Solving this problem that involves the area of triangle

It looks like you're trying to use the cross product formula for the area of a triangle, but that formula only applies to a triangle with vertices at the origin and the two points specified by the two vectors. The triangle in question has vertices at the origin, A, and B, which are specified by the vectors (-1, 1, 0) and (1/λ, 1/λ^2, 0). As I mentioned earlier, the problem can be solved without using vectors, cross products, or any fancy trig. The only thing you need to know is the formula for the area of a trapezoid and the Pythagorean theorem.
  • #1
chwala
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Homework Statement
See attached
Relevant Equations
Area of triangle
1651323882253.png


I am still looking at this question. One thing that i know is that the

distance ##AB=\dfrac {(λ+1)\sqrt {2λ^2-2λ+1}}{λ^2}##
distance ##OA=\sqrt 2##
distance ##OB=\dfrac{\sqrt{λ^2+1}}{λ^2}##
Perpendicular distance from point ##B## to the line ##OA=\dfrac{\sqrt{2(λ^4+2λ^3+2)}}{2λ^2}##
Therefore, the area of the triangle is given by;
##\dfrac {1}{2} ×OA ×\dfrac{\sqrt{2(λ^4+2λ^3+2)}}{2λ^2}##
##=\dfrac {1}{2} ×\sqrt 2 ×\dfrac{\sqrt{2(λ^4+2λ^3+2)}}{2λ^2}##
##=\dfrac {1}{2} ×\sqrt 2 ×\sqrt 2 ×\dfrac{\sqrt{λ^4+2λ^3+2}}{2λ^2}##
##=\dfrac{\sqrt{λ^4+2λ^3+2}}{2λ^2}##
Now as i look at this am in doubt on how they arrived at the height ##h=\dfrac {1}{\sqrt {2λ^2-2λ+1}}##...did they assume that ##O## will meet the line ##AB## at its mid point? If so, then the solution given may not be correct... otherwise, i do not see any other way...am still trying to figure this out.

Are they also assuming that the line ##AB## is parallel to the ##x- axis##? Point ##B## could be anywhere on the first quadrant since we are told that ##λ>0##.
 
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  • #2
By vector product the area is
[tex]\frac{1}{2}|\vec{OA} \times \vec{OB}|=\frac{1}{2}(\lambda^{-1}+\lambda^{-2})[/tex]
which does not coincide with you.
 
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  • #4
anuttarasammyak said:
By vector product the area is
[tex]\frac{1}{2}|\vec{OA} \times \vec{OB}|=\frac{1}{2}(\lambda^{-1}+\lambda^{-2})[/tex]
which does not coincide with you.
Thanks didn't think of vector calculus lol...

##A=\dfrac{1}{2}\left |||(-1,1,0)×({\dfrac{1}{λ}}, {\dfrac{1}{λ^2}},0)\right|||##
##A=\dfrac{1}{2}\sqrt\frac {(1+λ)^2}{λ^2}##
##A=\dfrac{1+λ}{2λ^2}##
 
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  • #5
As you have three perimeters, alternatively you can make use of heron's formula.
 
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  • #6
anuttarasammyak said:
As you have three perimeters, alternatively you can make use of heron's formula.
That would be quite long...i had initially thought of that approach...will nevertheless try and see what comes out...
 
  • #8
You and the other posters here are making the problem much more difficult than it needs to be. The problem can be done without invoking vectors or anything more complicated than plain old geometry. Not even trig. Triangle OAB is a portion of the trapezoid with vertices (-1, 0), A(-1, 1), B##(1/\lambda, 1/\lambda^2)##, and ##(1/\lambda, 0)##.

Calculate the area of the trapezoid and then subtract the areas of the two triangular pieces.
 
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  • #9
chwala said:
Are they also assuming that the line AB is parallel to the x−axis?
No.
 
  • #10
Mark44 said:
You and the other posters here are making the problem much more difficult than it needs to be. The problem can be done without invoking vectors or anything more complicated the plain old geometry. Not even trig. Triangle OAB is a portion of the trapezoid with vertices (-1, 0), A(-1, 1), B##(1/\lambda, 1/\lambda^2)##, and ##(1/\lambda, 0)##.

Calculate the area of the trapezoid and then subtract the areas of the two triangular pieces.
:biggrin: I will look at your approach @Mark44.
Having said that is my solution i.e Finding the area of the triangle in post ##1## also correct?
 
  • #11
chwala said:
Finding the area of the triangle in post 1 also correct?
No, since the value you got isn't equal to ##\frac{1 + \lambda}{2\lambda^2}##.
 
  • #12
Mark44 said:
You and the other posters here are making the problem much more difficult than it needs to be. The problem can be done without invoking vectors or anything more complicated than plain old geometry. Not even trig. Triangle OAB is a portion of the trapezoid with vertices (-1, 0), A(-1, 1), B##(1/\lambda, 1/\lambda^2)##, and ##(1/\lambda, 0)##.

Calculate the area of the trapezoid and then subtract the areas of the two triangular pieces.
Alternatively, a rectangle to which a right-angled triangle is added and a smaller one subtracted.

I did it with trig: quite straightforward
 
  • #13
Matcon said:
Alternatively, a rectangle to which a right-angled triangle is added and a smaller one subtracted.

I did it with trig: quite straightforward
This isn't an alternative approach -- it's exactly the same technique as what I described, except that I didn't mention that the triangles are right triangles. Drawing a sketch of the problem would show that the two triangles are right triangles. I agree that the problem is very straightforward.
 
  • #14
Mark44 said:
This isn't an alternative approach -- it's exactly the same technique as what I described, except that I didn't mention that the triangles are right triangles. Drawing a sketch of the problem would show that the two triangles are right triangles. I agree that the problem is very straightforward.
Probably the math required for your trapezoid method is also too simple.
 
  • #15
Mark44 said:
No, since the value you got isn't equal to ##\frac{1 + \lambda}{2\lambda^2}##.
True, i agree...My intention is to ask ...What if the question was specific i.e
'Find the area of the triangle' and not necessarily 'show'.
Would my steps to solution be correct?

distance ##OA=\sqrt 2##
Perpendicular distance from point ##B## to the line ##OA=\dfrac{\sqrt{2(λ^4+2λ^3+2)}}{2λ^2}##
Therefore, the area of the triangle is given by;
##A = \frac{1}{2} × base × height##
##A=\dfrac {1}{2} ×OA ×\dfrac{\sqrt{2(λ^4+2λ^3+2)}}{2λ^2}##
##A=\dfrac {1}{2} ×\sqrt 2 ×\dfrac{\sqrt{2(λ^4+2λ^3+2)}}{2λ^2}##
##A=\dfrac {1}{2} ×\sqrt 2 ×\sqrt 2 ×\dfrac{\sqrt{λ^4+2λ^3+2}}{2λ^2}##
##A=\dfrac{\sqrt{λ^4+2λ^3+2}}{2λ^2}##
 
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  • #16
neilparker62 said:
Height also using vector cross product.

$$h=\frac{|\vec{AO} \times \vec{AB}|}{|\vec{AB}|}$$

Edit to add reference for above method:

https://qc.edu.hk/math/Advanced Level/Point_to_line.htm
Ok i managed to find the height but not necessarily using the formula;
$$h=\frac{|\vec{AO} \times \vec{AB}|}{|\vec{AB}|}=\frac{|\vec{AO} \times \vec{AB}|}{|\vec{AB}|}$$
I used the distance from point ##O## to the midpoint of the line ##AB##.

I am calling this the line ##OM## therefore,

$$h=\frac{0.5×|\vec{OA} \times \vec{OB}|}{|\vec{OM}|}$$
##OM=\sqrt{\dfrac{(1-λ)^2}{(2λ)^2}+\dfrac{(1+λ^2)^2}{(2λ^2)^2}}##
##OM=\sqrt{\dfrac{λ^2(1-2λ+λ^2)+(1+2λ+λ^2)}{4λ^4}}##
...
##OM=\dfrac{(λ+1)\sqrt{(2λ^2-2λ+1)}}{2λ^2}##

Using,
$$h=\frac{0.5×|\vec{OA} \times \vec{OB}|}{|\vec{OM}|}$$
##h=\dfrac{1+λ}{2λ^2} ⋅\dfrac{2λ^2}{(λ+1)\sqrt{(2λ^2-2λ+1)}}=\dfrac{1}{\sqrt{(2λ^2-2λ+1)}}##
 
  • #17
chwala said:
Perpendicular distance from point B to the line
Something is wrong in the derivation. (a,b) and (-b,-a ) are symmetric as for the line OA. Thus perpendicular line length from (a,b) to line OA is |a+b|/sqrt2.
 
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  • #18
chwala said:
True, i agree...My intention is to ask ...What if the question was specific i.e
'Find the area of the triangle' and not necessarily 'show'.
Would my steps to solution be correct?
No. For the explanation of why it isn't, see below.
For this problem, the only difference between "Find the area" and "show that such-and-such is the area" is that in the first, you wouldn't have been given the solution.
chwala said:
distance ##OA=\sqrt 2##
Perpendicular distance from point ##B## to the line ##OA=\dfrac{\sqrt{2(λ^4+2λ^3+2)}}{2λ^2}##
Therefore, the area of the triangle is given by;
##A = \frac{1}{2} × base × height##
That's not the area of triangle AOB. That's the area of a right triangle ABC, where C is some point on the line through AO. And C is almost certainly different from O, the origin.
chwala said:
##A=\dfrac {1}{2} ×OA ×\dfrac{\sqrt{2(λ^4+2λ^3+2)}}{2λ^2}##
##A=\dfrac {1}{2} ×\sqrt 2 ×\dfrac{\sqrt{2(λ^4+2λ^3+2)}}{2λ^2}##
##A=\dfrac {1}{2} ×\sqrt 2 ×\sqrt 2 ×\dfrac{\sqrt{λ^4+2λ^3+2}}{2λ^2}##
##A=\dfrac{\sqrt{λ^4+2λ^3+2}}{2λ^2}##
If your alternate technique would have been correct, you would have gotten the same as the given answer.
 
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  • #19
Mark44 said:
You and the other posters here are making the problem much more difficult than it needs to be. The problem can be done without invoking vectors or anything more complicated than plain old geometry. Not even trig. Triangle OAB is a portion of the trapezoid with vertices (-1, 0), A(-1, 1), B##(1/\lambda, 1/\lambda^2)##, and ##(1/\lambda, 0)##.

Calculate the area of the trapezoid and then subtract the areas of the two triangular pieces.
You brilliant @Mark44 ...wow! eeissssh!

Using this we have;
##A_{trapezium} =\dfrac{1}{2} ×\left[ \dfrac{1+λ}{λ}\right]×\left[\dfrac{λ^2+1}{λ^2}\right]##
##A_{trapezium} =\dfrac{λ^3+λ^2+λ+1}{2λ^3}##
##A_{triangle1}=\dfrac{1}{2}× 1×1=\dfrac{1}{2}##
##A_{triangle2}=\dfrac{1}{2}×\dfrac{1}{λ} ×\dfrac{1}{λ^2} =\dfrac{1}{2λ^3}##
##A_{Required}=A_{trapezium}- [A_{triangle1}+A_{triangle2}]=\dfrac{λ^3+λ^2+λ+1}{2λ^3}-\dfrac{λ^3+1}{2λ^3}=\dfrac{λ(λ+1)}{2λ^3}=\dfrac{λ+1}{2λ^2}##
 
  • #20
Mark44 said:
You and the other posters here are making the problem much more difficult than it needs to be. The problem can be done without invoking vectors or anything more complicated than plain old geometry. Not even trig. Triangle OAB is a portion of the trapezoid with vertices (-1, 0), A(-1, 1), B##(1/\lambda, 1/\lambda^2)##, and ##(1/\lambda, 0)##.

Calculate the area of the trapezoid and then subtract the areas of the two triangular pieces.
What's 'difficult' about:

$$h=\frac{|\vec{AO} \times \vec{AB}|}{|\vec{AB}|}$$

or - for that matter:

$$Area=\frac{|\vec{AO} \times \vec{AB}|}{2}$$
 
  • #21
neilparker62 said:
What's 'difficult' about:

$$h=\frac{|\vec{AO} \times \vec{AB}|}{|\vec{AB}|}$$

or - for that matter:

$$Area=\frac{|\vec{AO} \times \vec{AB}|}{2}$$
@neilparker62 all variations are pretty good! I think @Mark44 was just pointing out a more straightforward approach to the problem... and of course on a light note...:smile:
 
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  • #22
neilparker62 said:
What's 'difficult' about:

$$h=\frac{|\vec{AO} \times \vec{AB}|}{|\vec{AB}|}$$

or - for that matter:

$$Area=\frac{|\vec{AO} \times \vec{AB}|}{2}$$
I'm not saying that's difficult, just that there is a much simpler solution that depends just on some basic geometry formulas. I'm all for doing things in the simplest possible way that will get the job done.
 
  • #23
Mark44 said:
I'm not saying that's difficult, just that there is a much simpler solution that depends just on some basic geometry formulas. I'm all for doing things in the simplest possible way that will get the job done.
Ok - let's do the Maths:
$$A=\frac{|\vec{AB} \times \vec{AO}|}{2}=0.5\;\left \lvert \left(\frac{\lambda+1}{\lambda},\frac{1-\lambda^2}{\lambda^2} \right) \times (-1,1)\right \rvert=\frac{\lambda+1}{2\lambda^2}$$
To find h, we already have the numerator of the given expression (twice area) so all we need do is apply Pythagoras to determine ##|\vec{AB}|##:
$$ |\vec{AB}|=\sqrt{ \left( \frac{1+\lambda}{\lambda} \right)^2+\left( \frac{1-\lambda^2}{\lambda^2} \right)^2 }=\frac{(1+\lambda) \sqrt{2\lambda^2-2\lambda+1}}{\lambda^2} $$ $$\implies h=\frac{|\vec{AB} \times \vec{AO}|}{|\vec{AB}|} = \frac{1}{\sqrt{2\lambda^2-2\lambda+1}}$$
 
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  • #24
Like I said, my approach is simpler. It doesn't rely on knowledge of vectors or cross products, topics that might not be known to students who have not been exposed to anything beyond very basic geometry; i.e., formulas for the areas of trapezoids and right triangles.
 
  • #25
Mark44 said:
Calculate the area of the trapezoid and then subtract the areas of the two triangular pieces.
The OP already did the area calculation using your method and that's all all good but I fail to see how it's simpler than the vector calculation above. And the determination of h is yet to be done.

The vector calculation could be 'disguised' as geometry/trig if required. But perhaps we should just agree that if the vector 'toolbox' is at your disposal, you should probably use it. We don't have to assume any particular "lowest common denominator" in respect of level of Maths.
 
  • #26
neilparker62 said:
I fail to see how it's simpler than the vector calculation above.
The work I showed was strictly for the area calculation of triangle AOB. My definition of "simpler" in regard to this calculation has to do with the order in which topics are presented to math students. That is, arithmetic, algebra, geometry, trigonometry, analytic geometry, calculus, and so on. The technique I provided could be understood by anyone who knows nothing more than how to find the area of a trapezoid and the area of a right triangle. For the calculation of h and ##\lambda## more sophisticated techniques than my area calculation would likely be required, but I didn't tackle either of those.
neilparker62 said:
The vector calculation could be 'disguised' as geometry/trig if required. But perhaps we should just agree that if the vector 'toolbox' is at your disposal, you should probably use it. We don't have to assume any particular "lowest common denominator" in respect of level of Maths.
I disagree. If A and B are two techniques for solving a problem, but technique A can be grasped by many more people, it is what I would call a simpler technique. I think if we could ask Richard Feynmann, he would also agree that invoking plain old geometry would be the more elegant approach.
 
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  • #27
We'll agree to disagree. You think your method is simplest by your definition of simple and I think my method is simplest by my definition of simple. I would say the same in respect of what constitutes an "elegant' solution. By the way I do teach at the level your method addresses so make no mistake I do appreciate 'plain old geometry' - "half the chord subtended by twice the angle" and all.
 
  • #28
neilparker62 said:
By the way I do teach at the level your method addresses so make no mistake I do appreciate 'plain old geometry'
And so did I for 20+ years, with two of those years teaching the entire high school math curriculum at a very small school, as well as all but one of the courses on offer at a much larger community college.
 
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  • #30
neilparker62 said:
You think your method is simplest by your definition of simple and I think my method is simplest by my definition of simple.
By my definition of "simpler" I think my 9-year-old grandniece could be able to successfully calculate the area part of this problem with only a brief lesson on the areas of trapezoids and right triangles. Last summer she was able to calculate the volume in her head of some solids made from an educational kit she was playing with. The solids could be formed out of 2D plastic pieces with magnetic edges. I think it would be much more difficult to get her to do the calculation using vectors and cross products. That's what I mean by simple versus not as simple.
 
  • #31
Mark, is this you?
:headbang:
 
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  • #32
Mark44 said:
By my definition of "simpler" I think my 9-year-old grandniece could be able to successfully calculate the area part of this problem with only a brief lesson on the areas of trapezoids and right triangles. Last summer she was able to calculate the volume in her head of some solids made from an educational kit she was playing with. The solids could be formed out of 2D plastic pieces with magnetic edges. I think it would be much more difficult to get her to do the calculation using vectors and cross products. That's what I mean by simple versus not as simple.
I have no doubt your grandniece will very shortly (if not already) have the capability to discover for herself the vector cross product formula (at least the 2D one if not 3D as well). And - for good measure - the quadratic formula which also has a geometric proof I believe. Once discovered I also have no doubt she will make good use of them in solving further related problems. And yes , she will benefit enormously from having derived them from first principles in the first instance - rather than by 'rote' learning.
 
  • #33
neilparker62 said:
I have no doubt your grandniece will very shortly (if not already) have the capability to discover for herself the vector cross product formula (at least the 2D one if not 3D as well). And - for good measure - the quadratic formula which also has a geometric proof I believe.
It's going to be a while -- she's only in 4th grade.
 
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  • #34
https://www.desmos.com/calculator/xhwii9id4h

Area of triangle by absolute value of (half) vector product ##A_1## and by geometry ##A_2##. The triangle is formed by the origin and the red and blue points which you can move around within the co-ordinate plane.

Unsurprisingly the respective formulas used must be equivalent since they yield the same result for area.
 
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  • #35
It seems that ##B=\left(\frac{1}{\lambda},\frac{1}{\lambda^2}\right)## for ##\lambda >0##
traces a parabola ## y=x^2## where ##x>0##.
Indeed, let ##X=\frac{1}{\lambda}## so that ##B=(X,X^2)##.
So,
chwala said:
Point ##B## could be anywhere on the first quadrant since we are told that ##λ>0##.
isn't quite true. It's not "anywhere in the first quadrant", but on this parabola in the first quadrant.

So, your triangle is made up of chords of the parabola ##y=x^2##
with fixed points at the parabola's vertex ##O## and another point at ##A=(-1,1)##
and the variable point ##B=(x,x^2)##, where ##x=\frac{1}{\lambda}##.

While this doesn't change the problem, it might change the way you attack and interpret the problem.
 
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