Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2016-09-21 07:12:11

mathattack
Member
Registered: 2016-03-07
Posts: 17

Vector Geometry

I've been working on this problem using vectors, but can't quite finish it.  I wrote out two equations, one from A to S, the other from B to T, and set them equal to each other.  I just can't figure out what the two parameters would be.  Can someone help?  Thanks!


In triangle ABC, S is a point on side BC such that BS:SC = 1:2, and T is a point on side AC such that AT:TC = 4:3. Let U be the intersection of AS and BT. Find AU:US.

Offline

#2 2016-09-21 07:13:41

mathattack
Member
Registered: 2016-03-07
Posts: 17

Re: Vector Geometry

One more, with similar issues.  Any help would be great, thanks!


Let G denote the centroid of triangle ABC. Let M and N be points on sides AB and AC, respectively, so that M, G, and N are collinear, and AM/MB = 5/2. Find AN/NC.

Offline

#3 2016-09-21 08:50:52

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Vector Geometry

Hi;

I failed to notice that you are trying this with vectors.

The answer is easily derived through the use of Geogebra, a piece of software that is free and should be part of everyone's math. With it, you will learn much coordinate geometry and will get your first lesson in integer detection algorithms.

What was your approach to the problem? Did you draw a good diagram?


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Offline

#4 2016-09-21 19:05:53

Bob
Administrator
Registered: 2010-06-20
Posts: 10,623

Re: Vector Geometry

hi mathattack

Q1.  If you are doing vector geometry, then the usual approach is to fix an origin and two base vectors.  Let's say A, AB = i and AC= j

To save time I'm not going to continue using bold for the vectors.

So BC = = -i + j, and BS = 1/3 . (-i + j)

So the line AS is r = i + lambda . 1/3 . (-i + j)

Do a similar thing for the equation for BT and find where they cross, U.

This will enable you to determine the ratio.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

Offline

#5 2016-09-21 19:48:11

thickhead
Member
Registered: 2016-04-16
Posts: 1,086

Re: Vector Geometry

(2) I get AN/NC=5/3


{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

Offline

#6 2016-09-28 07:53:01

mathattack
Member
Registered: 2016-03-07
Posts: 17

Re: Vector Geometry

Thank you both!  I was able to work out both solutions.  I hadn't realized before that once I got my equations for both vectors, after setting them equal to each other I could separate each variable and solve for when they equaled zero!  I didn't expect that to work but it did.  Thank you both for your insight and help.

Offline

Board footer

Powered by FluxBB