Problem 18 - Algebra and Trigonometry

Question

A plane has normal vector \(\mathbf{n}\) and passes through the point \(P\) (a) Find an equation for the plane. (b) Find the intercepts and sketch a graph of the plane. $$ \mathbf{n}=\left\langle-\frac{2}{3},-\frac{1}{3}, 1\right\rangle, \quad P(-6,0,-3) $$

Step-by-Step Solution

Answer

The plane equation is \(-\frac{2}{3}x - \frac{1}{3}y + z = 1\). Intercepts: \(x = -\frac{3}{2}\), \(y = -3\), \(z = 1\).

Step 1: Understand the Plane's Equation

A plane can be described by the equation: \( ax + by + cz = d \), where \( [a, b, c] \) are the components of the normal vector \( \mathbf{n} \). In this problem, the normal vector \( \mathbf{n} \) is given as \( \left\langle -\frac{2}{3}, -\frac{1}{3}, 1 \right\rangle \). Thus, the plane's equation starts as \( -\frac{2}{3}x - \frac{1}{3}y + z = d \).

Step 2: Use Point to Find d

To determine \( d \), substitute the coordinates of point \( P(-6, 0, -3) \) into the plane equation: \( -\frac{2}{3}(-6) - \frac{1}{3}(0) + (-3) = d \). Solving this gives:\[ d = 4 - 3 = 1 \].This results in the plane equation: \( -\frac{2}{3}x - \frac{1}{3}y + z = 1 \).

Step 3: Find x, y, and z Intercepts

To find intercepts, set two variables to zero and solve for the third:1. **x-intercept**: Set \( y = 0 \), \( z = 0 \): \( -\frac{2}{3}x = 1 \) \( \Rightarrow \) \( x = -\frac{3}{2} \).2. **y-intercept**: Set \( x = 0 \), \( z = 0 \): \( -\frac{1}{3}y = 1 \) \( \Rightarrow \) \( y = -3 \).3. **z-intercept**: Set \( x = 0 \), \( y = 0 \): \( z = 1 \) \( \Rightarrow \) \( z = 1 \).

Step 4: Sketch the Plane

Using the intercepts calculated, sketch a graph. Plot points \(( -\frac{3}{2}, 0, 0 )\), \(( 0, -3, 0 )\), and \(( 0, 0, 1 )\) on respective axes and draw the plane that connects these intercepts on a coordinate system to form a triangular plane in space.

Key Concepts

Normal Vector

In the world of coordinate geometry, the term "normal vector" refers to a vector that is perpendicular to a given surface or plane. For a plane, a normal vector is crucial in defining its orientation in three-dimensional space. This vector is typically represented as \( \mathbf{n} = \langle a, b, c \rangle \), where \( a, b, \) and \( c \) are its components along the x, y, and z axes respectively.

In our exercise, the plane's normal vector is provided as \( \mathbf{n} = \langle -\frac{2}{3}, -\frac{1}{3}, 1 \rangle \). This vector is perpendicular to every line that lies within the plane. Understanding the role of this vector helps simplify the task of finding the equation of the plane and analyzing its geometric properties. This is because the coefficients of the plane equation, \( ax + by + cz = d \), are directly derived from the components of the normal vector.

Intercepts

Intercepts provide critical points where a plane interacts uniquely with the coordinate axes. These are the points where the plane crosses a specific axis, and they help in visualizing the plane's orientation and position in space.

To find the intercepts of a plane, we individually set two of the three variables (x, y, z) to zero and solve for the third:

• **x-intercept**: Occurs when \( y = 0 \) and \( z = 0 \). For this plane: \( -\frac{2}{3}x = 1 \) leads to \( x = -\frac{3}{2} \).
• **y-intercept**: Occurs when \( x = 0 \) and \( z = 0 \). This gives us \( -\frac{1}{3}y = 1 \) so \( y = -3 \).
• **z-intercept**: Occurs when \( x = 0 \) and \( y = 0 \). Here, \( z = 1 \) is the solution.

These intercepts are handy not only in sketching the graph of the plane but also in finding out how the plane interacts with different parts of the space. Plotting these points helps to draw a triangular plane, which provides a visual comprehension of its positional attributes.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, merges algebraic principles with geometric explorations. It facilitates analyzing shapes and sizes in coordinate systems, using equations to represent geometric entities like lines, curves, and planes.

In the context of our problem, coordinate geometry helps in formulating the plane equation derived from the normal vector, \( \mathbf{n} = \langle -\frac{2}{3}, -\frac{1}{3}, 1 \rangle \), and the point \( P(-6, 0, -3) \) on the plane. Initially, the equation of a plane can be set as \( ax + by + cz = d \), where the coefficients \( a \), \( b \), and \( c \) stem from the normal vector.
Using the point \( P \), we substitute the x, y, z coordinates to calculate \( d \), completing the equation to \( -\frac{2}{3}x - \frac{1}{3}y + z = 1 \). This equation is vital for solving intercepts and sketching the plane, showcasing the intersection of algebraic calculations with geometric interpretations.