Find the area of the parallelogram with vertices A(-3, 0),

aspifsGak5u

aspifsGak5u

Answered question

2021-12-11

Find the area of the parallelogram with vertices A(-3, 0), B(-1 , 3), C(5, 2), and D(3, -1).

Answer & Explanation

Janet Young

Janet Young

Beginner2021-12-12Added 32 answers

From the given points, we may choose any three of them such and find a two vector that express the sides of the parallelogram "or one side and a diagonal of the parallelogram", we may choose for example the points B,C and D
Moreover, we can choose point B to be the common point for the two sides "the initial point of the two vector", thus we need to find vector BD and vector BC, using equation as follows
BC=<5+1,23>
=<6,1>
And, the other vector side BD is
BD=<3+1,13>
=<4,4>
And the cross product of both vectors is
BC×BD=<6,1>×<4,4>
=[i^j^k^610440]
=i^[1040]j^[6040]+k^[6144]
=i^0j^+k^((6)(4)(1)(4))
=20k^
Knowing, the cross product of the two vectors of the parallelogram we can use equation to find the area
Area=|20k^|
=20
Jenny Bolton

Jenny Bolton

Beginner2021-12-13Added 32 answers

Assume that the parallelogram plan is the xy plan
The four vertices are A(3,0), B(1,3), C(5,2) and D(3,1)
We can get the are by |AB×BC|
such that |AB| is the vector starting at the point A and ending with B and BC is the vector starting at the point B and ending with C
|AB|=(1(3))i+(30)j=2i+3j
|BC|=(5(1))i+(23)j=6ij
regarding that {i,j,k} are perpendicular unit vector on the {x,y,z}
So
|AB×|BC|=|(2i+3j)×(6ij)
|AB×|BC|=|(02k18k+0)|
|AB×|BC|=|20k=20
hence the area is 20 units

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