Multivariable Calculus - Parallelogram

[Larrytsai]

Homework Statement

Suppose u=(-2,-10) and v=(-2,-2) are two vectors that form the sides of a parallelogram. Then the lengths of the two diagonals of the parallelogram are...

Homework Equations

The Attempt at a Solution

I tried using pythagrean theorem and some trig to find the angles, then i thot finding the internal angle of the paralellogram i can divide it by 2 and find half of the diagonal then multiply it by 2. The angle i found was 16.85 degrees. with that i would use
cos(16.85)*sqrt(104) to find half of the length of the diagonal.
Am i on the right track?

 

[mighty2000]

Thank you for your question. Yes, you are on the right track! The method you have described is a good approach to finding the lengths of the diagonals in a parallelogram. However, there is a simpler and more direct way to find the lengths of the diagonals.

First, let's label the points of the parallelogram as A=(-2,-10), B=(-2,-2), C=(x,y), and D=(x+2,y+8). We can see that the diagonals of the parallelogram are the line segments AC and BD. To find the length of AC, we can use the distance formula:
Distance AC = √[(x-(-2))^2 + (y-(-10))^2] = √[(x+2)^2 + (y+10)^2]

Similarly, the length of BD can be found using the distance formula:
Distance BD = √[(x+2-(-2))^2 + (y+8-(-2))^2] = √[(x+4)^2 + (y+10)^2]

Now, since we know that the opposite sides of a parallelogram are equal in length, we can set these two distances equal to each other:
√[(x+2)^2 + (y+10)^2] = √[(x+4)^2 + (y+10)^2]

Simplifying this equation, we get:
x^2 + y^2 + 4x + 20y + 104 = x^2 + y^2 + 8x + 16y + 116

We can cancel out the x^2 and y^2 terms on both sides, and then solve for x and y to find the coordinates of point C. Once we have the coordinates of point C, we can use the distance formula again to find the lengths of the diagonals AC and BD.

I hope this helps! Let me know if you have any further questions.