Problem 20 - 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. $$ \mathbf{n}=\mathbf{i}+4 \mathbf{j}, \quad P(1,0,-9) $$

Step-by-Step Solution

Answer

Plane equation: \(x + 4y - 1 = 0\). Intercepts: x-intercept at 1, y-intercept at \(\frac{1}{4}\). No z-intercept.

Step 1: Find Plane Equation

To find the equation of the plane, we can use the point-normal form. The normal vector is \(\mathbf{n} = \langle 1, 4, 0 \rangle\) and the point \(P\) is \((1, 0, -9)\). The equation for the plane is given by: \[1(x - 1) + 4(y - 0) + 0(z + 9) = 0\] Simplify this to get the equation: \[x + 4y - 1 = 0\].

Step 2: Find x-Intercept

To find the x-intercept, set \(y = 0\) and \(z = 0\) in the plane equation: \[x + 4(0) - 1 = 0\] Solving gives the x-intercept as \(x = 1\).

Step 3: Find y-Intercept

To find the y-intercept, set \(x = 0\) and \(z = 0\) in the plane equation: \[0 + 4y - 1 = 0\] Solving gives the y-intercept as \(y = \frac{1}{4}\).

Step 4: Find z-Intercept

The normal vector \(\mathbf{n}\) has no z-component, which means the plane is parallel to the z-axis and does not intersect it, hence there is no z-intercept.

Step 5: Sketch the Plane

Draw a graph in the xy-plane, since the z-component has no influence due to plane parallelism to the z-axis. Mark the x-intercept at \((1,0)\) and the y-intercept at \(\left(0, \frac{1}{4}\right)\). Draw a line through these two points; the plane is represented by this line extended parallel to the z-axis.

Key Concepts

Normal Vector

A normal vector is a crucial concept when dealing with planes in three-dimensional space. It is a vector that is perpendicular to the plane, meaning it points directly away from the plane's surface at a 90-degree angle.
For a given plane, any vector \( \mathbf{n} = \langle a, b, c \rangle \) is called the normal vector. This vector determines the plane's orientation in space. The components \(a, b, \) and \(c\) of the vector represent the direction in which the plane is perpendicular.
Given the context of the exercise, with the normal vector being \( \mathbf{n} = \mathbf{i} + 4 \mathbf{j} \), it simplifies to \( \langle 1, 4, 0 \rangle \). In this case, since the z-component is zero, the plane does not tilt along the z-axis, effectively making it parallel to the z-axis.

Point-Normal Form

The point-normal form is a key formula to find the equation of a plane when a normal vector and a point on the plane are known. It is generally expressed as: \[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \] where \(\langle a, b, c \rangle\) is the normal vector and \((x_0, y_0, z_0)\) is the point on the plane.
In our specific case, given the normal vector \(\langle 1, 4, 0 \rangle\) and point \((1, 0, -9)\), the equation becomes: \[ 1(x - 1) + 4(y - 0) + 0(z + 9) = 0 \]
Simplifying this equation:

• Distribute to eliminate parentheses: \( x - 1 + 4y = 0 \)
• Rearrange to standard form: \( x + 4y - 1 = 0 \)

This form clearly defines the plane by showing that any point \((x, y, z)\) satisfying this equation lies on the plane.

x-intercept

The x-intercept of a plane is the point where the plane intersects the x-axis. To find it, we set the other variables, \(y\) and \(z\), to zero in the plane's equation.
Substituting into the equation \(x + 4y - 1 = 0\):

• Set \(y = 0\): \(x - 1 = 0\)
• Solving gives \(x = 1\)

Hence, the x-intercept of this plane is at the point \((1,0,0)\). At this point, the plane will touch or cross the x-axis.

y-intercept

The y-intercept is the point where the plane cuts across the y-axis. To locate this point, we set both \(x\) and \(z\) to zero and solve for \(y\).
For our plane equation \(x + 4y - 1 = 0\):

• Set \(x = 0\): \(4y - 1 = 0\)
• Solving 4y = 1 gives \(y = \frac{1}{4}\)

This tells us that the y-intercept of this plane is at \(\left(0, \frac{1}{4}, 0\right)\). The plane intersects the y-axis at this distinct point.