Question

Suppose \(\mathbf{u}=\left[\begin{array}{l}u_{1} \\ u_{2}\end{array}\right]\) and \(\mathbf{v}=\left[\begin{array}{l}v_{1} \\ v_{2}\end{array}\right]\) are not orthogonal; that is, \((\mathbf{u}, \mathbf{v}) \neq 0\) (a) Show that the quadratic equation $$ (\mathbf{u}, \mathbf{v}) k^{2}+\left(\|\mathbf{v}\|^{2}-\|\mathbf{u}\|^{2}\right) k-(\mathbf{u}, \mathbf{v})=0 $$ has a positive root \(k_{1}\) and a negative root \(k_{2}=-1 / k_{1}\). (b) Let \(\mathbf{u}_{1}^{(1)}=\mathbf{u}-k_{1} \mathbf{v}, \mathbf{v}_{1}^{(1)}=\mathbf{v}+k_{1} \mathbf{u}, \mathbf{u}_{1}^{(2)}=\mathbf{u}-k_{2} \mathbf{v},\) and \(\mathbf{v}_{1}^{(2)}=\mathbf{v}+k_{2} \mathbf{u},\) so that \(\left(\mathbf{u}_{1}^{(1)}, \mathbf{v}_{1}^{(1)}\right)=\left(\mathbf{u}_{1}^{(2)}, \mathbf{v}_{1}^{(2)}\right)=0,\) from the discussion given above. Show that $$ \mathbf{u}_{1}^{(2)}=\frac{\mathbf{v}_{1}^{(1)}}{k_{1}} \quad \text { and } \quad \mathbf{v}_{1}^{(2)}=-\frac{\mathbf{u}_{1}^{(1)}}{k_{1}} $$ (c) Let \(\mathbf{U}_{1}, \mathbf{V}_{1}, \mathbf{U}_{2},\) and \(\mathbf{V}_{2}\) be unit vectors in the directions of \(\mathbf{u}_{1}^{(1)}, \mathbf{v}_{1}^{(1)}, \mathbf{u}_{1}^{(2)},\) and \(\mathbf{v}_{1}^{(2)}\), respectively. Conclude from (a) that \(\mathbf{U}_{2}=\mathbf{V}_{1}\) and \(\mathbf{V}_{2}=-\mathbf{U}_{1},\) and that therefore the counterclockwise angles from \(\mathbf{U}_{1}\) to \(\mathbf{V}_{1}\) and from \(\mathbf{U}_{2}\) to \(\mathbf{V}_{2}\) are both \(\pi / 2\) or both \(-\pi / 2\).

Short answer

Answer: The counterclockwise angles between the unit vectors are both either $\pi/2$ or both $-\pi/2$.

Step-by-step solution

Find the roots of the quadratic equation

Given the quadratic equation, $$ (\mathbf{u}, \mathbf{v}) k^{2}+\left(\|\mathbf{v}\|^{2}-\|\mathbf{u}\|^{2}\right) k-(\mathbf{u}, \mathbf{v})=0 $$ Since it is a quadratic equation, then it must have two roots. Define these roots to be \(k_1\) and \(k_2\). By using the quadratic formula, \(k_1\) and \(k_2\) can be found by the relation: $$ k_{1,2}=-\frac{\left(\|\mathbf{v}\|^{2}-\|\mathbf{u}\|^{2}\right)\pm\sqrt{\left(\|\mathbf{v}\|^{2}-\|\mathbf{u}\|^{2}\right)^{2}+4(\mathbf{u}, \mathbf{v})^2}}{2(\mathbf{u}, \mathbf{v})} $$ Calculate the discriminant of the quadratic equation (within the square root): $$ \Delta=(\|\mathbf{v}\|^{2}-\|\mathbf{u}\|^{2})^2+4(\mathbf{u}, \mathbf{v})^2 $$ Notice that \((\mathbf{u},\mathbf{v}) \neq 0\) and the square of the norms is non-negative. Thus, \(\Delta > 0\), meaning there are two distinct roots. Moreover, since the product of the roots \(k_{1}k_{2}=-\frac{(\mathbf{u},\mathbf{v})}{(\mathbf{u},\mathbf{v})}=-1\), the roots must have opposite signs. Now we have found two roots \(k_1\) and \(k_2\) with opposing signs.

Define and prove the vector relationships

Given the vectors $$ \mathbf{u}_{1}^{(1)}=\mathbf{u}-k_{1} \mathbf{v}, \mathbf{v}_{1}^{(1)}=\mathbf{v}+k_{1} \mathbf{u}, \mathbf{u}_{1}^{(2)}=\mathbf{u}-k_{2} \mathbf{v}, \, \text{ and } \mathbf{v}_{1}^{(2)}=\mathbf{v}+k_{2} \mathbf{u}, $$ with both \((\mathbf{u}_{1}^{(1)},\mathbf{v}_{1}^{(1)})\) and \((\mathbf{u}_{1}^{(2)},\mathbf{v}_{1}^{(2)})\) equal to zero. First, we have to show $$ \mathbf{u}_{1}^{(2)}=\frac{\mathbf{v}_{1}^{(1)}}{k_{1}} $$ Since $$\mathbf{u}_{1}^{(2)}=\mathbf{u}-k_{2} \mathbf{v} = \mathbf{u} - \frac{1}{k_{1}}\mathbf{v},$$ multiplying both sides of this equation by \(k_1\) gives: $$ k_1\mathbf{u}_{1}^{(2)}=k_1\mathbf{u} - \mathbf{v} $$ And since \(\mathbf{v}_{1}^{(1)}=\mathbf{v}+k_{1} \mathbf{u},\) then $$ k_1\mathbf{u}_{1}^{(2)}=\mathbf{v}_{1}^{(1)}\implies \mathbf{u}_{1}^{(2)}=\frac{\mathbf{v}_{1}^{(1)}}{k_{1}} $$ Now, we must show $$ \mathbf{v}_{1}^{(2)}=-\frac{\mathbf{u}_{1}^{(1)}}{k_{1}} $$ Since $$\mathbf{v}_{1}^{(2)}=\mathbf{v}+k_{2} \mathbf{u} = \mathbf{v} -\frac{1}{k_1}\mathbf{u},$$ multiplying both sides of this equation by \(k_1\) gives: $$ k_1\mathbf{v}_{1}^{(2)}=-\mathbf{u}+k_1\mathbf{v} $$ And since \(\mathbf{u}_{1}^{(1)}=\mathbf{u}-k_{1} \mathbf{v},\) then $$ k_1\mathbf{v}_{1}^{(2)}=\mathbf{u}_{1}^{(1)}\implies \mathbf{v}_{1}^{(2)}=-\frac{\mathbf{u}_{1}^{(1)}}{k_{1}} $$

Investigate unit vectors and counterclockwise angles

Let the unit vectors \(\mathbf{U}_{1}, \mathbf{V}_{1}, \mathbf{U}_{2},\) and \(\mathbf{V}_{2}\) be in the directions of \(\mathbf{u}_{1}^{(1)}, \mathbf{v}_{1}^{(1)}, \mathbf{u}_{1}^{(2)},\) and \(\mathbf{v}_{1}^{(2)},\) respectively. Since \(\mathbf{U}_{2}=\frac{\mathbf{u}_{1}^{(2)}}{\Vert\mathbf{u}_{1}^{(2)}\Vert}\) and \(\mathbf{V}_{1}=\frac{\mathbf{v}_{1}^{(1)}}{\Vert\mathbf{v}_{1}^{(1)}\Vert}\), $$ \mathbf{U}_{2}=\frac{1}{\Vert\mathbf{u}_{1}^{(2)}\Vert}\frac{\mathbf{v}_{1}^{(1)}}{k_{1}}=\frac{1}{k_{1}\Vert\mathbf{u}_{1}^{(2)}\Vert}\mathbf{v}_{1}^{(1)}=\mathbf{V}_{1} $$ Similarly, since \(\mathbf{V}_{2}=\frac{\mathbf{v}_{1}^{(2)}}{\Vert\mathbf{v}_{1}^{(2)}\Vert}\) and \(-\mathbf{U}_{1}=-\frac{\mathbf{u}_{1}^{(1)}}{\Vert\mathbf{u}_{1}^{(1)}\Vert}\), $$ \mathbf{V}_{2}=\frac{1}{\Vert\mathbf{v}_{1}^{(2)}\Vert}\left(-\frac{\mathbf{u}_{1}^{(1)}}{k_{1}}\right)=\frac{-1}{k_{1}\Vert\mathbf{v}_{1}^{(2)}\Vert}\mathbf{u}_{1}^{(1)}=-\mathbf{U}_{1} $$ Therefore, the counterclockwise angles from \(\mathbf{U}_{1}\) to \(\mathbf{V}_{1}\) and from \(\mathbf{U}_{2}\) to \(\mathbf{V}_{2}\) must be both \(\pi / 2\) or both -\( \pi / 2\).

Key concepts

Orthogonal Vectors

Understanding the concept of orthogonal vectors is vital when diving into various fields of mathematics, including linear algebra and vector calculus. Two vectors are said to be orthogonal if the dot product between them is zero. The dot product, denoted \<\mathbf{u},\mathbf{v}\>, represents a scalar value obtained from multiplying the corresponding components of two vectors and summing the results.

In the context of our exercise, when we consider the vectors \(\mathbf{u}\) and \(\mathbf{v}\), being orthogonal means that their dot product would result in zero. However, as the problem specifies that they are not orthogonal, \(\left(\mathbf{u},\mathbf{v}\right) eq 0\), this information was used to construct a quadratic equation relating the vectors and the scalar \(k\). This gives rise to a situation where the vector manipulation seeks a particular scalar multiple that then results in the vectors becoming orthogonal.

Quadratic Equation

A quadratic equation is a second-degree polynomial equation in the variable \(x\), with the general form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). It can be solved using various techniques, including factoring, completing the square, or utilizing the quadratic formula: \[x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].

In our exercise, the quadratic equation formed from the relationship between the vectors \(\mathbf{u}\) and \(\mathbf{v}\) follows this pattern, with \(k\) acting as the variable, and the coefficients derived from the norms and dot product of the vectors. A key component of solving the quadratic is evaluating the discriminant \(\Delta\), which determines the nature and quantity of the roots. Here, we determine that the discriminant is positive, leading to the conclusion that there are two distinct real roots of opposite signs, which deeply impacts the subsequent vector analysis in the problem.

Vector Norms

In mathematics, the norm of a vector is a measure of its length or magnitude. For a vector \(\mathbf{v}\) with components \(v_1, v_2, \ldots, v_n\), the norm (often denoted as \(\|\mathbf{v}\|\)) is calculated using the formula: \[\|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + \ldots + v_n^2}\].

The norm is crucial in determining the scalar multiple in the discussed exercise, which affects the construction and solution of the quadratic equation. It also influences the process of normalizing a vector—turning it into a unit vector. This involves dividing the vector by its norm, a process seen in part (c) of our exercise, when finding the unit vectors in the directions of \(\mathbf{u}_{1}^{(1)}, \mathbf{v}_{1}^{(1)}, \mathbf{u}_{1}^{(2)}, \text{and} \mathbf{v}_{1}^{(2)}\). Knowing how to calculate and utilize vector norms is essential to explore and solve problems involving vector magnitudes and directions.