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.
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.
4 answers
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.
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
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
I messed up while factoring, because I forgot to reduce it. Thank you!