This is a cross product problem:
Given uxv=<2,-3,0>
find (u-5v)x(u+4v)
I appreciate your help Thank you!
(u-5v)x(u+4v)
= uxu - 5vxu + 4uxv - 20vxv
= 9uxv (why?)
Steve before you start asking why may I tell you that you completely did not understand the question this is the cross product between two vectors. x is not a variable it is a symbol of operation. So try again, you cannot just distribute.
To find the cross product of (u-5v) and (u+4v), we first need to calculate the individual components of (u-5v) and (u+4v) and then apply the cross product formulas:
Let's start by expanding (u-5v) and (u+4v):
(u-5v) = u - 5v = (2,-3,0) - 5(2,-3,0) = (2,-3,0) - (10,-15,0) = (-8,12,0)
(u+4v) = u + 4v = (2,-3,0) + 4(2,-3,0) = (2,-3,0) + (8,-12,0) = (10,-15,0)
Now, we can calculate the cross product of (-8,12,0) and (10,-15,0):
To find the cross product, we'll use the following formula:
(u-5v) x (u+4v) = (a1, a2, a3) x (b1, b2, b3) = (a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1)
Calculating the individual components:
a1 = -8, a2 = 12, a3 = 0
b1 = 10, b2 = -15, b3 = 0
Now, we can use the formula to find the cross product:
(-8,12,0) x (10,-15,0) = ((12*0) - (0*-15), (0*10) - (-8*0), (-8*-15) - (12*10))
= (0 - 0, 0 - 0, 120 + 120)
= (0,0,240)
Therefore, (u-5v) x (u+4v) = (0,0,240).
To find the cross product, we can use the following formula:
u x v = (u2v3 - u3v2)i - (u1v3 - u3v1)j + (u1v2 - u2v1)k
First, we need to find u - 5v and u + 4v separately.
Given:
u = <2, -3, 0>
v = <5, -1, 0>
u - 5v = <2, -3, 0> - 5<5, -1, 0>
= <2, -3, 0> - <25, -5, 0>
= <-23, 2, 0>
Similarly,
u + 4v = <2, -3, 0> + 4<5, -1, 0>
= <2, -3, 0> + <20, -4, 0>
= <22, -7, 0>
Now, let's calculate the cross product of (u - 5v) and (u + 4v).
(u - 5v) x (u + 4v) = ((-23)(-7) - (2)(0))i - ((-23)(0) - (2)(22))j + ((-7)(0) - (-23)(22))k
= (-161 - 0)i - (0 - 44)j + (0 + 506)k
= -161i - (-44)j + 506k
= <-161, 44, 506>
Therefore, (u - 5v) x (u + 4v) = <-161, 44, 506>.