`` find the area of the parallelogram spanned by vectors given, pls alo verify answer

u=`<< -4,1>>` and v=`<< 4,4>>`

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Use the cross product of the two vectors U X V to find the area of the parallelogram:

UxV = |u1v2-u2v1| = |-4*4 - 1*4| = 20

To verify:

Use b * h to find the area of the parallelogram:

|u|=`sqrt(1^2+4^2)=sqrt(17)`

|v|=`sqrt(4^2+4^2)=sqrt32`

Adding |u|+|v| = (0,5) and |u+v| = 5

Using law of cosines to find the angle between u and v:

cos A =`(17+32-25)/(2*sqrt17*sqrt32)` then A= 59.037

Area of parallelogram = b x h = `sqrt32*sqrt17*sin(59.037)=20`

**Sources:**

The area of a parallelogram enclosed by two vectors is given by the magnitude of the cross product of the vectors. Since the vectors are only in the plane, we can make vectors with the z-component zero for each vector.

Then

`u=(-4,1,0)` and

`v=(4,4,0)`

Now the cross product of two vectors `a=(a_1,a_2,a_3)` and `b=(b_1,b_2,b_3)` is:

`a times b=(a_2b_3-b_2a_3,b_1a_3-a_1b_3,a_1b_2-b_1a_2)`

but in this case, the z-components are zero so the cross product is

`u times v = (0,0,-4(4)-1(4))=(0,0,-20)`

which means that the area of the parallelogram is the magnitude of this, which is 20.

**The area of the parallelogram is 20.**

To verify that this is the area, we know that the area of a parallelogram is also ` `

`A=|u||v|sin theta` where `theta` is the angle between the two vectors.

`|u|=sqrt{(-4)^2+1^2}=sqrt17`

`|v|=sqrt{4^2+4^2}=sqrt32`

and by the dot product, the angle between two vectors is

`theta=cos^{-1}({u cdot v}/{|u||v|})`

`=cos^{-1}({-16+4}/{sqrt17sqrt32})approx 120.96`

when taking the sine of the angle and then multiplying those values together, we get 20.

**The area of the parallelogram is 20.**

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