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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
AskiitiansExpert-IITD
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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
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