Solving this problem that involves the area of triangle

[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.