Area of a parallelogram using 4 points

Question

 

I need some help using vectors to find the area of this parallelogram. I use three points to create two vectors with the same initial points and use a 2x2 determinant to compute the cross product then find it's magnitude. However, I keep getting the wrong answer. Thanks for the help!

 

 

Find the area of the parallelogram with vertices at (-2, -4), (-13, 8), (7, 7), and (-4, 19).

Answer 1

Gregg O is correct.

The vectors in Jesus's case should be

A = (9,11) not (9,8)

B = (11, -12)

 

AxB = 229 k

Answer 2

so who is right?

Answer 3

Well, in order to find the area of the parallelogram or the cross product magnitude, let's first do the cross product of two vectors. In this case I am going to chose the vector which connects the point (-13,8) to (-4,19) which I'll call vector A, and the vector which connects (-13,8) to (-2,4) which I'll call vector B.9

 

Now for Vector A, I can break it up into unit-vector form which will make it 9i + 8j and vector B it's unit-vector form will be 11i - 4j. (example: for Vector B I just need to find the change in x to get 11 for the the i-unit-vector and the change in y for the j-unit vector). Now, I used the right hand rule and made B my first vector on the matrix, but I don't need to do that, I just want a positive result yet it doesn't matter since it asks for the magnitude, which is the absolute value of the cross product. Thus, constructing the matrix we get:

 

| 11 -4 |

| 9    8 | 

 

where the all the i unit vectors are on the left column and all the j unit vector are at the right. All we have to do now is find the absolute value of the determinant of this matrix which is just: |(11*8) - (9*-4)| = |88 + 36| = 124. Thus the magnitude of the cross product and the area of the parallelogram is 124.

Answer 4

You have to make sure that none of the vectors you are using connects opposite corners (cuts through the interior of the parallelogram), so start out with a rough plot to determine which points you'll use for defining the vectors.

 

Let point A be (-2,-4), point B be (-13,8) and point C be (7,7).

vector AB is <-13-(-2), 8-(-4)> = <-11,12>

vector AC is <7-(-2),7-(-4)> = <9,11>

 

Take the cross product: 

|  i    j   k |

|-11 12 0  |

|9    11 0  |

 

= 0i + 0j + (-11*11-9*12)k

= -229k

 

The area is the magnitude of this vector, or

229 square units.