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deeksha sharma Grade: 11
        

 What is the area of triangle with vertices


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

7 years ago

Answers : (5)

AKASH GOYAL AskiitiansExpert-IITD
419 Points
										

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


AskiitiansExpert-IITD


 


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7 years ago
vikas askiitian expert
510 Points
										

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

7 years ago
rajan jha
49 Points
										

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


LIKE MY ANSWER?

7 years ago
Devasish Bindani
45 Points
										

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

7 years ago
FITJEE
43 Points
										

3.87

4 years ago
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