# Find the coordinates of a point in 3-space

• CaliforniaRoll88
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.
CaliforniaRoll88
Homework Statement
The vector from the origin to point ##A## is given as ##(6,-2,-4)##, and the unit vector directed from the origin toward point ##B## is ##(2,-2,1)/3##. If points ##A## and ##B## are ten units apart, find the coordinates of point ##B##.
Relevant Equations
##\left |\vec r\right |=10##
##\hat a_B=\frac {\vec B}{\left |\vec B\right |}={\frac {\vec B}{\sqrt {B^2_x+B^2_y +B^2_z}}}##
##\left| \vec r\right |=\sqrt {(x_B-x_A)^2+(y_B-y_A)^2+(z_B-z_A)^2}##
Ans:##(7.8,-7.8,3.9)##
##\hat a_B=\frac 2 3\hat a_x-\frac 2 3\hat a_y+\frac 1 3\hat a_z##
##\left| \vec r\right |=\sqrt {(x_B-6)^2+(y_B+2)^2+(z_B+4)^2}=10##
##\left |\vec A\right |=\sqrt {6^2+2^2+4^2}=\sqrt {5}{6}##

Source: Problem 1.3; Engineering Electromagnetics, 8th Edition, William Hayt, John Buck

Hi,
can you write an expression for the vector from A to B?
Edit: ah, your ##\vec r##.
Rewrite those x,y,z as ##|\vec B| \hat a_B##

CaliforniaRoll88
You are given that
##\vec A=\{6,-2,-4\}##
##\vec B=B\{\frac{2}{3},-\frac{2}{3},\frac{1}{3}\}##
##|\vec B-\vec A|=10## units
Does this help?

PeroK, CaliforniaRoll88 and BvU
kuruman said:
##B\{\frac{2}{3},-\frac{2}{3},\frac{1}{3}\}##
Why does this expression have a ##B## out front? Didn't it already get distributed inside?

Last edited:
CaliforniaRoll88 said:
Why does this expression have a ##B## out front? Didn't it already get distrubuted inside?
Nothing is distributed.

A vector ##\vec B## can always be written as its magnitude ##B## (a scalar) times a unit vector ##\hat b## in the same direction as ##\vec B##, i.e. ##\vec B=B~\hat b.## Then $$\vec B\cdot\vec B=(B~\hat b)\cdot(B~\hat b)=B^2(\hat b\cdot\hat b)=B^2(1)=B^2$$ as it should. Here ##\hat b=\{\frac{2}{3},-\frac{2}{3},\frac{1}{3}\}.## You can dot it with itself and verify that it is a unit vector.

PeroK and CaliforniaRoll88
BvU said:
Rewrite those x,y,z as ##|\vec B| \hat a_B##
##\vec B=|\vec B|\hat a_B=|\vec B|\frac 2 3\hat a_x-|\vec B|\frac 2 3\hat a_y+|\vec B|\frac 1 3\hat a_z##

Look at the last equation in post #3. Do you see that this equation has only one unknown? What is this unknown and how will its value help you find the answer to this problem?

CaliforniaRoll88 and berkeman
##\left ||\vec B|\hat a_B-\vec A\right |=10##

BvU
kuruman said:
Look at the last equation in post #3. Do you see that this equation has only one unknown? What is this unknown and how will its value help you find the answer to this problem?
The unknown is ##|\vec B|##. If I find out this value I can scale my unit vector to point B.
Is ##|\vec B|=3##?

CaliforniaRoll88 said:
The unknown is ##|\vec B|##. If I find out this value I can scale my unit vector to point B.
Is ##|\vec B|=3##?
Right. Go for it.

CaliforniaRoll88
CaliforniaRoll88 said:
##\left |\vec A\right |=\sqrt {6^2+2^2+4^2}=\sqrt {5}{6}##
Just a note on the LaTeX… you meant ##\sqrt {56}##.

CaliforniaRoll88
haruspex said:
Just a note on the LaTeX… you meant ##\sqrt {56}##.
I did, thanks.

kuruman said:
Right. Go for it.
##\vec B=|\vec B|\hat a_B=3\hat a_B=2\hat a_x-2\hat a_y+\hat a_z##
##B(2,-2,1)##
This isn't the answer.

CaliforniaRoll88 said:
##\vec B=|\vec B|\hat a_B=3\hat a_B=2\hat a_x-2\hat a_y+\hat a_z##
##B(2,-2,1)##
This isn't the answer.
I am not surprised that this is not the answer. How did you establish that ##|\vec B|=3## units? Just because there is a number ##3## in the denominator? A vector ##\vec C## that is a scalar number ##N## times ##\vec B## has the same unit vector and is written as ##\vec C=C\{\frac{2}{3},-\frac{2}{3},\frac{1}{3}\}.## Would you follow the same reasoning and say that ##|\vec C|=3## units?

Use the last equation in post #3.

CaliforniaRoll88
kuruman said:
Use the last equation in post #3.
##|\vec B-\vec A|=\left ||\vec B|\hat a_B+(-\vec A)\right |=10##
##\left< |\vec B|\frac {2}{3}-6,-|\vec B|\frac {2}{3}+2, |\vec B|\frac {1}{3}+4\right >##

What is an expression for the magnitude of a vector if you know its 3 components? Write it down and set it equal to 10 units. While you're at it, do yourself a favor and simplify from ##|\vec B|## to just ##B##. It will make the algebra look less daunting.

CaliforniaRoll88
kuruman said:
What is an expression for the magnitude of a vector if you know its 3 components?
##\sqrt {(B_x\frac {2}{3}-6)^2+(-B_y\frac {2}{3}+2)^2+(B_z\frac {1}{3}+4)^2}=10##
At this point it seems that I have 3 variables and 1 equation. Any hints of how to proceed?

Last edited:
The middle term under the radical doesn't look right.

After you correct it, can you solve the equation for ##B##?

CaliforniaRoll88
CaliforniaRoll88 said:
##\sqrt {(B_x\frac {2}{3}-6)^2+(-B_y\frac {2}{3}+2)^2+(B_z\frac {1}{3}+4)^2}=10##
At this point it seems that I have 3 variables and 1 equation. Any hints of how to proceed?
You have 3 variables because you introduced them when you edited the post which you shouldn't have done. Anyway, let's backtrack.
##|\vec B-\vec A|=\sqrt{(B_x-A_x)^2+(B_y-A_y)^2+(B_z-A_z)^2}.##

##B_x=\dfrac{2}{3}B##; ##B_y=-\dfrac{2}{3}B##; ##B_z=\dfrac{1}{3}B##.

##-A_x=-6##; ##-A_y=+2##; ##-A_z=+4##.

Substitute.

MatinSAR and CaliforniaRoll88
kuruman said:
Substitute.

##\sqrt {(B\frac {2}{3}-6)^2+(-B\frac {2}{3}+2)^2+(B\frac {1}{3}+4)^2}=10##
##B^2-\frac {28}{3}B-32=0##
Using the Quadratic Formula: ##B=12,-2\frac{2}{3}##.
Therefore, discarding the negative value and using ##B=12## I get ##B(2,-6,8)## which is wrong.
Do you know where I went wrong? Thank you.

CaliforniaRoll88 said:
##\sqrt {(B\frac {2}{3}-6)^2+(-B\frac {2}{3}+2)^2+(B\frac {1}{3}+4)^2}=10##
##B^2-\frac {28}{3}B-32=0##
Using the Quadratic Formula: ##B=12,-2\frac{2}{3}##.
Therefore, discarding the negative value and using ##B=12## I get ##B(2,-6,8)## which is wrong.
Do you know where I went wrong? Thank you.
You show the first and and last equations and nothing else. Since I don't know what you did in-between, all I can say is that you went wrong somewhere between the first equation and the last equation. The solutions of the quadratic are correct.

MatinSAR and CaliforniaRoll88
kuruman said:
You show the first and and last equations and nothing else. Since I don't know what you did in-between, all I can say is that you went wrong somewhere between the first equation and the last equation.
##\sqrt {(B\frac {2}{3}-6)^2+(-B\frac {2}{3}+2)^2+(B\frac {1}{3}+4)^2}=10##
##\left(B\frac{2}{3}-6\right)^2=\frac {4}{9}B^2-2(B\frac {2}{3})(6)+(-6)^2=\frac {4}{9}B^2-\frac{24}{3}B+36##
##\left(-B\frac{2}{3}+2\right)^2=\frac {4}{9}B^2-2(2)(B\frac {2}{3})+(4)^2=\frac {4}{9}B^2-\frac{8}{3}B+16##
##\left(B\frac{1}{3}+4\right)^2=\frac {1}{9}B^2-2(B\frac {1}{3})(4)+(4)^2=\frac {1}{9}B^2-\frac{4}{3}B+16##
##\frac {4}{9}B^2+\frac {4}{9}B^2+\frac {1}{9}B^2=B^2##
##-\frac {24}{3}B-\frac {8}{3}B+\frac {4}{3}B=-\frac {28}{3}B##
##36+16+16=68##
##B^2-\frac {28}{3}B+68=100##
##B^2-\frac {28}{3}B-32=0##
##B=12##
##B\frac {2}{3}-6=12*\frac {2}{3}-6=2##
##-B\frac {2}{3}+2=-12*\frac {2}{3}+2=-6##
##B\frac {1}{3}+4=12*\frac {1}{3}+4=8##
##\left<2,-6,8\right>##
Do you see where I went wrong now?

Last edited:
kuruman
kuruman said:
You are given that
##\vec A=\{6,-2,-4\}##
##\vec B=B\{\frac{2}{3},-\frac{2}{3},\frac{1}{3}\}##
##|\vec B-\vec A|=10## units
Does this help?
Note that the factor of ##\frac 1 3## is an unnecessary complication. Instead, we can look for:
$$\vec B=B\{2, -2, 1\}$$

CaliforniaRoll88, Steve4Physics and MatinSAR
CaliforniaRoll88 said:
##\left(-B\frac{2}{3}+2\right)^2=\frac {4}{9}B^2-2(2)(B\frac {2}{3})+(4)^2=\frac {4}{9}B^2-\frac{8}{3}B+16##
This is wrong! The ##2## became ##4##.

Last edited:
CaliforniaRoll88 and MatinSAR
An alternative approach is to use the scalar product of vectors. Suppose we have two vectors ##\vec a## and ##\vec b## and we want the distance from the point A (##\vec a##) and B (##B\vec b##) to be some number ##c##.

First, we have:$$\vec a \cdot \vec b = ab \cos \theta$$Then, we have the law of cosines:$$c^2 = a^2 + B^2b^2 - 2Bab \cos \theta = a^2 + B^2b^2 - 2B \vec a \cdot \vec b$$That should give the same quadratic equation for ##B##.

And, you can take ##\vec b## to be a unit vector or not, depending on what looks simpler.

Last edited:
CaliforniaRoll88 and MatinSAR
CaliforniaRoll88 said:
##\left(B\frac{1}{3}+4\right)^2=\frac {1}{9}B^2-2(B\frac {1}{3})(4)+(4)^2=\frac {1}{9}B^2-\frac{4}{3}B+16##
Try again.
CaliforniaRoll88 said:
##\left(-B\frac{2}{3}+2\right)^2=\frac {4}{9}B^2-2(2)(B\frac {2}{3})+(4)^2=\frac {4}{9}B^2-\frac{8}{3}B+16##
As @PeroK said, This is also wrong.

CaliforniaRoll88
@CaliforniaRoll88, to reduce the chance of algebraic and arithmetic mistakes, it helps if you can simplify the working.

##\vec B = |\vec B| ~\frac 13<2, -2, 1>##

Let ##k = \frac 13 |\vec B|## (e.g. see @PeroK, Post #23).

##\vec B = <2k, -2k, k>## and the coordinates of point B are ##(2k, -2k, k)##. Solve for k.

CaliforniaRoll88 and MatinSAR
CaliforniaRoll88 said:
##\sqrt {(B\frac {2}{3}-6)^2+(-B\frac {2}{3}+2)^2+(B\frac {1}{3}+4)^2}=10##
##\left(B\frac{2}{3}-6\right)^2=\frac {4}{9}B^2-2(B\frac {2}{3})(6)+(-6)^2=\frac {4}{9}B^2-\frac{24}{3}B+36##
##\left(-B\frac{2}{3}+2\right)^2=\frac {4}{9}B^2-2(2)(B\frac {2}{3})+(4)^2=\frac {4}{9}B^2-\frac{8}{3}B+16##
##\left(B\frac{1}{3}+4\right)^2=\frac {1}{9}B^2-2(B\frac {1}{3})(4)+(4)^2=\frac {1}{9}B^2-\frac{4}{3}B+16##
##\frac {4}{9}B^2+\frac {4}{9}B^2+\frac {1}{9}B^2=B^2##
##-\frac {24}{3}B-\frac {8}{3}B+\frac {4}{3}B=-\frac {28}{3}B##
##36+16+16=68##
##B^2-\frac {28}{3}B+68=100##
##B^2-\frac {28}{3}B-32=0##
##B=12##
##B\frac {2}{3}-6=12*\frac {2}{3}-6=2##
##-B\frac {2}{3}+2=-12*\frac {2}{3}+2=-6##
##B\frac {1}{3}+4=12*\frac {1}{3}+4=8##
##\left<2,-6,8\right>##
Do you see where I went wrong now?
Thanks for posting the details of your derivation. Yes, I do see where you went wrong. You subtracted the coordinates of the tip of ##\vec A## twice. Once when you found ##|\vec B-\vec A|## and then again at the very end after you found ##B##.

Once you have found the length of ##\vec B##, point B that you are looking for is at the tip of the arrow representing vector ##\vec B.~## If ## \vec B =12\{\frac{2}{3},-\frac{2}{3},\frac{1}{3}\}##, where is its tip?

Conceptually, you can think of reaching the solution as follows:
1. Construct given vector ##\vec A##.
2. At the tip of ##\vec A## construct a sphere of radius ##|\vec B-\vec A| = 10## units. Point B that you are looking for must lie on this sphere.
3. Draw unit vector ##\{\frac{2}{3},-\frac{2}{3},\frac{1}{3}\}## then extend it until it intersects the sphere. The point of intersection is what you are looking for.
Note that if you connect the ends of ##\vec A## and ##\vec B## with a straight line, you form a triangle to which you can apply the rule of cosines as suggested by @PeroK in post #25.

CaliforniaRoll88, MatinSAR and PeroK
There is another error here
##\left(B\frac{1}{3}+4\right)^2=\frac {1}{9}B^2-2(B\frac {1}{3})(4)+(4)^2=\frac {1}{9}B^2-\frac{4}{3}B+16.##
The sign in front of the linear term in ##B## should be ##+## not ##-##.

CaliforniaRoll88
MatinSAR said:
Try again.
kuruman said:
There is another error here
##\left(B\frac{1}{3}+4\right)^2=\frac {1}{9}B^2-2(B\frac {1}{3})(4)+(4)^2=\frac {1}{9}B^2-\frac{4}{3}B+16.##
The sign in front of the linear term in ##B## should be ##+## not ##-##.
##\left(B\frac{2}{3}-6\right)^2=\frac {4}{9}B^2-2(B\frac {2}{3})(6)+(-6)^2=\frac {4}{9}B^2-\frac{24}{3}B+36##
##\left(\frac{1}{3}B+4\right)^2=(\frac {1}{3}B)^2+2(\frac {1}{3}B)(4)+(4)^2=\frac {1}{9}B^2+\frac{4}{3}B+16##
##\left(-\frac{2}{3}B+2\right)^2=(\frac {2}{3}B)^2+2(\frac {2}{3}B)(2)+(2)^2=\frac {4}{9}B^2+\frac{8}{3}B+4##
##\frac {4}{9}B^2+\frac {4}{9}B^2+\frac {1}{9}B^2=B^2##
##-\frac {24}{3}B+\frac {8}{3}B+\frac {4}{3}B=-\frac {12}{3}B=4B##
##36+16+4=56##
##B^2-4B+56=100##
##B^2-4B-44=0##
##x = \frac {-b \pm \sqrt{b^2 -4ac}} {2a}##
##B = \frac {-(-4) \pm \sqrt{(-4)^2 -4(1)(-44)}} {2(1)}##
##B = \frac {4 \pm \sqrt{16 +176}} {2(1)}##
##B = \frac {4 \pm \sqrt{192}} {2}##

Last edited:
CaliforniaRoll88 said:
##\left(B\frac{2}{3}-6\right)^2=\frac {4}{9}B^2-2(B\frac {2}{3})(6)+(-6)^2=\frac {4}{9}B^2-\frac{24}{3}B+36##
##\left(\frac{1}{3}B+4\right)^2=(\frac {1}{3}B)^2+2(\frac {1}{3}B)(4)+(4)^2=\frac {1}{9}B^2+\frac{4}{3}B+16##
##\left(-\frac{2}{3}B+2\right)^2=(\frac {2}{3}B)^2+2(\frac {2}{3}B)(2)+(2)^2=\frac {4}{9}B^2+\frac{8}{3}B+4##
##\frac {4}{9}B^2+\frac {4}{9}B^2+\frac {1}{9}B^2=B^2##
##-\frac {24}{3}B+\frac {8}{3}B+\frac {4}{3}B=-\frac {12}{3}B=4B##
##36+16+4=56##
##B^2-4B+56=100##
##B^2-4B-44=0##
##x = \frac {-b \pm \sqrt{b^2 -4ac}} {2a}##
##B = \frac {-(-4) \pm \sqrt{(-4)^2 -4(1)(-44)}} {2(1)}##
##B = \frac {4 \pm \sqrt{16 +176}} {2(1)}##
##B = \frac {4 \pm \sqrt{192}} {2}##
##\left(\frac{1}{3}B+4\right)^2=(\frac {1}{3}B)^2+\color{red}{2(\frac {1}{3}B)(4)}+(4)^2=\frac {1}{9}B^2+\color{red}{\frac{4}{3}B}+16##

CaliforniaRoll88
There is another one that escaped my previous pass.
##\left(-\frac{2}{3}B+2\right)^2=(\frac {2}{3}B)^2\color{red}{+}2(\frac {2}{3}B)(2)+(2)^2=\frac {4}{9}B^2+\frac{8}{3}B+4##

I suggest that you redo and verify each equation before you typeset it in LaTeX.

CaliforniaRoll88
kuruman said:
##\left(\frac{1}{3}B+4\right)^2=(\frac {1}{3}B)^2+\color{red}{2(\frac {1}{3}B)(4)}+(4)^2=\frac {1}{9}B^2+\color{red}{\frac{4}{3}B}+16##
I think it should be ##\frac 8 3 B##.

CaliforniaRoll88
CaliforniaRoll88 said:
##\left(B\frac{2}{3}-6\right)^2=\frac {4}{9}B^2-2(B\frac {2}{3})(6)+(-6)^2=\frac {4}{9}B^2-\frac{24}{3}B+36##
##\left(\frac{1}{3}B+4\right)^2=(\frac {1}{3}B)^2+2(\frac {1}{3}B)(4)+(4)^2=\frac {1}{9}B^2+\frac{4}{3}B+16##
##\left(-\frac{2}{3}B+2\right)^2=(\frac {2}{3}B)^2+2(\frac {2}{3}B)(2)+(2)^2=\frac {4}{9}B^2+\frac{8}{3}B+4##
##\frac {4}{9}B^2+\frac {4}{9}B^2+\frac {1}{9}B^2=B^2##
##-\frac {24}{3}B+\frac {8}{3}B+\frac {4}{3}B=-\frac {12}{3}B=4B##
##36+16+4=56##
##B^2-4B+56=100##
##B^2-4B-44=0##
##x = \frac {-b \pm \sqrt{b^2 -4ac}} {2a}##
##B = \frac {-(-4) \pm \sqrt{(-4)^2 -4(1)(-44)}} {2(1)}##
##B = \frac {4 \pm \sqrt{16 +176}} {2(1)}##
##B = \frac {4 \pm \sqrt{192}} {2}##
If this was an exam, you would lose a lot of points for no particular reason.

You can use https://www.symbolab.com/ to check your calculations.
Edit :

CaliforniaRoll88
##\left(B\frac{2}{3}-6\right)^2=\frac {4}{9}B^2-2(B\frac {2}{3})(6)+(-6)^2=\frac {4}{9}B^2-\frac{24}{3}B+36##
##\left(\frac{1}{3}B+4\right)^2=(\frac {1}{3}B)^2+2(\frac {1}{3}B)(4)+(4)^2=\frac {1}{9}B^2+\frac{8}{3}B+16##
##\left(-\frac{2}{3}B+2\right)^2=(-\frac {2}{3}B)^2+2(-\frac {2}{3}B)(2)+(2)^2=\frac {4}{9}B^2-\frac{8}{3}B+4##
##\frac {4}{9}B^2+\frac {4}{9}B^2+\frac {1}{9}B^2=B^2##
##-\frac {24}{3}B+\frac {8}{3}B-\frac {8}{3}B=-\frac {24}{3}B=-8B##
##36+16+4=56##
##B^2-8B+56=100##
##B^2-8B-44=0##
##x = \frac {-b \pm \sqrt{b^2 -4ac}} {2a}##
##B = \frac {-(-8) \pm \sqrt{(-8)^2 -4(1)(-44)}} {2(1)}##
##B = \frac {8 \pm \sqrt{64+176}}{2}##
##B = \frac {8 \pm \sqrt{240}}{2}##
##B = \frac {8 \pm \sqrt{15*16}}{2}##
##B = \frac {8 \pm 4\sqrt{15}}{2}=##
##B = 4 \pm 2\sqrt{15}=11.74597,-3.74597##
##\vec B =11.74597\{\frac{2}{3},-\frac{2}{3},\frac{1}{3}\}##
##B(7.83,-7.83,3.915)##
Finally got the answer with this method.

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