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. Find a general expression for a nonzero vector orthogonal to the plane containing the curve. $$\begin{aligned}\mathbf{r}(t)=&(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j} \\ &+(e \cos t+f \sin t) \mathbf{k},\end{aligned}$$ where \(\langle a, c, e\rangle \times\langle b, d, f\rangle \neq \mathbf{0}\).

Short answer

Question: Find a general expression for a non-zero vector orthogonal to the plane containing the curve defined by the position vector \(\mathbf{r}(t) = (a\cos t + b\sin t) \mathbf{i} +(c\cos t + d\sin t) \mathbf{j} + (e\cos t + f\sin t) \mathbf{k}\). Answer: \(\mathbf{v} = (cf-de) \mathbf{i} + (af-be) \mathbf{j} + (ad-bc) \mathbf{k}\)

Step-by-step solution

Tangent vector computation

First, differentiate the position vector \(\mathbf{r}(t)\) with respect to the parameter \(t\) in order to find the tangent vector \(\mathbf{dr/dt}\): $$\begin{aligned} \frac{d\mathbf{r}}{dt}(t) &= \left(-a \sin t + b \cos t\right) \mathbf{i} + \left(-c \sin t + d \cos t\right) \mathbf{j} + \left(-e \sin t + f \cos t\right) \mathbf{k}\end{aligned}$$

Vector orthogonal to the plane

Now that we have the tangent vector, let's compute the cross product of the position vector \(\mathbf{r}(t)\) and the tangent vector \(\mathbf{dr/dt}\). By using the general formula for the cross product: $$\mathbf{v} = \mathbf{r} \times \frac{d\mathbf{r}}{dt} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ a\cos t+b\sin t & c\cos t+d\sin t & e\cos t+f\sin t\\ -a\sin t+b\cos t & -c\sin t+d\cos t & -e\sin t+f\cos t \end{vmatrix} $$

Computing cross product

Calculate the determinant of the matrix above for \(\mathbf{v}\): $$\mathbf{v} = (cf-de) \mathbf{i} + (af-be) \mathbf{j} + (ad-bc) \mathbf{k}$$ Now, \(\mathbf{v}\) is a general expression for a nonzero vector orthogonal to the plane containing the curve. Therefore, the answer is: $$\mathbf{v} = (cf-de) \mathbf{i} + (af-be) \mathbf{j} + (ad-bc) \mathbf{k}$$

Key concepts

Cross Product

The cross product is a mathematical operation that can be performed on two vectors in three-dimensional space. It results in a third vector that is orthogonal (perpendicular) to the plane formed by the two original vectors.

The resulting vector's magnitude is equal to the area of the parallelogram spanned by the original vectors, and its direction is given by the right-hand rule. If we have two vectors \(\mathbf{a}\) and \(\mathbf{b}\), their cross product \(\mathbf{a} \times \mathbf{b}\) is calculated using the determinant of a matrix formed by the unit vectors \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) and the components of the vectors \(\mathbf{a}\) and \(\mathbf{b}\).

In the context of the textbook exercise, the cross product is used to find a vector orthogonal to the plane in which a curve lies. By taking the cross product of the position vector \(\mathbf{r}(t)\) and its derivative with respect to time \(\mathbf{dr/dt}\), we get a new vector that is orthogonal to the tangent of the curve at any point, and hence, orthogonal to the plane of the curve itself.

Tangent Vector

A tangent vector to a curve at a given point is a vector that represents the direction of the curve at that point. Mathematically, it's obtained by differentiating the position vector of the curve with respect to its parameter.

In the exercise, we calculate the tangent vector by differentiating the position vector \(\mathbf{r}(t)\) with respect to \(t\). The function \(\mathbf{r}(t)\) describes the position of a point on the curve at any given time, and its derivative \(\mathbf{dr/dt}\) represents the instantaneous rate of change of the position, which is the tangent vector. It points in the direction of the curve's motion and is tangent to the curve, giving us essential information about the curve's trajectory at any point \(t\).

Parametric Equations

Parametric equations represent a set of related quantities as explicit functions of an independent variable, known as the parameter. These equations are often used in physics and engineering to describe the trajectories of objects, where time is a common parameter.

In our exercise, \(\mathbf{r}(t)\) is given in parametric form, where \(t\) is the parameter, and the equations for \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) components depend on trigonometric functions of \(t\). Parametric equations are particularly useful for representing curves in space, as they can easily incorporate the complex motions that result from combinations of basic functions like sines and cosines.

Determinant of a Matrix

The determinant of a matrix is a special number that provides important properties about the matrix and the linear transformation it represents. For a three-dimensional matrix, it can be interpreted as the scaled volume of the parallelepiped determined by its column or row vectors.

When calculating a cross product using the determinant of a matrix, as in our exercise, the values within the matrix help determine the components of the resulting vector. The formula for the determinant of a 3x3 matrix involves subtraction and multiplication of diagonal products, leading to the coefficients seen in the orthogonal vector \(\mathbf{v}\). It's important to be precise when calculating determinants, as errors can lead to incorrect vectors and thus incorrect interpretations of geometric relationships.