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
Solution in the book
In the book , first a - b was found .
a - b = -3i +2j -4k
then c - b was found.
c - b = -i + 2j + k
Then (a - b)X (a - b ) =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 (a - b) and (c - b ) 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.
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. $\vec{b} = \vec{0}$.