91Ó°ÊÓ

A normal vector is central to defining a plane in three-dimensional space. In our exercise, we deal with a normal vector denoted as \(\mathbf{n} = \langle 3, 0, -\frac{1}{2}\rangle\). This vector acts like an arrow pointing straight out from the surface of the plane.
Think of it as a rule that tells you how the plane is tilted or rotated in space. By nature, a normal vector is perpendicular to every line within the plane itself.

• It is crucial for constructing the plane's equation.
• It determines the plane's orientation.

Knowing the normal vector and a point through which the plane passes, we can easily find the equation of the plane. This is done using a formula where the components of the normal vector \(\langle a, b, c \rangle\) are plugged in along with the coordinates of the given point \((x_0, y_0, z_0)\).
This gives us a reliable method to describe any plane concretely in 3D space.

Intercepts are key points where the plane crosses the x, y, or z axes. Let's break down each intercept from our specific plane example with the equation: \(6x - z = 4\).

• X-intercept: To find where the plane meets the x-axis, set both \(y = 0\) and \(z = 0\), leaving \(6x = 4\). Solving this, we get \(x = \frac{2}{3}\). Therefore, the x-intercept is \(\left(\frac{2}{3}, 0, 0\right)\).
• Y-intercept: Notice that our plane's equation lacks the \(y\) component, meaning it doesn't technically pierce the y-axis in the traditional sense. It's more accurate to say there's no defined y-intercept here.
• Z-intercept: Set \(x = 0\) and \(y = 0\) to find this intercept, simplifying the equation to \(-z = 4\). Thus, \(z = -4\), placing the z-intercept at \((0, 0, -4)\).

These intercepts are crucial for visualizing and sketching the plane, showing where it penetrates the three axes.

Sketching a 3D graph of a plane might seem overwhelming at first, but understanding intercepts makes it easier.
Once you've calculated the intercepts, visualize the plane as a surface extending through these points on the graph.To start, mark the intercepts on the 3D coordinate system. In our case, the x-intercept is at \(\left(\frac{2}{3}, 0, 0\right)\) and the z-intercept is at \((0, 0, -4)\). Although we don't have a y-intercept due to the absence of \(y\) in the equation, the plane still exists and is defined by its tilt from the normal vector.

• The plane will stretch infinitely, but visually focuses between and beyond these intercept points.
• No y-intercept emphasizes that the plane is parallel to the y-axis.

By considering both the normal vector and these critical intercepts, you can generate a clear picture of how the plane exists in 3D space, even if only on paper or a digital graphing tool.