Problem 31 - Thomas Calculus

Question

Find a plane through \(P_{0}(2,1,-1)\) and perpendicular to the line of intersection of the planes \(2 x+y-z=3, x+2 y+z=2\)

Step-by-Step Solution

Answer

The equation of the plane is \(3x - 5y + 3z + 2 = 0\).

Step 1: Understand Intersection of Planes

To find a plane through a point and perpendicular to the line of intersection of two planes, we first need to find the direction vector of that line. The line of intersection of two planes is perpendicular to the normal vectors of both planes.

Step 2: Find Normal Vectors

The normal vector of the first plane, \(2x + y - z = 3\), is \(\mathbf{n_1} = (2, 1, -1)\). The normal vector of the second plane, \(x + 2y + z = 2\), is \(\mathbf{n_2} = (1, 2, 1)\).

Step 3: Calculate Direction Vector of Intersection

The direction vector of the line of intersection can be found by taking the cross product of the normal vectors, \(\mathbf{n_1} \times \mathbf{n_2}\). Compute it: \[(2, 1, -1) \times (1, 2, 1) = ((1 \times 1) - (-1 \times 2), (-1 \times 1) - (2 \times 2), (2 \times 2) - (1 \times 1)) = (3, -5, 3)\].

Step 4: Determine Plane's Normal Vector

A plane perpendicular to the line of intersection will have the same direction vector as its normal vector. Thus, the normal vector of our new plane is \(\mathbf{n} = (3, -5, 3)\).

Step 5: Use Point-Normal Form of a Plane

The equation of a plane with a normal vector \((a, b, c)\) passing through point \((x_0, y_0, z_0)\) is \(a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\). Plugging in our point \(P_0(2, 1, -1)\) and normal vector \((3, -5, 3)\), we get the equation: \[3(x - 2) - 5(y - 1) + 3(z + 1) = 0\].

Step 6: Simplify the Plane Equation

Simplify the equation: \(3x - 6 - 5y + 5 + 3z + 3 = 0\). This becomes \(3x - 5y + 3z + 2 = 0\).

Key Concepts

Direction Vector

When discussing the intersection of planes, one key concept is the direction vector. This vector indicates the direction of a line. It's like an arrow pointing along the line's path. If you think of the intersection of two planes as a line, the direction vector shows you the path that line takes.

To find a line of intersection between two planes, you need a direction vector that lies along that intersection. This vector is crucial because it helps define the orientation of the line. It's determined by the interaction of the planes' normal vectors, which we'll explore shortly.

In our problem, the direction vector is calculated using the cross product of the normal vectors of the given planes. This approach is essential whenever dealing with intersecting planes.

Cross Product

The cross product is a mathematical operation used to find a vector that is perpendicular to two given vectors. In the context of planes, it's used to derive the direction vector for the line of intersection.

The formula for the cross product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is: \[\mathbf{a} \times \mathbf{b} = (a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1)\]. This yields a new vector that is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\).

In the exercise, the cross product helps us find the direction vector of the line of intersection between two planes, ensuring it's perpendicular to both normal vectors. For the given planes, the cross product of their normal vectors \((2, 1, -1)\) and \(1, 2, 1\) produced a resulting vector of \(3, -5, 3\). This is used as the direction vector for our solution.

Normal Vector of a Plane

A normal vector of a plane is a vector that is perpendicular to the surface of the plane. It plays a vital role in defining the plane's orientation in space. In simple terms, it's this vector that tells you which "way" the plane is facing.

For any plane equation of the form \(ax + by + cz = d\), the normal vector is \( (a, b, c) \). In our exercise, the normal vectors of the given planes are identified as \(\mathbf{n_1} = (2, 1, -1)\) and \(\mathbf{n_2} = (1, 2, 1)\).

These vectors are essential because the line of intersection between these planes is perpendicular to both normal vectors. By finding the cross product of these vectors, we can determine the direction vector of the intersection line. Consequently, identifying and understanding the normal vector aids in constructing the equation for the plane through a given point, perpendicular to this line.