Find the coordinates of a point in 3-space

[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?

 

[CaliforniaRoll88]

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?

 

[Steve4Physics]

$d^2=(x_Q-x_P)^2+(y_Q-y_P)^2$
$100=(3k-1)^2+(4k-2)^2$
$100=(3k-1)^2+(4k-2)^2$

You have repeated a line, but no matter.

$(a-b)^2=a^2-2ab+b^2$
$(3k-1)^2=(3k)^2-2(3k)(-1)+(-1)^2=9k^2+6k+1$

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.

$(4k-2)^2=(4k)^2-2(4k)(-2)+(-2)^2=16k^2+16k+4$

No. Same mistake as described above.

$(3k-1)^2+(4k-2)^2=25k^2+22k+5$

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.

@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.