Find the coordinates of a point in 3-space

In summary, the conversation is about finding the vector from point A to point B and using various equations and formulas to solve for its magnitude and components. After some backtracking and substitution, we find that the correct solution is B(2, -6, 8). However, it is unclear where the mistake was made in the process.
  • #36
PeroK said:
Note that the factor of ##\frac 1 3## is an unnecessary complication. Instead, we can look for:
$$\vec B=B\{2, -2, 1\}$$
How do you disregard the ##1/3##?
 
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  • #37
CaliforniaRoll88 said:
How do you disregard the ##1/3##?
Why do you include it?
 
  • #38
CaliforniaRoll88 said:
How do you disregard the ##1/3##?
Let's look at two questions. Let A be any point:

a) What are the coordinates of a point a distance ##c## units from ##A## in the direction of the vector ##\vec b##?

b) What are the coordinates of a point a distance ##c## units from ##A## in the direction of the vector ##\hat b##?
 
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  • #39
CaliforniaRoll88 said:
Finally got the answer with this method.
It can be done much more easily using the Post #27 and @PeroK's suggestions!
 
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  • #40
PeroK said:
a) What are the coordinates of a point a distance ##c## units from ##A## in the direction of the ##\vec b##?
Let ##B## be any point not ##A##:
Then the coordinates of ##B## away from ##A## by ##c## units is ##B(cA_x,cA_y,cA_z)##
Is this right?
 
  • #41
CaliforniaRoll88 said:
Let ##B## be any point not ##A##:
Then the coordinates of ##B## away from ##A## by ##c## units is ##B(cA_x,cA_y,cA_z)##
Is this right?
That's not right at all. The point of my question was that the answers to a) and b) are the same. A direction is a direction.
 
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  • #42
CaliforniaRoll88 said:
Let ##B## be any point not ##A##:
Then the coordinates of ##B## away from ##A## by ##c## units is ##B(cA_x,cA_y,cA_z)##
Is this right?
I suggest that you do some 2D problems so that you can draw a diagram. I think you are struggling with 3D geometry because you are struggling with the basic concepts.
 
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  • #43
CaliforniaRoll88 said:
Let ##B## be any point not ##A##:
Then the coordinates of ##B## away from ##A## by ##c## units is ##B(cA_x,cA_y,cA_z)##
Is this right?
Hi @CaliforniaRoll88. That's not right. As suggested by @PeroK, maybe it would help you to do a 2D problem first - a diagram is then easy to draw so you can see what's happening. How about trying this:

"The vector from the origin to point ##P## is given as ##<1, 2>##, and the unit vector directed from the origin towards point ##Q## is ##{\frac 15}{<3,4>}## . If points ##P## and ##Q## are ten units apart, find the coordinates of point ##Q##.

And you might want to re-read Post #27!

Edited.
 
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  • #44
Steve4Physics said:
Hi @CaliforniaRoll88. That's not right. As suggested by @PeroK, maybe it would help you to do a 2D problem first - a diagram is then easy to draw so you can see what's happening. How about trying this:

"The vector from the origin to point ##P## is given as ##<1, 2>##, and the unit vector directed from the origin towards point ##Q## is ##{\frac 15}{<3,4>}## . If points ##P## and ##Q## are ten units apart, find the coordinates of point ##Q##.

And you might want to re-read Post #27!

Edited.
##\vec P=P\left<1,2\right>##
##\vec Q=Q\frac{1}{5}\left<3,4\right>##
Let ##k=\frac{1}{5}Q##
##\vec Q=k\left<3,4\right>=\left<3k,4k\right>##
##\vec P\cdot\vec Q=1\cdot3k+2\cdot4k=11k##
##\vec R## is the resultant of ##\vec P## and ##\vec Q##
Law of Cosines:
##R^2=P^2+Q^2-2\vec P\cdot\vec Q##
##10^2=5+25k-22k##
Am I headed in the right direction?
 
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  • #45
CaliforniaRoll88 said:
##\vec P=P\left<1,2\right>##
##\vec Q=Q\frac{1}{5}\left<3,4\right>##
Let ##k=\frac{1}{5}Q##
##\vec Q=k\left<3,4\right>=\left<3k,4k\right>##
##\vec P\cdot\vec Q=1\cdot3k+2\cdot4k=11k##
##\vec R## is the resultant of ##\vec P## and ##\vec Q##
Law of Cosines:
##R^2=P^2+Q^2-2\vec P\cdot\vec Q##
##10^2=5+25k-22k##
Am I headed in the right direction?
Looks good!

PS except, should have a quadratic in ##k## with a ##25k^2## term. I missed that.
 
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  • #46
CaliforniaRoll88 said:
##\vec R## is the resultant of ##\vec P## and ##\vec Q##
A couple of points (pun intended).

1) We want to make ##|\vec {PQ}|=10##. Knowing the resultant ##\vec P + \vec Q## doesn't help. It's possible that you might be thinking ##\vec {PQ} = \vec P + \vec Q## but that's wrong! A suitable diagram will make this clear.

2) There is (IMO) a much simpler approach. The distance between points P and Q must be 10. You are given point P(1, 2). What are coordinates of point Q in terms of k? The rest is simple!
 
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  • #47
##10^2=5+25k-22k##
##95=3k##
##k=\frac {95}{3}##
##\vec Q=k\left<3,4\right>=\left<3k,4k\right>=\left<3\frac {95}{3},4\frac {95}{3}\right>=\left<95,126\frac {2}{3}\right>##
Is this right @Steve4Physics, @PeroK?
Edit: I am sure it's wrong.
 
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  • #48
CaliforniaRoll88 said:
##10^2=5+25k-22k##
##95=3k##
##k=\frac {95}{3}##
##\vec Q=k\left<3,4\right>=\left<3k,4k\right>=\left<3\frac {95}{3},4\frac {95}{3}\right>=\left<95,126\frac {2}{3}\right>##
Is this right @Steve4Physics, @PeroK?
Edit: I am sure it's wrong.
I missed that you had ##25k## instead of ##25k^2##. You need to find some way of reducing the number of simple algebraic errors. And, you need a way of spotting them yourself.
 
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  • #49
CaliforniaRoll88 said:
##\vec P=P\left<1,2\right>##
As a completely separate correction, it’s probably worth noting that the above is wrong. ##\vec P = \left< 1, 2 \right>##. There is no "##P##" on the right-hand side.

The magnitude of ##\vec P## is ##P = \sqrt {1^2 +2^2} = \sqrt 5##. By writing “##P\left< 1, 2 \right>##” you are multiplying ##\vec P## by ##\sqrt 5##; this gives a new vector ##\sqrt 5## times bigger than ##\vec P##.
 
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  • #50
CaliforniaRoll88 said:
##10^2=5+25k-22k##
##95=3k##
##k=\frac {95}{3}##
##\vec Q=k\left<3,4\right>=\left<3k,4k\right>=\left<3\frac {95}{3},4\frac {95}{3}\right>=\left<95,126\frac {2}{3}\right>##
Is this right @Steve4Physics, @PeroK?
Edit: I am sure it's wrong.
Hi @CaliforniaRoll88. Yes, it's wrong!

You seem to be struggling and we are not looking at the original homework problem. So maybe a bit of extra help is justifiable.

Hopefully by now you have drawn a clear diagram (xy axes) with points and vectors marked, estimating the approximate position of point Q ‘by eye’.

You have correctly written (in Post #44): ##\vec Q=k\left<3,4\right>=\left<3k,4k\right>##

##\vec Q## is an ‘arrow’ from the origin to point Q. Therefore the coordinates of point Q are ##(3k, 4k)## where ##k## is some as yet unknown value.

You have 2 points P(1,2) and Q(3k, 4k). You require the distance between them to be 10. So what equation can you now write down and solve?
 
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  • #51
Steve4Physics said:
You have 2 points P(1,2) and Q(3k, 4k). You require the distance between them to be 10. So what equation can you now write down and solve?
##d^2=(x_Q-x_P)^2+(y_Q-y_P)^2##
##100=(3k-1)^2+(4k-2)^2##
##100=(3k-1)^2+(4k-2)^2##
##(a-b)^2=a^2-2ab+b^2##
##(3k-1)^2=(3k)^2-2(3k)(-1)+(-1)^2=9k^2+6k+1##
##(4k-2)^2=(4k)^2-2(4k)(-2)+(-2)^2=16k^2+16k+4##
##(3k-1)^2+(4k-2)^2=25k^2+22k+5##
@Steve4Physics, I can see that I am using the quadratic formula method. How do I employ the Law of Cosines here?
 
Last edited:
  • #52
CaliforniaRoll88 said:
##d^2=(x_Q-x_P)^2+(y_Q-y_P)^2##
##100=(3k-1)^2+(4k-2)^2##
##100=(3k-1)^2+(4k-2)^2##
You have repeated a line, but no matter.

CaliforniaRoll88 said:
##(a-b)^2=a^2-2ab+b^2##
##(3k-1)^2=(3k)^2-2(3k)(-1)+(-1)^2=9k^2+6k+1##
No. You have used b=-1. But b=1 because the minus sign is already taken into account in the term ##a-b##.

Here's a simple example, to illustrate what you have done. We know ##(3-1)^2 = 2^2 = 4##. What you have done is:
##(3 - 1)^2 = 3^2 - (2)(3)(-1) + (-1)^2 = 9 + 6 + 1 = 16##
This is wrong! It should be:
##(3 - 1)^2 = 3^2 - (2)(3)(1) + 1^2 = 9 - 6 + 1 = 4##.

This is very important basic algebra, so you need to make sure you master it.

CaliforniaRoll88 said:
##(4k-2)^2=(4k)^2-2(4k)(-2)+(-2)^2=16k^2+16k+4##
No. Same mistake as described above.

CaliforniaRoll88 said:
##(3k-1)^2+(4k-2)^2=25k^2+22k+5##
This expression (when corrected) must be equal to 100, This gives you a quadratic equation you can solve to find ##k##. Then you know Q##(3k,4k)## and the problem is done.

CaliforniaRoll88 said:
@Steve4Physics, I can see that I am using the quadratic formula method. How do I employ the Law of Cosines here?
You do not need the law of cosines if you use the above method.
 
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