Thread: Please help with this question about vectors

Hi

I have no idea how to go about this question
I have attached the question and all of my working so far.
I was trying to do the bit about finding the point of intersection of the line joining A to the midpoint of BC, and the line
joining B to the midpoint of AC but it wasn't working out.
I equated the two line and compared x coords and y coords but then I got two equations with 4 unknowns.
I really don't think I was doing it right.
Could someone please point me in the right direction.
Thanks in advance for replies.

James

find first the equation of the median AM it is the line that connects the vertex A with the mid point of BC .
This line has equation as per your coordinates of the vertices A(p1,q1), B(p2,q2),C(p3,q3).
the equation of the median :
AM IS: (q2+q3-2q1)x-(p2+p3-2p1)y+(p2q1-q2p1)+(p3q1-q3p1)=0

the median BE is : (q1+q3-2q2)x-(p1+p3-2p2)y+(p3q2-q3p2)+(p1q2-q1p2)=0

the median CZ is : (q1+q2-2q3)x-(p1+p2-2p3)y+(p1q3-q1p3)+(p2q3-q2p3)=0

if these 3 lines meet in one point then they form a pencil of lines.
the condition that 3 lines form a pencil of lines is that the determinant of their coefficients = 0
ex given if the lines a1x+b1y+C1=0 , a2x+b2y+c2=0 and a3x+b3y+c3=0 form a pencil of lines then the determinant of the coefficients a1,b2,c1,...etc =0
solve the system of the 3 equations I have given you to find their common point. WHOSE COORDINATES MUST BE (p1,+p2+p3)/3 AND (q1+q2+q3)/3
this is the barycenter of the triangle ABC...
I hope you understand all the above...
MINOAS

James

find first the equation of the median AM it is the line that connects the vertex A with the mid point of BC .
This line has equation as per your coordinates of the vertices A(p1,q1), B(p2,q2),C(p3,q3).
the equation of the median :
AM IS: (q2+q3-2q1)x-(p2+p3-2p1)y+(p2q1-q2p1)+(p3q1-q3p1)=0

the median BE is : (q1+q3-2q2)x-(p1+p3-2p2)y+(p3q2-q3p2)+(p1q2-q1p2)=0

the median CZ is : (q1+q2-2q3)x-(p1+p2-2p3)y+(p1q3-q1p3)+(p2q3-q2p3)=0

if these 3 lines meet in one point then they form a pencil of lines.
the condition that 3 lines form a pencil of lines is that the determinant of their coefficients = 0
ex given if the lines a1x+b1y+C1=0 , a2x+b2y+c2=0 and a3x+b3y+c3=0 form a pencil of lines then the determinant of the coefficients a1,b2,c1,...etc =0
solve the system of the 3 equations I have given you to find the common their point. WHOSE COORDINATES MUST BE (p1,+p2+p3)/3 AND (q1+q2+q3)/3
this is the barycenter of the triangle ABC...
I hope you understand all the above...
MINOAS

Originally Posted by MINOANMAN

James

find first the equation of the median AM it is the line that connects the vertex A with the mid point of BC .
This line has equation as per your coordinates of the vertices A(p1,q1), B(p2,q2),C(p3,q3).
the equation of the median :
AM IS: (q2+q3-2q1)x-(p2+p3-2p1)y+(p2q1-q2p1)+(p3q1-q3p1)=0

the median BE is : (q1+q3-2q2)x-(p1+p3-2p2)y+(p3q2-q3p2)+(p1q2-q1p2)=0

the median CZ is : (q1+q2-2q3)x-(p1+p2-2p3)y+(p1q3-q1p3)+(p2q3-q2p3)=0

if these 3 lines meet in one point then they form a pencil of lines.
the condition that 3 lines form a pencil of lines is that the determinant of their coefficients = 0
ex given if the lines a1x+b1y+C1=0 , a2x+b2y+c2=0 and a3x+b3y+c3=0 form a pencil of lines then the determinant of the coefficients a1,b2,c1,...etc =0
solve the system of the 3 equations I have given you to find the common their point. WHOSE COORDINATES MUST BE (p1,+p2+p3)/3 AND (q1+q2+q3)/3
this is the barycenter of the triangle ABC...
I hope you understand all the above...
MINOAS

Hey Minoas

How are you getting the equation of the median AM IS: (q2+q3-2q1)x-(p2+p3-2p1)y+(p2q1-q2p1)+(p3q1-q3p1)=0??

I thought you would do AM=M-A which is the way I have done on my attachment?

Also I may be wrong here but aren't the p's meant to go with x and the q's with y?

Originally Posted by MINOANMAN

James

find first the equation of the median AM it is the line that connects the vertex A with the mid point of BC .
This line has equation as per your coordinates of the vertices A(p1,q1), B(p2,q2),C(p3,q3).
the equation of the median :
AM IS: (q2+q3-2q1)x-(p2+p3-2p1)y+(p2q1-q2p1)+(p3q1-q3p1)=0

the median BE is : (q1+q3-2q2)x-(p1+p3-2p2)y+(p3q2-q3p2)+(p1q2-q1p2)=0

the median CZ is : (q1+q2-2q3)x-(p1+p2-2p3)y+(p1q3-q1p3)+(p2q3-q2p3)=0

if these 3 lines meet in one point then they form a pencil of lines.
the condition that 3 lines form a pencil of lines is that the determinant of their coefficients = 0
ex given if the lines a1x+b1y+C1=0 , a2x+b2y+c2=0 and a3x+b3y+c3=0 form a pencil of lines then the determinant of the coefficients a1,b2,c1,...etc =0
solve the system of the 3 equations I have given you to find the common their point. WHOSE COORDINATES MUST BE (p1,+p2+p3)/3 AND (q1+q2+q3)/3
this is the barycenter of the triangle ABC...
I hope you understand all the above...
MINOAS

Hey Please ignore my previous quote post since I have now realised that it is simple geometry not vectors!
But I do have a question. What do you mean by the determinant of the coeefficients?
Also can't you just solve them simultaneously?
Also what do you mean by the lines forming a pencil?
I would have thought a pencil is straight but to me it seems that there is no way these three lines should form a straight line?

James

A pencil of lines is just a set of lines that all meet at a certain point called the centre of the pencil.
if 3 lines L1,L2 and L3 are lines of the same pencill then there exist numbers k1,k2 and k3 such that k1xL1+k2xL2+k3xL3 =0
in our case k1=k2=k3=1 ie. if you just add all the three equations of the lines I gave you they will be =0 this means that the three lines are members of the same pencil
in other words they meet in the same point all of them this point is the well known point called the barycenter of the trangle ABC and its coordinates are the ones I mntioned above.
Now go here and check what i mean by determinant.
Determinant - Wikipedia, the free encyclopedia
if 3 lines meet at a point their coefficients form a 3x3 determinant that is = 0
sent me a private message if you need more explanations.

Minoas

Originally Posted by MINOANMAN

James

A pencil of lines is just a set of lines that all meet at a certain point called the centre of the pencil.
if 3 lines L1,L2 and L3 are lines of the same pencill then there exist numbers k1,k2 and k3 such that k1xL1+k2xL2+k3xL3 =0
in our case k1=k2=k3=1 ie. if you just add all the three equations of the lines I gave you they will be =0 this means that the three lines are members of the same pencil
in other words they meet in the same point all of them this point is the well known point called the barycenter of the trangle ABC and its coordinates are the ones I mntioned above.
Now go here and check what i mean by determinant.
Determinant - Wikipedia, the free encyclopedia
if 3 lines meet at a point their coefficients form a 3x3 determinant that is = 0
sent me a private message if you need more explanations.

Minoas

Hey Minoas when you're doing this adding the coeficients determinant thing; do you have to add the x coeficients then the y ones then the ones at the end separately?
IE Do you deal with them separately? Or do you just add them all together?
Also one thing I don't understand is you said that since these lines form a pencil then then there exists K1 K2 and K3 which are then multiplied by the coeficients.
How do you know that in this case they are equal to 1? Wouldn't it kind of muck things up if they were all different?
Are there cases where they are all different?
If they are all different and you add the coefficients does the sum still come to zero?
Also I have to find the coordinates of the point of intersection so will I have to solve two of the equations simultaneously
or will the determinant thing show me the coordinates somehow? I think I will have to solve two of them simultaneously?
Am I right?

James
Pls follow me carefully.

1. if 3 lines L1,L2,L3 are members of a pencil then there exist 3 real numbers k1,k2,k3, such that K1XL1+K2XL2+K3XL3=0 this is a well known theorem of coordinate Geometry for pencils.
In our case since there is a cyclic symmetry of the parameters p1,p2,p3,q1,q2,q3 if you ad the 3 equations you goet 0 that is k1=k2=k3=1.

Since this might be difficult for you to compile...there is another way to verify if 3 lines are members of the same pencil...
get the 3 equations and get the coefficients...only not x not the y...form a 3x3 determinant and check if it is 0 if it is zero then your lines intersect at a point. as simple as such.
for example is 3x+2y-5=0 , 3x+5y+2=0 and 4x-2y-7=0 then the determinat of the coefficient is first raw" 3,2,-5" second raw "3,5,2" and third "4,-2,-7"
find the determinant which is 95 thus different from zero and consequently the 3 lines are not forming a pencil. you may verify this by drawing the three lines.
good luck
if you need any more guidance pls send me a private msg
MINOAS

Originally Posted by MINOANMAN

James
Pls follow me carefully.

1. if 3 lines L1,L2,L3 are members of a pencil then there exist 3 real numbers k1,k2,k3, such that K1XL1+K2XL2+K3XL3=0 this is a well known theorem of coordinate Geometry for pencils.
In our case since there is a cyclic symmetry of the parameters p1,p2,p3,q1,q2,q3 if you ad the 3 equations you goet 0 that is k1=k2=k3=1.

Since this might be difficult for you to compile...there is another way to verify if 3 lines are members of the same pencil...
get the 3 equations and get the coefficients...only not x not the y...form a 3x3 determinant and check if it is 0 if it is zero then your lines intersect at a point. as simple as such.
for example is 3x+2y-5=0 , 3x+5y+2=0 and 4x-2y-7=0 then the determinat of the coefficient is first raw" 3,2,-5" second raw "3,5,2" and third "4,-2,-7"
find the determinant which is 95 thus different from zero and consequently the 3 lines are not forming a pencil. you may verify this by drawing the three lines.
good luck
if you need any more guidance pls send me a private msg
MINOAS

Hey Minoas thanks for showing me this pencil thing but I have never studied this before and since the paper is meant to be done using methods I have studied I do not think I am meant to use this method. I think I am simply meant to solve the equations simultaneously. I have checked the markscheme and indeed all they do is solve them simultaneously.
I am not quite sure how to solve these first two simultaneously but I am working on it.
Thanks for your help anyway.