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Parametric equations are expressions that define a curve using parameters. Instead of defining a function as y in terms of x or z, we use parameters (usually denoted by \( t \)) to give each variable a separate expression. This is particularly helpful in three-dimensional space, where curves aren't limited to straightforward algebraic expressions.
To create parametric equations for a line, you need a point on the line and a direction vector. The parametric form involves the initial point and the direction vector scaled by the parameter. Here's the basic setup:

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

Where \((x_0, y_0, z_0)\) is the initial point and \(\langle a, b, c \rangle\) is the direction vector. In the exercise, the point given is \((\sqrt{2}, \sqrt{2}, 4)\) and the direction is derived from the cross product of the gradient vectors.

When dealing with problems involving the intersection of 3D surfaces, it often results in a curve. To find this intersection curve, we generally solve the equations of the two surfaces simultaneously.
In this scenario, we have two surfaces:

• A cylinder described by \(x^2 + y^2 = 4\)
• A plane described by \(x^2 + y^2 - z = 0\)

By substituting the equation of the cylinder into the plane equation, we can isolate variables or reveal more information about the intersection. In our example, this yielded \(z = 4\), revealing that the intersection is a 2D circle lying in a plane. When surfaces like these overlap, they create what is called a curve of intersection in 3D space. Understanding the nature of these curves is critical for finding tangent lines and other geometric properties.

The gradient vector is a fundamental concept in multivariable calculus. It gives the direction of the steepest ascent on a surface and is perpendicular to the level curves (or surfaces). For a function \(f(x, y, z)\), its gradient is given by \(abla f = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \rangle\).
In our example, we calculate the gradients for each surface. At the point \((\sqrt{2}, \sqrt{2}, 4)\), the two surfaces have gradients:

• \(abla f_1 = \langle 2x, 2y, 0 \rangle = \langle 2\sqrt{2}, 2\sqrt{2}, 0 \rangle\)
• \(abla f_2 = \langle 2x, 2y, -1 \rangle = \langle 2\sqrt{2}, 2\sqrt{2}, -1 \rangle\)

These vectors indicate how each surface 'slopes' at that point. The tangent line to their intersection is perpendicular to both gradients, hence the necessity to compute their cross product.

The cross product of two vectors is a cornerstone operation in vector calculus, especially in 3D geometry. Given two vectors \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\) and \(\mathbf{b} = \langle b_1, b_2, b_3 \rangle\), their cross product \(\mathbf{a} \times \mathbf{b}\) is:\[\mathbf{a} \times \mathbf{b} = \left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{array}\right|\]
This results in a vector perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\) and is useful for determining directions.In the exercise, we use the cross product of \(abla f_1\) and \(abla f_2\) to find the direction vector for the tangent line to the curve formed by the surfaces' intersection. The cross product helps us discover a new vector \(\langle -2\sqrt{2}, 2\sqrt{2}, 0 \rangle\) that is orthogonal to the gradient vectors, and hence, parallel to the tangent line we sought to parametrize.