Question

Vectors \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) for lines a. If \(\mathbf{r}(t)=\langle a t, b t, c t\rangle\) with \(\langle a, b, c\rangle \neq\langle 0,0,0\rangle,\) show that the angle between \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) is constant for all \(t>0\) b. If \(\mathbf{r}(t)=\left\langle x_{0}+a t, y_{0}+b t, z_{0}+c t\right\rangle,\) where \(x_{0}, y_{0},\) and \(z_{0}\) are not all zero, show that the angle between \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) varies with \(t\) c. Explain the results of parts (a) and (b) geometrically.

Short answer

In summary, the key difference between parts (a) and (b) is that the line in part (a) passes through the origin, while the line in part (b) does not. This affects the position vector, \(\mathbf{r}\), as the angle between it and the tangent vector, \(\mathbf{r'}\), remains constant for part (a) and varies with \(t\) for part (b). Geometrically, this means that for the line passing through the origin, the rate of change of the position vector remains constant, while it depends on the position of the point in part (b).

Step-by-step solution

Part (a): Show that the angle between \(\mathbf{r}\) and \(\mathbf{r'}\) is constant for all \(t>0\)

We are given that \(\mathbf{r}(t)=\langle a t, b t, c t\rangle\). To find \(\mathbf{r'}\), we differentiate \(\mathbf{r}(t)\) with respect to \(t\): \(\mathbf{r'}(t)=\frac{d\mathbf{r}}{dt}=\langle a, b, c\rangle.\) Now, let's find the angle between \(\mathbf{r}\) and \(\mathbf{r'}\). The dot product of two vectors \(\mathbf{u}\) and \(\mathbf{v}\) is given by \(\mathbf{u}\cdot\mathbf{v}=|\mathbf{u}||\mathbf{v}|\cos{\theta}\), where \(\theta\) is the angle between them. In our case, the angle between \(\mathbf{r}\) and \(\mathbf{r'}\) is given by: \(\cos{\theta} = \frac{\mathbf{r}\cdot\mathbf{r'}}{|\mathbf{r}||\mathbf{r'}|}\) Now, calculate the dot product and magnitudes: \(\mathbf{r}\cdot\mathbf{r'} = (a t)(a)+(b t)(b)+(c t)(c)=t(ab+bc+ca)\) \(|\mathbf{r}|=\sqrt{(a t)^{2}+(b t)^{2}+(c t)^{2}}=t\sqrt{a^{2}+b^{2}+c^{2}}\) \(|\mathbf{r'}|=\sqrt{a^{2}+b^{2}+c^{2}}\) So, the angle between \(\mathbf{r}\) and \(\mathbf{r'}\) is given by: \(\cos{\theta} = \frac{t(ab+bc+ca)}{t^{2}\sqrt{(a^{2}+b^{2}+c^{2})(a^{2}+b^{2}+c^{2)}}}\) After canceling out some terms: \(\cos{\theta} = \frac{ab+bc+ca}{a^{2}+b^{2}+c^{2}}\) We can see that the angle \(\theta\) does not depend on \(t\), so the angle between \(\mathbf{r}\) and \(\mathbf{r'}\) is constant for all \(t>0\).

Part (b): Show that the angle between \(\mathbf{r}\) and \(\mathbf{r'}\) varies with \(t\)

We are given that \(\mathbf{r}(t)=\left\langle x_{0}+a t, y_{0}+b t, z_{0}+c t\right\rangle\). Differentiate \(\mathbf{r}(t)\) with respect to \(t\): \(\mathbf{r'}(t)=\frac{d\mathbf{r}}{dt}=\langle a, b, c\rangle.\) Now, let's find the angle between \(\mathbf{r}\) and \(\mathbf{r'}\) as we did in part (a): \(\cos{\theta} = \frac{(x_{0}+a t)(a)+(y_{0}+b t)(b)+(z_{0}+c t)(c)}{\sqrt{(x_{0}+a t)^{2}+(y_{0}+b t)^{2}+(z_{0}+c t)^{2}}\sqrt{a^{2}+b^{2}+c^{2}}}.\) We can see that the angle \(\theta\) depends on \(t\), so the angle between \(\mathbf{r}\) and \(\mathbf{r'}\) varies with \(t\).

Part (c): Explain the results geometrically

Part (a) represents a line passing through the origin, as \(\mathbf{r}(t)=\langle a t, b t, c t\rangle\) is in the direction of the vector \(\langle a, b, c\rangle\). Since vector \(\mathbf{r'}(t)=\langle a, b, c\rangle\) is constant and the angle between \(\mathbf{r}\) and \(\mathbf{r'}\) is constant, it means that as the point moves along the line, the rate of change of the position vector with respect to time remains constant. On the other hand, in part (b), the line does not pass through the origin and has an initial position of \(\langle x_{0}, y_{0}, z_{0}\rangle\). In this case, the angle between \(\mathbf{r}(t)\) and \(\mathbf{r'}(t)\) depends on \(t\), which means that the rate of change of the position vector depends on the position of the point. This reflects that the vector \(\mathbf{r'}\), which is tangent to the curve, does not remain constant with the position of the point on the curve.

Key concepts

Angle Between Vectors

When dealing with vectors, understanding the angle between them can be crucial in applications across physics, engineering, and geometry. The angle between two vectors, \( \mathbf{u} \) and \( \mathbf{v} \), is determined using the dot product formula: \[\cos{\theta} = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}||\mathbf{v}|}\]Here, \( \theta \) is the angle between the vectors, \( \mathbf{u} \cdot \mathbf{v} \) is the dot product, and \(|\mathbf{u}|\) and \(|\mathbf{v}|\) are the magnitudes of the vectors. The value of \( \theta \) defines how aligned the vectors are:

• \( \theta = 0 \) degrees: vectors are pointing in the same direction, \( \cos{\theta} = 1 \).
• \( \theta = 90 \) degrees: vectors are perpendicular, \( \cos{\theta} = 0 \).
• \( \theta = 180 \) degrees: vectors are pointing in opposite directions, \( \cos{\theta} = -1 \).

Understanding these angles helps in analyzing the projection, alignment, and resultant forces or directions in multi-dimensional spaces.

Vector Derivatives

Vector derivatives explain how a vector-valued function changes as it moves through its domain. In plain terms, it's like taking the derivative in calculus, but applied to vectors. When you have a vector function such as \( \mathbf{r}(t) = \langle a(t), b(t), c(t) \rangle \), you derive each component with respect to \( t \), which gives: \[\mathbf{r}'(t) = \left\langle \frac{da}{dt}, \frac{db}{dt}, \frac{dc}{dt} \right\rangle\]The derivative \( \mathbf{r}'(t) \) not only helps in finding the tangent vector at any point \( t \) but also reveals how the vector or path changes over time. This is particularly useful:

• In motion analysis to determine velocity and acceleration.
• For understanding the curvature or the direction that a path is turning towards.

By understanding vector derivatives, modeling dynamic systems becomes more approachable, whether you're tracking a satellite or designing paths in robotics.

Parametric Equations

Parametric equations provide a way to represent curves using parameters, most commonly through time \( t \). While traditional equations offer a direct relationship (like \( y = f(x) \)), parametric equations express each coordinate individually as a function of \( t \). Consider a line \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle = \left\langle x_0 + at, y_0 + bt, z_0 + ct \right\rangle \). The parameter \( t \) tells us which point on the line we're at. This representation is highly flexible:

• It allows us to describe complex geometries easily
• It is useful in computer graphics to animate scenes over time

In practice, parametric equations simplify where direct expressions would become too complex, especially in 3D modeling and physics simulations where path and position are tied to time.

Geometric Interpretation of Vectors

Vectors are more than just numbers—geometric interpretation places them into our physical world. Each vector is a directed line segment characterized by both magnitude and direction. Consider a vector \( \mathbf{v} = \langle a, b, c \rangle \), where:

• \( a, b, c \) are the components representing dimensions in space (x, y, z).
• The vector's length, or magnitude, \( |\mathbf{v}| = \sqrt{a^2 + b^2 + c^2} \), gives the distance the vector covers.

Geometrically, vectors are visualized as arrows that originate from a point. Their direction shows the line of action, and the length is proportional to their magnitude. Using vectors:

• You can represent forces acting on objects.
• They help in determining directions and velocities.

Understanding vectors geometrically thus provides intuitive insight into various phenomena, be it navigating a 3D space or describing the interaction of forces.