Finding Coordinates of Point B - Positional Vectors

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In summary, the conversation was about finding the coordinates of point B using a system of equations and circular geometry. The initial problem was provided, along with the steps taken to solve it and the final answer. The conversation also included a clarification on units and a thank you for the solution.
  • #1
Adhil
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I need help finding the coordinates of point B. I have the answer from the textbook but I have no clue how they got it. I also cannot go any further until I solve it.

Note: I am studying at home so I have no lectures to go to for help.

Questions is 3.4
https://uploads.tapatalk-cdn.com/20170615/8ee21893380739b1b87312ef1e6fc34f.jpg
https://uploads.tapatalk-cdn.com/20170615/326c5ea861736cdabf5dd13dfa373f31.jpg
 
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  • #2
Welcome to MHB! We need a lot more information than this to help you out. Please give us the complete problem statement, and any work you've done on the problem so far. If you don't know how to start the problem, that's fine, too. Just let us know where you're stuck.
 
  • #3
Sorry the images did not upload at first
 
  • #4
http://uploads.tapatalk-cdn.com/20170615/326c5ea861736cdabf5dd13dfa373f31.jpg

point A has coordinates $(400,800)$

let the coordinates of point B be $(a,b)$

$|r_{AB}| = \sqrt{(a-400)^2+(b-800)^2} = 400$

$|r_{OA}+r_{AB}| = \sqrt{a^2+b^2} = 1200$

solve the system for $(a,b)$ ...
 
  • #5
Okay I did that but I only got the first pair of answers... B(785,907)...I think my working maybe be wrong.
 
  • #6
Adhil said:
Okay I did that but I only got the first pair of answers... B(785,907)...I think my working maybe be wrong.

your first ordered pair is good ... I got the 2nd pair to be (255,1173)
 
  • #7
Can you please show me the process of how you obtained it. I can get the basis of it from your diagram but I am unfamiliar with that method... Like I said, I'm homeschooling
 
  • #8
to make the values manageable, let the positions and distances be in units of $10^2$ meters

point A $(4,8)$

point B $(a,b)$

$|r_{OA}+r_{AB}| = |r_{OB}| \implies a^2+b^2=12^2$

$|r_{AB}| = 4 \implies (a-4)^2 + (b-8)^2 = 4^2$

expanding the left side ...

$a^2 -8a + 16 + b^2 - 16b + 64 = 16$

combine like terms & rearrange ...

$a^2 + b^2 = (8a+16b) - 64$

substitute $12^2$ for $a^2+b^2$ ...

$208 = 8a+16b \implies 26 = a+2b \implies a = 26-2b$

substitute for $a$ in the equation $a^2+b^2=12^2$ ...

$(26-2b)^2 + b^2 = 12^2$

resulting quadratic in standard form is ...

$5b^2 - 104b + 532 = 0$

$b = \dfrac{104 \pm \sqrt{(-104)^2 - 4(5)(532)}}{10}$

two values for $b$, then use $a = 26-2b$ to get the corresponding values for $a$ ... don't forget to multiply both sets of coordinates by $10^2$ at the end to get the values in meters.
 
  • #9
skeeter said:
to make the values manageable, let the positions and distances be in units of $10^2$ meters

point A $(4,8)$

point B $(a,b)$

$|r_{OA}+r_{AB}| = |r_{OB}| \implies a^2+b^2=12^2$

$|r_{AB}| = 4 \implies (a-4)^2 + (b-8)^2 = 4^2$

expanding the left side ...

$a^2 -8a + 16 + b^2 - 16b + 64 = 16$

combine like terms & rearrange ...

$a^2 + b^2 = (8a+16b) - 64$

substitute $12^2$ for $a^2+b^2$ ...

$208 = 8a+16b \implies 26 = a+2b \implies a = 26-2b$

substitute for $a$ in the equation $a^2+b^2=12^2$ ...

$(26-2b)^2 + b^2 = 12^2$

resulting quadratic in standard form is ...

$5b^2 - 104b + 532 = 0$

$b = \dfrac{104 \pm \sqrt{(-104)^2 - 4(5)(532)}}{10}$

two values for $b$, then use $a = 26-2b$ to get the corresponding values for $a$ ... don't forget to multiply both sets of coordinates by $10^2$ at the end to get the values in meters.
Thank you so much! I never thought about using circular geometry to solve it (even though the signs where obvious) [emoji1]
 

FAQ: Finding Coordinates of Point B - Positional Vectors

What is the difference between position vectors and displacement vectors?

Position vectors represent the location of a point relative to the origin, while displacement vectors represent the change in position of an object from one point to another.

How do I find the coordinates of point B using position vectors?

To find the coordinates of point B using position vectors, you will need the position vector of the origin (usually denoted as O) and the position vector of point B (usually denoted as P). The coordinates of point B can be found by adding the components of these two vectors together.

Can position vectors be negative?

Yes, position vectors can have negative components. This indicates that the point is located in the negative direction from the origin on the corresponding axis.

What is the formula for calculating the magnitude of a position vector?

The magnitude of a position vector can be calculated using the Pythagorean theorem, where the magnitude is equal to the square root of the sum of the squares of the vector's components. In equation form, it is written as |OP| = √(x² + y² + z²), where x, y, and z are the components of the vector.

How can I use position vectors in real-life applications?

Position vectors are commonly used in physics and engineering to describe the position of an object in space. They can also be used in navigation systems, such as GPS, to determine the location of a point on Earth's surface. Additionally, position vectors are used in computer graphics to represent the position of objects in a three-dimensional space.

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