Solving this problem that involves the area of triangle

[SammyS]

Mark, is this you?

 

[neilparker62]

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.

 

[Mark44]

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.

 

[neilparker62]

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.

 

[robphy]

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,

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.

 

[neilparker62]

Area of Triangle ABC:
$$=a_j(a_i-b_i)-\frac{1}{2} [a_ia_j-b_ib_j +(a_j-b_j)(a_i-b_i)]$$ $$ =a_ja_i-a_jb_i -\frac{1}{2}[a_ia_j-b_ib_j+a_ja_i-a_jb_i-b_ja_i+b_jb_i]$$ $$=\frac{1}{2}(b_ja_i-b_ia_j)$$ Or: $$=\frac{1}{2}|a||b| \sin(\theta_{B}-\theta_{A})\\=\frac{1}{2}|a||b| (\sin\theta_{B}\cos\theta_{A}-\cos\theta_{B}\sin\theta_{A})$$ $$=\frac{1}{2}|a||b|\left( \frac{b_{j} a_{i} }{|b||a|} - \frac{b_{i} a_{j} }{|b||a|}\right)$$ $$=\frac{1}{2}(b_{j}a_{i}-b_{i}a{j}) $$ All roads lead to Rome!

 

[Prof B]

Calculate the area of the trapezoid and then subtract the areas of the two triangular pieces.

Why don't you teach her about vectors and then ask her which is simpler? When one of the three points is the origin, cross-product is the clear winner.

 

[Mark44]

Mark, is this you?

Sorry, I missed this one from you, Sammy. Yes, that's me all right.

 

[Mark44]

Why don't you teach her about vectors and then ask her which is simpler?

Are you referring to my 9-year-old grand niece? I'm confident that vectors would go right over her head.

 

[chwala]

Sorry, I missed this one from you, Sammy. Yes, that's me all right.

Nice humour @Mark44

 

[neilparker62]

Why don't you teach her about vectors and then ask her which is simpler? When one of the three points is the origin, cross-product is the clear winner.

All roads lead to Rome and in this case Rome is indeed (half of) the vector cross product.