What is the area of triangle with vertices

(1,-1,2), (2,1,-1), (3,-1,2)

Dear Deeksha

First calculate the sides of the triangle

let A=(1,-1,2), B=(2,1,-1), C=(3,-1,2)

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

AB=i + 2j - 3k

Similarly

side AC= 2i

side BC=i - 2j + 3k

area of triangle= 0.5 |(cross product of AB and AC)|

                    = 0.5|ABxAC|

                   =0.5|-4k-6j|

                   =0.5(√42 + 62)

                   =√15

                 = 3.87

All the best.

AKASH GOYAL

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let A,B,C are the vertices of triangle

A,B,C are (1,-1,2) , (2,1,-1) , (3,-1,2) respectively

position vector of A is i-j+2k,

position vector of B is 2i+j-k & 

position vector of C is 3i-j+2k

now ,

             AB =B-A=i+2j-3k  &  BC=B-C=i-2j+3k

area of triangle is (AB)*(BC)= -6j -4k              (* represents cross product)

                                      =sqrt(36+16)      in magnitude

                                      =2sqrt13 square units

THE AREA OF THE TRIANGLE WILL BE 1/2(DETERMINANTOF THE COOORDINATES);

LIKE MY ANSWER?

let A=(1,-1,2)

    B=(2,1,-1)

    C=(3,-1,2)

by distance formula AB=(14)^1/2  BC=(14)^1/2  CA=2

height=(15)^1/2

we know that area of triangle is 1/2*(base)*(height)

thus area ABC=1/2*2*(15)^1/2 =(15)^1/2

3.87