Question

find the area of the triangle whose vertices are A(1,2,3), B(2,3,1) and C(3,1,2)​

Answer 1

Answer:

3/2

Step-by-step explanation:

Apply Heron's method

Answer 2

Answer:

≈ 2.6 square units

Step-by-step explanation:

Given the vertices of the Triangle ABC are A(1,2,3), B(2,3,1) and C(3,1,2)​

so, the Area of the Triangle is given by,

= 1/2 * ║AB→ X AC→║

Here,

AB→ = ( 2-1 , 3-2 , 1-3 )  = (1,1,-2)

AC→ = ( 3-1, 1-2 , 2-3 )  = (2,-1,-1)

║AB→ X AC→║ = ║ i→  j→  k→ ║

                            ║1     1    -2 ║

                            ║2    -1    -1 ║

                         =  i→(-1 - 2) - j→(-1+4) + k→(-1-2)

                         = -3i→ -3j→ -3k→

∴AB→ X AC→ = (-3,-3,-3)

so, the area of the ΔABC = 1/2║(-3,-3,-3)║

                                          = 1/2 * √ [(-3)²+(-3)²+(-3)²]

                                          = 1/2 * √27

                                          = 1/2 * 3√3

                                          = 0.5 * 3*1.732

                                         = 2.598 sq.units

                                         ≈ 2.6 square units