Final answer:
To find vector ~u satisfying both the cross product ~v × ~u = < 4, 2, 5 > and the magnitude condition ||~u|| = 3, we solve the system of equations derived from the cross product and verify the magnitude condition for each possible vector option.
Step-by-step explanation:
To solve for vector ~u given the vector ~v = < 3, −1, −2 > and the vector product ~v × ~u = < 4, 2, 5 >, along with the condition ||~u|| = 3, we utilize the properties of cross products and vector magnitudes.
First, ensure that the magnitude of ~u gives us 3, which is the square root of the sum of the squares of its components. Thus, if ~u = < x, y, z >, we must have ∙(x^2 + y^2 + z^2) = 3^2 = 9.
Then, calculating the cross product of ~v and a general vector ~u = < x, y, z >, we must get the components of < 4, 2, 5 >. The cross product in terms of the determinants of a matrix composed by unit vectors i, j, k and the components of vectors ~v and ~u is:
|i j k ||3 -1 -2||x y z |
This results in < 3z + 2y, -3x + 2z, xy -3y > which must equal < 4, 2, 5 >. So, we have a system of equations:
3z + 2y = 4 (1),
-3x + 2z = 2 (2),
xy -3y = 5 (3).
Solving these equations and checking against each vector option A, B, C, D, only one will satisfy both the cross product result and the magnitude condition.