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Reply 1
franks
Is there a way of finding whether 2 vectors are perpendicular ?
For example if a triangle is right angled? You see, I know how to work out if they are parallel, or lie on a straight line but just not perpendicular !!!!
For example if a triangle is right angled? You see, I know how to work out if they are parallel, or lie on a straight line but just not perpendicular !!!!
Two vectors are perpendicular if a.b=0
Reply 2
iluvcheesecake
Two vectors are perpendicular if a.b=0
(where a and b are the 2 vectors of course)
err nope
I have a triangle AOB and I have told that vector AB is 6a and vector OB is 6b , I also know that M is midpoint of OA, P lies on AB so that AP:PB is 2:1 and N is the midpoint of OP.
I want to know if this is a right angled triangle, cos I have worked out the correct answer to one of the questions assuming this - but I need to prove it...
I really dont see how the dot product has anything to do with the above - please enlighten me !!
I have a triangle AOB and I have told that vector AB is 6a and vector OB is 6b , I also know that M is midpoint of OA, P lies on AB so that AP:PB is 2:1 and N is the midpoint of OP.
I want to know if this is a right angled triangle, cos I have worked out the correct answer to one of the questions assuming this - but I need to prove it...
I really dont see how the dot product has anything to do with the above - please enlighten me !!
In triangle OAB
OA = 6a
OB = 6b
Midpoint of OA is M
P lies on AB so AP:PB = 2:1
The midpoint of OP is N
Show that the area of the quadrilateral AMNP is half the area of triangle OAB
I dont actually think that I have done the question the correct way, my method just happens to work. My friend has done it a completely different way (also assuming various things) and also got the 'correct' answer.... any more ideas now the question is up ?
I got my answer by working out area of AOB , OMN and OBP ....
OA = 6a
OB = 6b
Midpoint of OA is M
P lies on AB so AP:PB = 2:1
The midpoint of OP is N
Show that the area of the quadrilateral AMNP is half the area of triangle OAB
I dont actually think that I have done the question the correct way, my method just happens to work. My friend has done it a completely different way (also assuming various things) and also got the 'correct' answer.... any more ideas now the question is up ?
I got my answer by working out area of AOB , OMN and OBP ....
franks
err nope
I have a triangle AOB and I have told that vector AB is 6a and vector OB is 6b , I also know that M is midpoint of OA, P lies on AB so that AP:PB is 2:1 and N is the midpoint of OP.
I want to know if this is a right angled triangle, cos I have worked out the correct answer to one of the questions assuming this - but I need to prove it...
I really dont see how the dot product has anything to do with the above - please enlighten me !!
I have a triangle AOB and I have told that vector AB is 6a and vector OB is 6b , I also know that M is midpoint of OA, P lies on AB so that AP:PB is 2:1 and N is the midpoint of OP.
I want to know if this is a right angled triangle, cos I have worked out the correct answer to one of the questions assuming this - but I need to prove it...
I really dont see how the dot product has anything to do with the above - please enlighten me !!
Lets try listing the facts which we can gather from your information.
OA = OB+BA = OB-AB = 6b-6a=OA=6(b-a)
OB = 6b.
AB = 6a.
M is the midpoint of OA so OM = 3(b-a).
P lies on AB so AP:PB is 2:1. AP=2PB. OP-OA=2[OB-OP]. OP-OA = 2OB - 2OP. 3OP = 2OB + OA. 3OP = 12b + 6b-6a = 18b-6a. OP = 6b - 2a.
N is the midpoint of OP. ON = 3b-a
Now lets just think..
The triangle is AOB.
(OA)^2 = 36(b-a)^2 = 36(b^2 - 2ba + a^2)
(OB)^2 = 36b^2
(AB)^2 = 36a^2
But
(OA)^2 + (OB)^2 NOT equal to (AB)^2
(OA)^2 + (AB)^2 NOT equal to (OB)^2
(OB)^2 + (AB)^2 NOT equal to (OA)^2
Either i have made an error somewhere or the triangle is not right angled.
To be honest i don't see why we were given information about N and M, as they aren't part of the supposedly right angled triangle.
Where is the question from?
Ar = area (of triangle)
a = |a|, etc.
let ON = NP = d
let AB = 6c
let MON = θ
let OAP = OAB = ø
OM = MA = 3a
OB = 6b
AP = 4c
Ar(OMN) = ½(3a)dsinθ
Ar(OAP) = ½(6a)(2d)sinθ
=> Ar(OMN) = ¼Ar(OAP)
Ar(AMNP) = Ar(OAP) - Ar(OMN) = Ar(OAP) - ¼Ar(OAP)
Ar(AMNP) = (3/4)Ar(OAP)
===================
Ar(OAP) = ½(6a)(4c)sinø
Ar(OAP) = 12acsinø
Ar(OAB) = ½(6a)(6c)sinø
Ar(OAB) = 18acsinø
=> Ar(OAP) = (2/3)Ar(OAB)
=>Ar(AMNP) = (3/4)(2/3)Ar(OAB)
Ar(AMNP) = ½Ar(OAB)
=================
a = |a|, etc.
let ON = NP = d
let AB = 6c
let MON = θ
let OAP = OAB = ø
OM = MA = 3a
OB = 6b
AP = 4c
Ar(OMN) = ½(3a)dsinθ
Ar(OAP) = ½(6a)(2d)sinθ
=> Ar(OMN) = ¼Ar(OAP)
Ar(AMNP) = Ar(OAP) - Ar(OMN) = Ar(OAP) - ¼Ar(OAP)
Ar(AMNP) = (3/4)Ar(OAP)
===================
Ar(OAP) = ½(6a)(4c)sinø
Ar(OAP) = 12acsinø
Ar(OAB) = ½(6a)(6c)sinø
Ar(OAB) = 18acsinø
=> Ar(OAP) = (2/3)Ar(OAB)
=>Ar(AMNP) = (3/4)(2/3)Ar(OAB)
Ar(AMNP) = ½Ar(OAB)
=================
Maybe you don't need the number for this question:
I think another solution:
For the triangle OAB, as M is the mid-point of OA and N is the mid-point of OP so MN is parallel to AB; the distance from O to AB (called OH) is double distance from MN to AB ( accroding to Tarlet's theory).
In triangle OAP ; MN is parallel to AP and equal to a hafl of AP. But AP = 2/3 AB
So MN = 1/3 AB.
The area of triangle OAB = 1/2 *OH*AB
The area of quadrilateral AMNP = 1/2*(1/2*OH)*(MN+AP)
= 1/4*OH*AB
Therefore the area of (AMNP) is a half of (AOB)
Is that right??
I think another solution:
For the triangle OAB, as M is the mid-point of OA and N is the mid-point of OP so MN is parallel to AB; the distance from O to AB (called OH) is double distance from MN to AB ( accroding to Tarlet's theory).
In triangle OAP ; MN is parallel to AP and equal to a hafl of AP. But AP = 2/3 AB
So MN = 1/3 AB.
The area of triangle OAB = 1/2 *OH*AB
The area of quadrilateral AMNP = 1/2*(1/2*OH)*(MN+AP)
= 1/4*OH*AB
Therefore the area of (AMNP) is a half of (AOB)
Is that right??
ok, take three - i have just deleted this message twice 
thanks soo much for helping me with that question guys, just to clear something up - I only wanted the triangle to be rightangled so that I could justify my method (as I assumed the traingle was right angled) ... but now I see how to do it - similar triangles etc. thanks for explaining it
There is another part of the question which I have got an answer to (12/5) , but I dont think is right :
If the line AN is extended it crosses OB at point C
OC = kb
find k
Thank you all so much for all your help
thanks soo much for helping me with that question guys, just to clear something up - I only wanted the triangle to be rightangled so that I could justify my method (as I assumed the traingle was right angled) ... but now I see how to do it - similar triangles etc. thanks for explaining it
There is another part of the question which I have got an answer to (12/5) , but I dont think is right :
If the line AN is extended it crosses OB at point C
OC = kb
find k
Thank you all so much for all your help
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