MHB Finding Coordinates of Point B - Positional Vectors

[Adhil]

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

 

[Ackbach]

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.

 

[Adhil]

Sorry the images did not upload at first

 

[skeeter]

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 $(a,b)$ ...

 

[Adhil]

Okay I did that but I only got the first pair of answers... B(785,907)...I think my working maybe be wrong.

 

[skeeter]

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)

 

[Adhil]

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

 

[skeeter]

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.

 

[Adhil]

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]