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To describe a line in three-dimensional space, we often use parametric equations. Parametric equations express the coordinates of the points on the line as functions of a parameter, usually denoted by \( t \). This method is particularly useful for defining paths and trajectories. For example, if you know a point on the line \((x_0, y_0, z_0)\) and the direction vector \((a, b, c)\), the parametric equations can be written as:

• \( x = x_0 + at \)
• \( y = y_0 + bt \)
• \( z = z_0 + ct \)

In the provided exercise, the line tangent to the curve of intersection is represented by these equations, with the point \((1, 1, 3)\) and direction vector \((60, -90, 0)\). Parametric equations allow us to easily understand how each coordinate changes with the parameter \( t \). This versatile approach is a cornerstone of calculus and vector algebra.

The gradient vector is a fundamental tool in multivariable calculus, primarily because it provides insights into the surface's behavior at given points. For a function \( f(x, y, z) \), the gradient, denoted as \( abla f \), is a vector of partial derivatives:

• Its components are \( \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \).
• The gradient vector points in the direction of the steepest ascent of the function.
• At any given point, it is normal to the level surface defined by the function value at that point.

In the exercise, two surfaces are described by functions \( f \) and \( g \). Calculating their gradients at the intersection point \((1, 1, 3)\) gives us normal vectors to these surfaces. Evaluating these gradients helps us determine the tangent line's direction through their cross product. Gradient vectors are crucial for defining normal lines and understanding surface orientation.

The cross product is a vector operation used to find a vector perpendicular to two given vectors in three-dimensional space. If you have vectors \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \), the cross product \( \mathbf{a} \times \mathbf{b} \) is calculated as follows:

• \( \mathbf{i}(a_2b_3 - a_3b_2) \)
• \(-\mathbf{j}(a_1b_3 - a_3b_1) \)
• \(\mathbf{k}(a_1b_2 - a_2b_1) \)

In the step-by-step solution, the tangent line's direction vector is found by taking the cross product of gradient vectors from two surfaces at the intersection point. This cross product is crucial because it provides a direction that is tangent to both surfaces simultaneously, allowing us to write the line's parametric equations. Since the cross product yields a vector orthogonal to the original vectors, it naturally fits into problems involving tangents and perpendicularly oriented vectors.

When two surfaces intersect, they form a curve called the curve of intersection. To find characteristics like tangent lines to this curve, it is crucial to understand the nature and positioning of this curve:

• The intersection of two surfaces can mathematically be defined by solving their equations simultaneously.
• This curve can have special properties at specific points, making its analysis important in both geometry and calculus.
• Tangent lines to these curves are significant in applications involving surface intersections, such as in optimizing trajectories in physics and engineering.

The process of finding the tangent line to the curve involves using derivatives and vectored calculations. In the exercise, the mentioned surfaces intersect at the point \((1, 1, 3)\), and calculating the gradients and using the cross product help in determining the tangent's direction, illustrating how these mathematical tools interconnect to describe physical phenomena effectively.