Determine the value(s) of x such that the area of the parallelogram formed by the vectors a=(x+1, 1, -2) and b=(x,3,0) is the square root of 41.
My cross-product was < 6, -2x, 2x+3 >
So the magnitude is √(36 + 4x^2 + 4x^2 + 12x + 9 )
= √(8x^2 + 12x + 45)
√(8x^2 + 12x + 45) = √41
square both sides:
8x^2 + 12x + 4 = 0
2x^2 + 3x + 1 = 0
(2x + 1)(x + 1) = 0
x = -1/2 or x = -1
check the x = -1
your two vectors are < 0, 1, -2 > and < -1, 3, 0 >
cross-product is < 6, 2, 1>
magnitude or length = √(36+4+1) = √41
The area of a parallelogram is the magnitude of the cross products of the vectors forming it.
give it a shot, let me know what you did.
I got -4 and -8 as my answers, but when is substituted them in the original vectors and found the area I didn't get root 41.
I messed up while factoring, because I forgot to reduce it. Thank you!
To find the area of a parallelogram formed by two vectors, we can use the formula:
Area = |a x b|
where "a x b" represents the cross product of vectors a and b.
Let's calculate the cross product of vectors a and b:
a x b = ((1)(0) - (1)(3), (-2)(0) - (x + 1)(0), (x + 1)(3) - (x)(1))
= (-3, 0, 3x + 3 - x)
= (-3, 0, 2x + 3)
Now we can find the magnitude of vector a x b:
|a x b| = sqrt((-3)^2 + 0^2 + (2x + 3)^2)
= sqrt(9 + 4x^2 + 12x + 9)
= sqrt(4x^2 + 12x + 18)
We want the area to be the square root of 41. So, we can set up the equation:
sqrt(4x^2 + 12x + 18) = sqrt(41)
To solve for x, we can square both sides of the equation:
4x^2 + 12x + 18 = 41
Rearranging the equation:
4x^2 + 12x - 23 = 0
Now we can solve this quadratic equation using the quadratic formula:
x = (-b ± √(b^2 - 4ac))/2a
Plugging in the values: a = 4, b = 12, c = -23
x = (-12 ± √(12^2 - 4(4)(-23)))/(2(4))
x = (-12 ± √(144 + 368))/8
x = (-12 ± √512)/8
x = (-12 ± 16√2)/8
Simplifying the expression:
x = (-3 ± 4√2)/2
Therefore, the values of x such that the area of the parallelogram formed by the vectors a and b is the square root of 41 are:
x = (-3 + 4√2)/2 and x = (-3 - 4√2)/2