Problem 6 - Calculus: Early Transcendentals

Question

Find an equation of the plane containing the line of intersection of \(x+y+z=1\) and \(x-y+2 z=2,\) and perpendicular to the xy-plane.

Step-by-Step Solution

Answer

The equation of the plane is \(z = 0\).

Step 1: Identify Direction Vectors

The direction vectors of the intersecting line can be found by solving the system of equations provided. The direction vectors for the line of intersection of the planes are obtained by the cross product of gradients of the plane equations.

Step 2: Find the Direction Vector of First Plane

The first plane equation is given by \(x+y+z=1\). The gradient vector of this plane is \(abla f_1 = \langle 1, 1, 1 \rangle\).

Step 3: Find the Direction Vector of Second Plane

The second plane equation is given as \(x-y+2z=2\). The gradient vector of this plane is \(abla f_2 = \langle 1, -1, 2 \rangle\).

Step 4: Calculate Cross Product

Compute the cross product of the two gradient vectors: \( abla f_1 = \langle 1, 1, 1 \rangle \) and \( abla f_2 = \langle 1, -1, 2 \rangle \). This gives the direction vector of the intersection line, calculated as follows:\[ \mathbf{v} = abla f_1 \times abla f_2 = \langle 1, 1, 1 \rangle \times \langle 1, -1, 2 \rangle = \langle 3, -1, -2 \rangle \]

Step 5: Determine Normal Vector of Plane

To find the normal vector of the new plane, since it must be perpendicular to the xy-plane, we can use the vector \( \mathbf{n} = \langle 0, 0, 1 \rangle \).

Step 6: Find a Point of Intersection

Choose an arbitrary value for one variable to find a point that lies on both planes. Let \(z=0\), solve the system for \(x\) and \(y\):- From \(x+y+0 = 1\), we have: \(x+y=1\).- From \(x-y+0=2\), we have: \(x-y=2\).Solving these two equations yields \(x=\frac{3}{2}, y=-\frac{1}{2}\), so the point \(\left(\frac{3}{2}, -\frac{1}{2}, 0\right)\) lies on the line of intersection.

Step 7: Write Equation of the Plane

Using the point \(\left(\frac{3}{2}, -\frac{1}{2}, 0\right)\) and the normal vector \( \mathbf{n} = \langle 0, 0, 1 \rangle \), the equation of the plane is:\[ 0(x - \frac{3}{2}) + 0(y + \frac{1}{2}) + 1(z - 0) = 0 \]Simplifying, we get the plane equation: \(z = 0\).

Key Concepts

Direction Vectors

When dealing with plane intersection problems, direction vectors play a key role, especially when finding the line of intersection of two planes. A direction vector essentially provides the path in which a line advances through space. Imagine it as the arrow showing the "way" of a line. In the context of two intersecting planes, each plane has its own unique direction vectors.
To determine the precise direction of the intersecting line, we utilize the cross product of the gradient vectors from each of the planes involved. This process essentially gives us the direction vector of the line where both planes meet. Knowing this, we can further determine other aspects like the equation of planes formed by these intersecting lines.

Cross Product

The cross product is a crucial mathematical operation in finding a perpendicular vector to two given vectors. In the context of plane intersections, using cross product enables us to determine the direction vector for the line where the planes intersect.
For example, given two gradient vectors obtained from the plane equations — say \(abla f_1 = \langle 1, 1, 1 \rangle\) and \(abla f_2 = \langle 1, -1, 2 \rangle\) — the cross product \(abla f_1 \times abla f_2\) results in a new vector (for this case, \(\langle 3, -1, -2 \rangle\)). This vector essentially indicates the direction that the line of intersection follows.
The cross product is particularly advantageous here because it ensures that the resulting vector is orthogonal to both input vectors, which aligns with the geometry of intersecting planes.

Gradient Vectors

Gradient vectors are derived from the equations of planes. These vectors point in the direction of the greatest increase of the function. For plane equations, they serve a crucial role in intersection problems.
When you examine a plane represented by the equation such as \(x + y + z = 1\), its gradient vector \(abla f_1 = \langle 1, 1, 1 \rangle\) can be thought of as the "normal" to that plane. It provides the necessary components for identifying direction when comparing with another plane's gradient.

• To find a gradient vector from a plane equation, simply take the partial derivatives of each variable to discover the vector components.
• Gradient vectors not only help define intersection direction, but also provide insight into the nature of the plane itself.

Normal Vector

A normal vector is integral in defining the orientation of a plane in space. It is a vector that is perpendicular, or "normal," to the surface of the plane. In the exercise discussed, determining the normal vector was crucial to define the plane perpendicular to the xy-plane.
The normal vector for this specific plane is \(\langle 0, 0, 1 \rangle \). This indicates that the plane is perpendicular to the xy-plane and is horizontal, quite tangibly the plane \(z=0\).

• Normal vectors are vital in calculating angles between planes and determining their orientation in three-dimensional space.
• They also help write plane equations, acting as coefficients in the standard plane equation form \(Ax + By + Cz = D\).

By understanding and using normal vectors effectively, we can explore complex intersection relationships and better grasp spatial relationships between different geometric entities.