Question-28
Let \(u\) and \(v\) be two vectors in \(\mathbb{R}^{2}\) that satisfy \(||u|| = 2 ||v||\). What is the value of \(\alpha\) such that \(w = u + \alpha v\) bisects the angle between \(u\) and \(v\)?
\(2\)
If we have two vectors \(x\) and \(y\) such that \(||x|| = ||y||\), then the parallelogram formed by these two vectors turns out to be a rhombus. By the paralellogram law of vector addition, we know that \(x\) and \(y\) form two sides of a parallelogram whose diagonal is \(x + y\). When \(||x|| = ||y||\), this parallelogram is actually a rhombus, which is a parallelogram all of whose sides are equal. In a rhombus, the diagonals bisect the angle between the sides. Hence \(x + y\) is the angle bisector of \(x\) and \(y\).
We can now use this idea to solve the given problem. We are given that \(w = u + \alpha v\). If \(\alpha = 2\), then we have \(w = u + 2v\), in which the vectors being summed have equal norm. This comes from the data given to us: \(||u|| = 2||v||\). This guarantees that \(w\) is an angle bisector of \(u\) and \(2v\), which is the same as the angle bisector of \(u\) and \(v\). Recall that scaling a vector by a positive constant doesn’t change its direction. Hence, \(\boxed{\alpha = 2}\).