Question

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

Answer 1

We will use Herons formula to find area  formed by the vectors
So,
A=√(s(s-a)(s-b)(s-c))
where s is semiperimeter
s=(a+b+c)/2
Let x,y,z be the described vectors instead of a,b,c
Now,
a=║x-y║
a=║i-j-3k - ( 4i-3j+k)║
a=║-3i + 2j - 4k║
a= √29
a=5.385

b=║y-z║
b=║4i-3j+k - (3i-j + 2k)║
b=║i -2j -k║
b=√6
b=2.449

c=║z-x║
c=║3i-j + 2k - (i-j-3k )║
c=║2i+5k║
c=√29
c=5.385

So,
s=(5.385+5.385+2.449)/2
s=6.6095
Now,
Area will be
A=√(s(s-a)(s-b)(s-c))
A=√6.6095(6.6095-5.385)(6.6095-2.449)(6.6095-5.385)
A=6.4120

So, the area is 6.4120 square units

Answer 2

VECTOR METHOD :

Final Answer : √165 /2 sq. units = 6.42 sq. units

Steps:
1) We have,
A = (1,-1,-3)
B = (4,-3,1)
C = (3,-1,2)

We know Area of Triagle ABC in vector Form:
= 1/2 (|AB x AC|)

2) AB = ( 4,-3,1)-(1,-1,-3) = (3, -2,4)
AC = (3,-1,2)-(1,-1,-3) = (2,0,4)

Now,
See Pic :
AB x AC = (-10,-7,4)
| AB x AC | =
$\sqrt{ {( - 10)}^{2} + {( - 7)}^{2} + ( {4}^{2} ) } = \sqrt{165}$

Therefore,
Area of Triangle = 1/2 (| AB, x AC | )
= 1/2 (√165 )
= 12.845/2 = 6.42 sq units