Find the coordinates of a point in 3-space

[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:$(7.8,-7.8,3.9)$

$\hat a_B=\frac 2 3\hat a_x-\frac 2 3\hat a_y+\frac 1 3\hat a_z$
$\left| \vec r\right |=\sqrt {(x_B-6)^2+(y_B+2)^2+(z_B+4)^2}=10$
$\left |\vec A\right |=\sqrt {6^2+2^2+4^2}=\sqrt {5}{6}$

Not sure where to go from here. Please help!

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

 

[BvU]

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

 

[kuruman]

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?

 

[CaliforniaRoll88]

$B\{\frac{2}{3},-\frac{2}{3},\frac{1}{3}\}$

Why does this expression have a $B$ out front? Didn't it already get distributed inside?

 

[kuruman]

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.

 

[CaliforniaRoll88]

Rewrite those x,y,z as $|\vec B| \hat a_B$

$\vec B=|\vec B|\hat a_B=|\vec B|\frac 2 3\hat a_x-|\vec B|\frac 2 3\hat a_y+|\vec B|\frac 1 3\hat a_z$

 

[kuruman]

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]

$\left ||\vec B|\hat a_B-\vec A\right |=10$

 

[CaliforniaRoll88]

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 $|\vec B|=3$?

 

[kuruman]

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.

 

[haruspex]

$\left |\vec A\right |=\sqrt {6^2+2^2+4^2}=\sqrt {5}{6}$

Just a note on the LaTeX… you meant $\sqrt {56}$.

 

[CaliforniaRoll88]

Just a note on the LaTeX… you meant $\sqrt {56}$.

I did, thanks.

 

[CaliforniaRoll88]

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.

 

[kuruman]

$\vec B=|\vec B|\hat a_B=3\hat a_B=2\hat a_x-2\hat a_y+\hat a_z$
$B(2,-2,1)$
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]

Use the last equation in post #3.

$|\vec B-\vec A|=\left ||\vec B|\hat a_B+(-\vec A)\right |=10$
$\left< |\vec B|\frac {2}{3}-6,-|\vec B|\frac {2}{3}+2, |\vec B|\frac {1}{3}+4\right >$
I am stuck at this point. Please help!

 

[kuruman]

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]

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?

 

[kuruman]

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

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

 

[kuruman]

$\sqrt {(B_x\frac {2}{3}-6)^2+(-B_y\frac {2}{3}+2)^2+(B_z\frac {1}{3}+4)^2}=10$
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.

 

[CaliforniaRoll88]

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.

 

[kuruman]

$\sqrt {(B\frac {2}{3}-6)^2+(-B\frac {2}{3}+2)^2+(B\frac {1}{3}+4)^2}=10$
$B^2-\frac {28}{3}B-32=0$
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.

 

[CaliforniaRoll88]

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?

 

[PeroK]

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:
$$\vec B=B\{2, -2, 1\}$$

 

[PeroK]

$\left(-B\frac{2}{3}+2\right)^2=\frac {4}{9}B^2-2(2)(B\frac {2}{3})+(4)^2=\frac {4}{9}B^2-\frac{8}{3}B+16$

This is wrong! The $2$ became $4$.

 

[PeroK]

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.

 

[MatinSAR]

$\left(B\frac{1}{3}+4\right)^2=\frac {1}{9}B^2-2(B\frac {1}{3})(4)+(4)^2=\frac {1}{9}B^2-\frac{4}{3}B+16$

Try again.

$\left(-B\frac{2}{3}+2\right)^2=\frac {4}{9}B^2-2(2)(B\frac {2}{3})+(4)^2=\frac {4}{9}B^2-\frac{8}{3}B+16$

As @PeroK said, This is also wrong.

 

[Steve4Physics]

@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.

 

[kuruman]

$\sqrt {(B\frac {2}{3}-6)^2+(-B\frac {2}{3}+2)^2+(B\frac {1}{3}+4)^2}=10$
$\left(B\frac{2}{3}-6\right)^2=\frac {4}{9}B^2-2(B\frac {2}{3})(6)+(-6)^2=\frac {4}{9}B^2-\frac{24}{3}B+36$
$\left(-B\frac{2}{3}+2\right)^2=\frac {4}{9}B^2-2(2)(B\frac {2}{3})+(4)^2=\frac {4}{9}B^2-\frac{8}{3}B+16$
$\left(B\frac{1}{3}+4\right)^2=\frac {1}{9}B^2-2(B\frac {1}{3})(4)+(4)^2=\frac {1}{9}B^2-\frac{4}{3}B+16$
$\frac {4}{9}B^2+\frac {4}{9}B^2+\frac {1}{9}B^2=B^2$
$-\frac {24}{3}B-\frac {8}{3}B+\frac {4}{3}B=-\frac {28}{3}B$
$36+16+16=68$
$B^2-\frac {28}{3}B+68=100$
$B^2-\frac {28}{3}B-32=0$
$B=12$
$B\frac {2}{3}-6=12*\frac {2}{3}-6=2$
$-B\frac {2}{3}+2=-12*\frac {2}{3}+2=-6$
$B\frac {1}{3}+4=12*\frac {1}{3}+4=8$
$\left<2,-6,8\right>$
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.

 

[kuruman]

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]

Try again.

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 $-$.

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

 

[kuruman]

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

$\left(\frac{1}{3}B+4\right)^2=(\frac {1}{3}B)^2+\color{red}{2(\frac {1}{3}B)(4)}+(4)^2=\frac {1}{9}B^2+\color{red}{\frac{4}{3}B}+16$

 

[kuruman]

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.

 

[MatinSAR]

$\left(\frac{1}{3}B+4\right)^2=(\frac {1}{3}B)^2+\color{red}{2(\frac {1}{3}B)(4)}+(4)^2=\frac {1}{9}B^2+\color{red}{\frac{4}{3}B}+16$

I think it should be $\frac 8 3 B$.

 

[MatinSAR]

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

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 :
Link

 

[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.

 

[CaliforniaRoll88]

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 $1/3$?

 

[PeroK]

How do you disregard the $1/3$?

Why do you include it?

 

[PeroK]

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 $\hat b$?

 

[Steve4Physics]

Finally got the answer with this method.

It can be done much more easily using the Post #27 and @PeroK's suggestions!

 

[CaliforniaRoll88]

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?

 

[PeroK]

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.

 

[PeroK]

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.

 

[Steve4Physics]

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.

 

[CaliforniaRoll88]

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?

 

[PeroK]

$\vec P=P\left<1,2\right>$
$\vec Q=Q\frac{1}{5}\left<3,4\right>$
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.

 

[Steve4Physics]

$\vec R$ 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!

 

[CaliforniaRoll88]

$10^2=5+25k-22k$
$95=3k$
$k=\frac {95}{3}$
$\vec Q=k\left<3,4\right>=\left<3k,4k\right>=\left<3\frac {95}{3},4\frac {95}{3}\right>=\left<95,126\frac {2}{3}\right>$
Is this right @Steve4Physics, @PeroK?
Edit: I am sure it's wrong.

 

[PeroK]

$10^2=5+25k-22k$
$95=3k$
$k=\frac {95}{3}$
$\vec Q=k\left<3,4\right>=\left<3k,4k\right>=\left<3\frac {95}{3},4\frac {95}{3}\right>=\left<95,126\frac {2}{3}\right>$
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.

 

[Steve4Physics]

$\vec P=P\left<1,2\right>$

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 $\vec P$.

 

[Steve4Physics]

$10^2=5+25k-22k$
$95=3k$
$k=\frac {95}{3}$
$\vec Q=k\left<3,4\right>=\left<3k,4k\right>=\left<3\frac {95}{3},4\frac {95}{3}\right>=\left<95,126\frac {2}{3}\right>$
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?