Let u,v∈R3. Suppose that u is a unit vector,

u⋅v=−3

and

u×v=[1,1,5]

What is the length of the vector v?

Question

Let u,v∈R3.   Suppose that u is a unit vector,

u⋅v=−3

and

u×v=[1,1,5]

What is the length of the vector v?

BlackTom AI · Accepted Answer

Let me break down the given information and use vector identities step by step.

First, note that u is a unit vector, so |u| = 1. The dot product u·v = -3 tells us about the component of v along u. In particular, the parallel component of v to u is a_parallel = (u·v) u = (-3) u, and its magnitude is |a_parallel| = |u·v| = 3.

Next, consider the cross product u × v. Its magnitude is |u × v| = |u| |v_perp| sin(theta), but since v_perp is defined as the component of v perpendicular to u, we have |u × v| = |v_perp| when |u| = 1. Given u × v = [1, 1, 5], its magnitude is |u × v| = sqrt(1^2 + 1^2 + 5^2) = sqrt(27) = 3*sqrt(3). Therefore, the magnitude of the perpendicular component of v to u is |v_perp| = 3*sqrt(3).

Now we can relate the total length of v to its components along and perpendicular to u. Since a_parallel is parallel to u and v_perp is perpendicular to u, they form an orthogonal decomposition of v, and by the Pythagorean theorem:
|v|^2 = |a_parallel|^2 + |v_perp|^2 = 3^2 + (3*sqrt(3))^2 = 9 + 27 = 36.

Finally, taking the square root gives |v| = sqrt(36) = 6.

Thus, the length (magnitude) of v is 6.

Let u,v∈R3. Suppose that u is a unit vector, u⋅v=−3 and u×v=[1,1,5] What is the length of the vector v?数值题

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