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Chapter 18: Problem 8

Draw a sketch of the given plane and find two unit vectors which are normal to the plane. $$ 4 x-4 y-2 z-9=0 $$

Short Answer

Expert verified
The unit normal vectors are \(\left(\frac{2}{3}, -\frac{2}{3}, -\frac{1}{3}\right)\) and \(\left(-\frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right)\).

Step by step solution

01

Understand the Plane Equation

The given plane equation is linear and can be written as: 4x - 4y - 2z - 9 = 0. This equation is in the standard form Ax + By + Cz + D = 0 where A, B, and C are the coefficients that form the normal vector to the plane.
02

Find the Normal Vector

The normal vector \(\mathbf{N}\) to the plane is given by the coefficients of x, y, and z in the plane equation. Thus, \(\mathbf{N} = (4, -4, -2)\).
03

Calculate the Magnitude of the Normal Vector

Find the magnitude of the normal vector \(\mathbf{N}\): \[ |\mathbf{N}| = \sqrt{4^2 + (-4)^2 + (-2)^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6 \].
04

Find the Unit Normal Vectors

Normalize the normal vector \(\mathbf{N}\) to find the unit normal vectors: \[ \mathbf{n_1} = \frac{1}{6}(4, -4, -2) = \left(\frac{4}{6}, -\frac{4}{6}, -\frac{2}{6}\right) = \left(\frac{2}{3}, -\frac{2}{3}, -\frac{1}{3}\right) \]. The other unit normal vector is the negative of \(\mathbf{n_1}\), which is: \[ \mathbf{n_2} = -\mathbf{n_1} = \left(-\frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right) \].
05

Sketch the Plane

To sketch the plane, note that it intersects the coordinate axes. Set one variable to zero at a time to find intercepts: For x-intercept, set y and z to 0: \[ 4x - 9 = 0 \implies x = \frac{9}{4} = 2.25 \]. For y-intercept, set x and z to 0: \[ -4y - 9 = 0 \implies y = -2.25 \]. For z-intercept, set x and y to 0: \[ -2z - 9 = 0 \implies z = -4.5 \]. Draw a 3D coordinate system and plot these intercepts. Then, sketch the plane passing through those intercepts.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

plane equation
A plane equation in three dimensions is an essential mathematical tool, represented in the form: Ax + By + Cz + D = 0. This equation describes a flat, two-dimensional surface in a three-dimensional space. Each coefficient (A, B, and C) corresponds to the direction and orientation of the plane. In the equation 4x - 4y - 2z - 9 = 0, the coefficients 4, -4, and -2 determine the plane's slant.
  • A represents the x-component.
  • B is the y-component.
  • C corresponds to the z-component.

Understanding this structure helps break down complex problems into more manageable parts by focusing on x, y, and z separately.
normal vector
The normal vector is a fundamental concept in understanding planes. It is perpendicular to the plane鈥檚 surface and can be derived from the coefficients in the plane equation. Given the plane equation 4x - 4y - 2z - 9 = 0, the normal vector \(\mathbf{N}\)\ is: \[\mathbf{N} = (4, -4, -2)\]\
This vector points perpendicularly away from the plane in three-dimensional space.

The coordinates of the normal vector (4, -4, -2) indicate its direction. For visuailzation:
  • 4 shows the x-direction component.
  • -4 shows the y-direction component.
  • -2 shows the z-direction component.
Knowing how to find and interpret the normal vector is critical in many areas such as physics, geometry, and beyond.
unit vector
A unit vector has a magnitude (length) of 1 and shows direction. To find a unit normal vector from our normal vector \(\mathbf{N} = (4, -4, -2)\)\, we need to normalize it. First, calculate the magnitude \(\left|\mathbf{N}\right|\)\ using: \[\left|\mathbf{N}\right| = \sqrt{4^2 + (-4)^2 + (-2)^2} = \sqrt{36} = 6\]\
With this magnitude, we can find the unit normal vectors by dividing each component of \(\mathbf{N}\)\ by 6:
\[\mathbf{n_1} = \left(\frac{2}{3}, -\frac{2}{3}, -\frac{1}{3}\right)\]\ The reverse, \(\mathbf{n_2} = -\mathbf{n_1} = \left(-\frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right)\)\, is also a unit vector.
  • Step 1: Find the magnitude of \(\mathbf{N}\)\.
  • Step 2: Divide each component by the magnitude.
These unit vectors help standardize direction.
intercepts
Intercepts are points where the plane intersects the coordinate axes. They provide reference points for sketching the plane. For the plane 4x - 4y - 2z - 9 = 0:
  • x-intercept: Set y and z to zero, \(4x - 9 = 0\)\, giving \(x = \frac{9}{4} = 2.25\)\.
  • y-intercept: Set x and z to zero, \(-4y - 9 = 0\)\, giving \(y = -2.25\)\.
  • z-intercept: Set x and y to zero, \(-2z - 9 = 0\)\, giving \(z = -4.5\)\.

These intercept points (2.25, 0, 0), (0, -2.25, 0), and (0, 0, -4.5) are crucial to visualizing and sketching the plane in 3D space.
Plot these specific points on the coordinate system and draw a plane through them for a proper visualization.

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Most popular questions from this chapter

Find equations of the line through the point \((3,6,4)\), intersecting the \(z\) axis, and parallel to the plane \(x-3 y+5 z-6=0\).

Find an equation in cartesian coordinates for the surface whose equation is given in cylindrical coordinates. \(r^{2} \cos 2 \theta=z^{3}\)

Draw a sketch of the graph of the given equation and name the surface. $$ 4 x^{2}-9 y^{2}-z^{2}=36 $$

Find an equation of the plane containing the given intersecting lines. \(\frac{x-2}{4}=\frac{y+3}{-1}=\frac{z+2}{3}\) and \(\left\\{\begin{array}{r}3 x+2 y+z+2=0 \\ x-y+2 z-1=0\end{array}\right.\)

Find a unit vector whose representations are perpendicular to the plane containing \(\overrightarrow{P Q}\) and \(\overrightarrow{P R}\) if \(\overrightarrow{P Q}\) is a representation of the vector \(\mathbf{i}+3 \mathbf{j}-2 \mathbf{k}\) and \(\overrightarrow{P R}\) is a representation of the vector \(2 \mathbf{i}-\mathbf{j}-\mathbf{k}\).
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