Question

The equation of plane passing through the intersection of the planes \(\mathrm{x}-\mathrm{y}+\mathrm{z}=1\) and \(\mathrm{x}+\mathrm{y}-\mathrm{z}=1\) and perpendicular to \(\mathrm{x}-2 \mathrm{y}+\mathrm{z}=2\) is (A) \(x+3 y+z=3\) (B) \(3 x+y-z=3\) (C) \(x-3 y-z=3\) (D) \(x-3 y+z=3\)

Short answer

The equation of the plane passing through the intersection of the given planes and perpendicular to the third plane is \(x - 3y + z = 3\).

Step-by-step solution

Find a line for the intersection of the given two planes

First, we need to find a common line for the intersection of the given two planes. The given planes are: \(\) \[ \begin{cases} x - y + z = 1 & (1) \\ x + y - z = 1 & (2) \end{cases} \] Add (1) and (2) together: \[ 2x=2 \] Divide by 2: \[ x=1 \] Now subtract (2) from (1): \[ 2y + 2z = 0 \] Divide by 2: \[ y + z = 0 \] So the common line for the intersection of these two planes is: \(x=1\), \(y+z=0\).

Determine the direction vector

Now, let's find the direction vector (or normal vector) of our plane. We know that our plane is perpendicular to the plane with equation \(x - 2y + z = 2\). Now, let's rewrite the equation in normal vector form: \[ (1, -2, 1) \cdot (x - 1, y, z) = k \] Where k is a constant.

Plug the intersection line into the normal vector equation

Now, we will plug the intersection line into the normal vector equation: \[ (1, -2, 1) \cdot (1 - 1, y, -y) = k \] Simplifying and solving for k yields: \[ k=(-2)(y)+1(-y)= -3y \]

Find the equation of the desired plane

Now, let's find the equation of the desired plane using the normal vector, constant k, and intersection line: \[ (1, -2, 1) \cdot (x - 1, y, z) = -3y \] Expanding and simplifying, we have: \[ (x - 1) - 2(y) + (z) = -3y \] Solving for x, we have: \[ x - 3y + z = -2y + 1 \] \[ x - 3y + z = 3 \] So the equation of the desired plane is \(x - 3y + z = 3\). Thus, the correct answer is (D).

Key concepts

Intersection of Planes

When two planes intersect in three-dimensional space, they typically intersect along a line. This line is formed at the points that satisfy both plane equations. Let's say we have two plane equations: \(x - y + z = 1\) and \(x + y - z = 1\). To find their line of intersection, we solve these two equations simultaneously.

• Add the equations to eliminate one variable, such as adding them results in \(2x = 2\), giving us \(x = 1\).
• Subtract one equation from the other to eliminate another variable, such as subtracting gives \(2y + 2z = 0\), leading to \(y + z = 0\).

So, the line of intersection has the equations \(x = 1\) and \(y + z = 0\), describing a line in the space where these planes meet.

Perpendicular Planes

For a plane to be perpendicular to another, the normal vector of the first must be perpendicular to the normal vector of the second. In simpler terms, the dot product of their normal vectors must be zero. This property is utilized to find a specific plane among various possibilities.

In this exercise, the desired plane must be perpendicular to the plane given by \(x - 2y + z = 2\). The normal vector for this plane is \((1, -2, 1)\). This plane's orientation acts as a reference for establishing orthogonality with our plane because if two planes are perpendicular:

• The direction vector (or normal vector) must equally balance the influence of its components in a way that maintains perpendicularity.
• The dot product between vectors of planes results in zero.

This understanding aids in determining the correct equation of the desired perpendicular plane.

Direction Vector

A plane's equation is often linked to its normal vector, which dictates its orientation in space. This vector is perpendicular to any direction vector lying in the plane, making it foundational in defining the plane's orientation.

For our specific situation, the direction of the common intersection line of planes is influenced by the problem's requirement for the plane to be perpendicular to another. Understanding the direction vector involves:

• Determining a vector perpendicular to the given direction (from plane \(x - 2y + z = 2\): \((1, -2, 1)\)).
• This vector then serves as a base in solving for the desired plane's equation.

The project becomes solving algebraic expressions ensuring the perpendicular requirement is met.

Plane Equation

A plane in a three-dimensional space can be represented by an equation of the form: \(ax + by + cz = d\), where \(a\), \(b\), and \(c\) are the components of the plane's normal vector. This equation provides a full description of the plane's spatial behavior through the coefficients and constant term.

To derive the equation of the plane passing through the intersection of two given planes that is also perpendicular to another plane, we must include both the intersection information and orthogonal condition:

• Identify the intersection line \(x=1\) and \(y+z=0\).
• Utilize the given perpendicular condition against \((1, -2, 1)\).
• Plug these into an expression \((1, -2, 1) \cdot (x - 1, y, z) = k\), solve for \(k\).

This leads us to the final plane equation \(x - 3y + z = 3\), fulfilling all conditions required by the original exercise context.