[Math] Doubt on finding the area of a triangle formed by the tips of vectors

vectors

Question :

Find the area of a triangle formed by the tips of vectors a=
i-j-3k , b= 4i-3j+k , c= 3i-j+ 2k.

Fig 1 is the representation of the triangle.

Fig 2 and 3 show my solution

enter image description here

Solution in the book

In the book , first ab was found .

ab = -3i +2j -4k

then cb was found.

cb = –i + 2j + k

Then (ab)X (ab )
=10i + 7j – 4k

In the book , the magnitude of this cross product was stated as the area ofthe
parallelogram formed in Fig 1.Halving the magnitude gave the area of the triangle.
Final answer : 6.4 sq unit

But I cannot understand why was the cross product of (ab) and (cb ) found .

My solution is as under :

Fig 2 shows my opinion.

The area of the parallelogram in fig 1 is the same as the area of the
parallelogram formed between vectors a and c as in fig 2.

This is shown in Fig 3.

a X c = -5i -8j – 4k

Then the magnitude of cross product of a and c should give
the area of the parallelogram.

But I know my solution is incorrect.

I cannot understand why is it incorrect.

Please provide me the explanation.

THANK YOU VERY MUCH.

Best Answer

The vectors $\vec{a}$ and $\vec{b}$ start from the origin. So what you have designated as $\vec{a}$ and $\vec{c}$ in the second figure are actually the vectors $\vec{a} - \vec{b}$ and $\vec{c}-\vec{b}$. Your method is correct, provided that the bottom left hand corner of the parallelogram is actually the origin, i.e. $\vec{b} = \vec{0}$.