Let \(u\) and \(v\) be two vectors in \(\mathbf{R}^{2}\) whose Euclidean norms…
2017
Let \(u\) and \(v\) be two vectors in \(\mathbf{R}^{2}\) whose Euclidean norms satisfy ‖\(u\)‖=2‖\(v\)‖. What is the value of \(\alpha\) such that \(w = u + \alpha v\) bisects the angle between \(u\) and \(v\)?
- A.
2
- B.
1/2
- C.
1
- D.
-1/2
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Correct answer: A
Key idea: A vector w bisects the angle between u and v if it makes equal angles with u and v. Equivalently, (w·u)/||u|| = (w·v)/||v||.
Let w = u + α v. Using the angle-bisector condition and ||u|| = 2||v||, we have (w·u)/||u|| = (w·v)/||v||, which simplifies to w·u = 2 w·v.
Substitute w = u + α v: (u + α v)·u = 2 (u + α v)·v.
Expand: u·u + α v·u = 2 u·v + 2α v·v. Rearranged: ||u||^2 - 2 u·v + α( u·v - 2||v||^2 ) = 0.
Use ||u||^2 = 4||v||^2 to get 4||v||^2 - 2 u·v + α( u·v - 2||v||^2 ) = 0, so α( u·v - 2||v||^2 ) = 2( u·v - 2||v||^2 ).
If u·v ≠ 2||v||^2 (the nondegenerate case), divide both sides to obtain α = 2.
Degenerate case: if u·v = 2||v||^2 then u and v are collinear and point in the same direction, so any positive scalar multiple of v added to u keeps the same direction and the bisector notion is not unique.
Therefore, in the general (nondegenerate) case, α = 2.