular to this position vector in lecture 7 since 46 min. 00 sec.: http://ocw.mit.edu/courses/mathematics/ ... amreview/ ?? In my opinion it's evident that the normal vector <1,1,2> in the plane equation x+y+2z = (any number) is not perpendicular to the position vector <2,1,0> (doesn't matter which number is on RHS).
By doing this operation he gets a plane equation x+y+2z = 3 but I'm sorry the normal vector <1,1,2> is NOT perpendicular to the position vector <2,1,0>. Is such a plane equation is VALID (please, answer "yes" or "not" and if "yes" then why)?
Thanks a lot.
Re: Force Vector Component?
Explanation: as I said before, we DON'T require the normal vector to be orthogonal (orthogonal just means perpendicular) to the position vector of points on the plane. What we require is that the normal vector be orthogonal to any vector connecting two points in the plane. See my previous posts for more about this.
As an example, two points on the plane x + y + 2z = 3 are P(2,1,0) and Q(0,3,0). The vector connecting P to Q is PQ = q — p = <2, 2, 0>. THIS vector must be orthogonal to the normal vector, for the plane equation to be 'valid' as you say. And indeed it is, as you can check by computing the dot product n•PQ and showing that it equals 0. This relationship will hold for ANY other vector connecting two points in the plane.
As an example, two points on the plane x + y + 2z = 3 are P(2,1,0) and Q(0,3,0). The vector connecting P to Q is PQ = q — p = <2, 2, 0>. THIS vector must be orthogonal to the normal vector, for the plane equation to be 'valid' as you say. And indeed it is, as you can check by computing the dot product n•PQ and showing that it equals 0. This relationship will hold for ANY other vector connecting two points in the plane.
Last edited by a.s.h. on Tue Jul 07, 2015 11:41 pm, edited 1 time in total.
Re: Force Vector Component?
You are right that the dot product of the position vector with the normal equals the RHS (see my post about pointnormal form of a plane: n•x = n•a ).Boris Lagutin wrote:P.S. Where my doubts come from. They come from a statement that if a number on RHS is not "0" it means that a plane is shifted by this number with respect to the origin. I understand it in the following way. If we insert any vector components into a plane equation in which RHS is some number (nonzero) and we get a dotproduct equal to this number on RHS it means that this vector is in this plane and perpendicular to this plane equation normal vector. Maybe I mistake?
However, recall that the dot product is 0 if the two vectors are orthogonal, and nonzero if they are NOT orthogonal. So if the RHS of the plane equation is nonzero, the dot product of the position vector and the normal is nonzero, which means the position vector is NOT orthogonal to the normal vector!