Question

calculate the area of triangle whose vertices are given by (3,-1,2),(1,-1,2)and (4,-2,1)

Answer 1

Answer:

Say A --> (3,-1,2), B--> (1,-1,2) and C--> (4,-2,1)

__

AB = (1-3)i + (-1-(-1))j + (2-2)k = -2i

__

AC = (4-3)i + ((-2)-(-1))j + (1-2)k = i-j-k

__ __

So area of triangle is given by ∆ = 1/2|AB × AC|

∆ = 1/2 |i j k|

|-2 0 0|

|1 -1 -1|

Solving the determinant gives, ∆= (1/2)[i(0-0)-j(2-0)+k(2-0)] = -j+k

So area is the modulus of the vector, which is √{(-1)^2+1^2} = √2