Question

The vertices of a triangle are given by i+3j+2k , 2i-j+k and -i+2j+3k , then the area of the triangle is

Answer 1

Answer:

The vertices are A (1 3 2), B(2 -1 1) C(-1 2 3) AB =(1 -4 -1) AC=(-2 -1 1)

ABxAC=[ i j k ]

[ 1-4 -1]

[-2 -1 1]

=(-5)i -(-1)j+(-9)k

=-5i+j-9k

|ABxAC|=√(25+1+81)=√(107)

Area=1/2(√(107)

Answer 2

Given,

Vertices of triangle are i+3j+2k , 2i-j+k and -i+2j+3k

To Find,

Area of the triangle.

Solution,

Now the point of vertices are

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

So, AB =(1 -4 -1),

AC=(-2 -1 1)

So,

ABxAC=[ i j k ]

[ 1-4 -1]×[-2 -1 1]

=(-5)i -(-1)j+(-9)k

=-5i+j-9k

For area first we have to find out |ABxAC|.

|ABxAC|

=√(25+1+81)

=√(107)unit

Then, area will be=1/2|ABxAC|

                               =1/2√(107)sq. unit

Hence, area of the triangle is 1/2√(107)sq. unit