Question

Consider the curve. $$\mathbf{r}(t)=\langle a \cos t+b \sin t, c \cos t+d \sin t, e \cos t+f \sin t\rangle$$ 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 $$\mathbf{r}(t)=\langle a \cos t+b \sin t, c \cos t+d \sin t, e \cos t+f \sin t\rangle$$where \(\langle a, c, e\rangle \times\langle b, d, f\rangle \neq 0\)

Short answer

Answer: The vector orthogonal to the plane containing the curve is given by \(\mathbf{n} = \langle de + fc, fa + be, -ac + bd\rangle\).

Step-by-step solution

Find the tangent vector to the given curve

To find the tangent vector to the curve, we take the derivative of the position vector function \(\mathbf{r}(t)\) with respect to \(t\). $$\mathbf{r}'(t)=\langle -a \sin t+b \cos t, -c \sin t+d \cos t, -e \sin t+f \cos t\rangle$$

Create two tangent vectors

To make our problem more general, we can choose two different values of \(t\) to find two linearly independent tangent vectors. Let's choose \(t=0\) and \(t=\frac{\pi}{2}\). Tangent Vector 1: $$\mathbf{t}_1 = \mathbf{r}'(0)=\langle b, d, f\rangle$$ Tangent Vector 2: $$\mathbf{t}_2 = \mathbf{r}'(\frac{\pi}{2})=\langle -a, -c, -e\rangle$$

Calculate the cross product

Now, we calculate the cross product of the two tangent vectors we found above. The cross product will be our normal vector to the plane containing the curve. $$\mathbf{n} = \mathbf{t}_1 \times \mathbf{t}_2 = \langle b, d, f\rangle \times \langle -a, -c, -e\rangle$$ $$\mathbf{n} = \langle d(-e) - f(-c), f(-a) - b(-e), b(-c) - a(-d)\rangle$$ $$\mathbf{n} = \langle de + fc, fa + be, -ac + bd\rangle$$ So, our normal vector to the plane is given by: $$\mathbf{n} = \langle de + fc, fa + be, -ac + bd\rangle$$

Key concepts

Cross Product

The cross product is a fundamental operation in vector calculus that finds a vector that is perpendicular to two given vectors in three dimensions. It's an important tool for finding a normal vector to a plane, which can be essential for various applications like physics, engineering, and computer graphics.

When given two vectors, say \textbf{A} = \(\begin{bmatrix} a_1 \ a_2 \ a_3 \$\) and \textbf{B} = \(\begin{bmatrix} b_1 \ b_2 \ b_3 \end{bmatrix}\), their cross product \textbf{A} \textbf{B} = \(\begin{bmatrix} a_2b_3- a_3b_2 \ a_3b_1- a_1b_3 \ a_1b_2- a_2b_1 \end{bmatrix}\) results in a vector that is orthogonal to both \textbf{A} and \textbf{B}. The direction of the resulting vector is determined by the right-hand rule.

One interesting aspect of the cross product is its relationship to the area of a parallelogram formed by the two vectors; the magnitude of the cross product equals the area of that parallelogram. This also means that if two vectors are parallel or if one is the zero vector, their cross product will be the zero vector, as there's no parallelogram to speak of.

Tangent Vectors

Tangent vectors offer a great deal of insight into the behavior of curves, especially in three-dimensional space. These vectors are tangent to a curve at a particular point, which means they touch the curve exactly at that point and point in the direction in which the curve is heading.

For a differentiable curve described by a position vector \(\textbf{r}(t)\), the tangent vector at any point \(t\) is the derivative of that position vector with respect to \(t\). In the given exercise, the tangent vectors are determined by differentiating the position vector of the curve, resulting in \(\textbf{t}_1\) and \(\textbf{t}_2\) for different values of \(t\). These vectors represent the directions of the curve at those specific instances and hence are crucial for determining the plane in which the curve lies.

By selecting tangent vectors at different points, we can ensure they are linearly independent, which allows us to use the cross product to find a vector orthogonal to the plane they define.

Three-dimensional Curves

Three-dimensional curves, such as the one in our exercise, are represented by vector functions of a single parameter, typically time \(t\), and occupy a unique position within 3D space. The curve \(\textbf{r}(t)\) is traced by the vector function as \(t\) varies, creating a path through space that can twist and turn in all directions.

The representation of a curve through its vector function, like the one in the question \(\textbf{r}(t) = \langle a \cos t+b \sin t, c \cos t+d \sin t, e \cos t+f \sin t\rangle\), captures not only its path but also how fast and in what manner the curve moves through space. For example, the coefficients \(a, b, c, d, e,\) and \(f\) adjust the stretching and compressing of the curve along each axis.

Knowing the properties of these 3D curves helps in many fields such as physics, representing particle paths, or in animation and graphics, designing complex motion. It is essential to understand not only the position but also the orientation and the shape of the curve within its ambient space, which is where the concepts like tangent vectors and the cross product become indispensable.