Plotting Technique & Perspective Guidelines:

• Draw tick marks parallel to the adjacent axes to maintain proper visual perspective.
• Plotting \(P(3, 2, 5)\): 🎬 [00:04:52] Move \(3\) units along the \(x\)-axis, \(2\) units parallel to the \(y\)-axis, and then \(5\) units parallel upward along the \(z\)-axis.
• Plotting \(Q(4, -2, -1)\): 🎬 [00:05:54] Move \(4\) units along the \(x\)-axis, extend the \(y\)-axis negatively to move \(-2\) units, and move \(1\) unit downward along the negative \(z\)-axis.

• \(x = 2\): 🎬 [00:07:26]
  In \(\mathbb{R}^2\), this is a vertical line. In \(\mathbb{R}^3\), since \(y\) and \(z\) can be any value, it represents a plane parallel to the \(YZ\)-plane passing through \((2,0,0)\).
• \(y = 3\): 🎬 [00:09:05]
  In \(\mathbb{R}^2\), this is a horizontal line. In \(\mathbb{R}^3\), because \(x\) and \(z\) are unrestricted, it represents a plane parallel to the \(XZ\)-plane passing through \((0,3,0)\).
• \(y = x\): 🎬 [00:10:49]
  In \(\mathbb{R}^2\), this is a line through the origin with slope \(1\) making a \(45^\circ\) angle with the \(x\)-axis. In \(\mathbb{R}^3\), projecting this line parallel to the unrestricted \(z\)-axis creates a vertical plane containing the \(z\)-axis.
• \(xy = 0\): 🎬 [00:12:24]
  By the Zero Product Property, \(xy = 0 \implies x = 0\) or \(y = 0\). In \(\mathbb{R}^3\), \(x=0\) is the \(YZ\)-plane and \(y=0\) is the \(XZ\)-plane. Thus, the equation represents the union of both the \(YZ\)-plane and the \(XZ\)-plane.

General Method for Cylinders: Identify the missing variable, sketch the 2D curve in the plane of the remaining variables, and project the curve parallel to the missing variable’s axis.

• \(y = x^2\): 🎬 [00:13:56]
  The variable \(z\) is missing. Sketch the parabola \(y = x^2\) in the \(XY\)-plane, then project the curve endlessly parallel to the \(z\)-axis to form a parabolic cylinder.
• \(x^2 + z^2 = 16\): 🎬 [00:16:42]
  The variable \(y\) is missing. In the \(XZ\)-plane, this represents a circle centered at the origin with radius \(R = 4\). Projecting this circle along the \(y\)-axis yields a circular cylinder centered along the \(y\)-axis.

1. Find the Boundary: The boundary is the equality plane \(y = x\) (or \(x = y\)). Since the inequality is inclusive (\(\le\)), sketch a solid vertical plane passing through the origin.
2. Determine the Region: In 2D, \(y \ge x\) includes all points above the line. In 3D, this translates to all points on and behind the vertical plane \(y = x\).

Substitute \(h = -1\), \(k = 2\), \(l = 4\), and \(R = \frac{1}{2}\) into standard form:

\[ (x – (-1))^2 + (y – 2)^2 + (z – 4)^2 = \left(\frac{1}{2}\right)^2 \] \[ (x + 1)^2 + (y – 2)^2 + (z – 4)^2 = \frac{1}{4} \]

Step 1: Divide by the common leading coefficient \(2\):

\[ x^2 + y^2 + z^2 + 2y – z = \frac{1}{2} \]

Step 2: Group variable terms and set up completion of squares:

\[ x^2 + (y^2 + 2y + \underline{\quad}) + \left(z^2 – z + \underline{\quad}\right) = \frac{1}{2} \]

Step 3: Add completing constants to both sides:

• For \(y\): \(\left(\frac{2}{2}\right)^2 = 1\)
• For \(z\): \(\left(\frac{-1}{2}\right)^2 = \frac{1}{4}\)

\[ x^2 + (y^2 + 2y + 1) + \left(z^2 – z + \frac{1}{4}\right) = \frac{1}{2} + 1 + \frac{1}{4} \]

Step 4: Factor into standard form:

\[ (x – 0)^2 + (y + 1)^2 + \left(z – \frac{1}{2}\right)^2 = \frac{7}{4} \]

Conclusion:

• Center: \(\left(0, -1, \frac{1}{2}\right)\)
• Radius: \(R = \sqrt{\frac{7}{4}} = \frac{\sqrt{7}}{2}\)

• \(x^2 + y^2 + z^2 = 1\) defines a sphere centered at the origin with radius \(R_1 = 1\).
• \(x^2 + y^2 + z^2 = 25\) defines a sphere centered at the origin with radius \(R_2 = 5\).
• Because the inequalities are inclusive (\(\le\)), the region consists of all points on and between the two concentric spheres centered at the origin with radii \(1\) and \(5\) (a solid spherical shell).