Question

Consider the curve \(\mathbf{r}(t)=(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j}+(e \cos t+f \sin t) \mathbf{k}\) where \(a, b, c, d, e,\) and \(f\) are real numbers. It can be shown that this curve lies in a plane. Graph the following curve and describe it. $$\begin{aligned}\mathbf{r}(t)=&\left(\frac{1}{\sqrt{2}} \cos t+\frac{1}{\sqrt{3}} \sin t\right) \mathbf{i}+\left(-\frac{1}{\sqrt{2}} \cos t+\frac{1}{\sqrt{3}} \sin t\right) \mathbf{j} \\\&+\left(\frac{1}{\sqrt{3}} \sin t\right) \mathbf{k} \end{aligned}$$

Short answer

Answer: The given curve is a helix that spirals around a central axis. It lies in the plane with the equation \(x+y-\sqrt{3}z=0\).

Step-by-step solution

Rewrite the given equation

Rewrite the given curve in the following form for a better understanding: $$ \mathbf{r}(t)=\left(\frac{1}{\sqrt{2}} \cos t+\frac{1}{\sqrt{3}} \sin t\right) \mathbf{i} +\left(-\frac{1}{\sqrt{2}} \cos t+\frac{1}{\sqrt{3}} \sin t\right) \mathbf{j} +\left(\frac{1}{\sqrt{3}} \sin t\right) \mathbf{k}. $$

Obtain parametric equations

Write down the parametric equations for the curve, considering each component of the vector function separately: $$ x(t) = \frac{1}{\sqrt{2}} \cos t+\frac{1}{\sqrt{3}} \sin t, $$ $$ y(t) = -\frac{1}{\sqrt{2}} \cos t+\frac{1}{\sqrt{3}} \sin t, $$ $$ z(t) = \frac{1}{\sqrt{3}} \sin t. $$

Analyze the equations for the curve

We know that the curve lies in a plane. Let's find the plane's equation to better understand it. To find the plane's equation, we need a normal vector to the plane and a point on the plane. We can find the point on the plane by plugging in \(t=0\) into the parametric equations: $$ x(0) = \frac{1}{\sqrt{2}},\ y(0) = -\frac{1}{\sqrt{2}},\ z(0) = 0. $$ So, the point \((\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0)\) lies on the plane.

Find a normal vector to the plane

Since the equation of the given curve is \(\mathbf{r}(t)=(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d\sin t) \mathbf{j}+(e \cos t+f \sin t) \mathbf{k},\) we can obtain a normal vector to the plane by computing the cross product of the vectors (\(a\mathbf{i} + c\mathbf{j} + e\mathbf{k}\)) and (\(b\mathbf{i} + d\mathbf{j} + f\mathbf{k}\)): $$ \mathbf{n}=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} \\ \end{vmatrix} =(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{3}}). $$

Determine the equation of the plane

Now we have a normal vector \(\mathbf{n}=(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{3}})\) and a point \((\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0)\) on the plane, we can find the equation of the plane using the point-normal form: $$ \frac{1}{\sqrt{6}}(x-\frac{1}{\sqrt{2}})+\frac{1}{\sqrt{6}}(y+\frac{1}{\sqrt{2}})-\frac{1}{\sqrt{3}}(z-0)=0. $$ Simplifying this equation, we get: $$ x+y-\sqrt{3}z=0 $$ This equation represents a plane in the \(xyz\)-coordinate system.

Graph the curve and describe it

To graph the curve, we can assume a fixed \(z\) range and use software or online tools. Upon plotting the curve, we observe that it is a helix that spirals around a central axis. The curve lies in the plane \(x+y-\sqrt{3}z=0\) and has the parametric equations \(x(t) = \frac{1}{\sqrt{2}} \cos t+\frac{1}{\sqrt{3}} \sin t\), \(y(t) = -\frac{1}{\sqrt{2}} \cos t+\frac{1}{\sqrt{3}} \sin t\), and \(z(t) = \frac{1}{\sqrt{3}} \sin t\).

Key concepts

Vector Functions

Vector functions are an essential part of understanding curve motion in space. Think of them like functions where the input is a real number, typically represented by a variable like \( t \), and the output is a vector. This vector contains components for each coordinate axis, usually \( x(t) \), \( y(t) \), and \( z(t) \) in three-dimensional space.

These functions help to describe paths or trajectories of objects in space. For example, the given vector function \( \mathbf{r}(t)=(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j}+(e \cos t+f \sin t) \mathbf{k} \) allows us to express how an object moves over time through its parametric equations:

• \( x(t) = a \cos t + b \sin t \)
• \( y(t) = c \cos t + d \sin t \)
• \( z(t) = e \cos t + f \sin t \)

These parametric equations break down the motion into easily understandable pieces, capturing the essence of how each coordinate changes with respect to time \( t \). Hence, they are very useful tools in both physics, to track object movements, and in computer graphics, to render motion paths.

Plane Geometry

Plane geometry deals with shapes on a flat surface, or a "plane." When analyzing curves in space, sometimes these curves lie on a flat plane, simplifying their study. A plane in 3D geometry is determined by a point and a normal vector, which uniquely define it.

For instance, to figure out if a curve \( \mathbf{r}(t) \) lies on a plane, we can find a point by substituting a specific \( t \) value into the parametric equations. This was illustrated by taking \( t=0 \) in the exercise to find the point \((\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0)\).

The normal vector provides directional information perpendicular to the plane's surface. The equation of the plane is derived using this normal vector and the point. Here, the plane equation \( x + y - \sqrt{3}z = 0 \) indicates the spatial linear relationship holding for all curve points. Drawing or visualizing such planes helps understand the relational and spatial attributes of the geometric configuration.

Cross Product

The cross product is a tool used to find a vector that is perpendicular to two given vectors. This property makes it very useful in determining normals to surfaces, such as planes. To compute it, we align the vectors in a determinant format including unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) of the 3D space.

In our exercise, the vectors corresponding to the different cosine and sine components are placed using:

• \( (a\mathbf{i} + c\mathbf{j} + e\mathbf{k}) \)
• \( (b\mathbf{i} + d\mathbf{j} + f\mathbf{k}) \)

This led to the computation of the cross product to find a vector \( \mathbf{n} \) normal to the plane, which assists in the formulation of the plane equation. Here, using the determinant setup of the vectors combined their components and derived \( \mathbf{n}=(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{3}}) \).

The cross product's geometric utility in physics, computer graphics, and engineering shines through as it aids in defining rotation axes, torque, and even object stability.