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Chapter 2: Problem 10
The equation of plane through intersection of the planes \(x+2 y+z-1=0\) and \(2x+y+3 z-2=0\) and perpendicular to plane \(x+y+z=1\) is (1) \(x+4 y-3 z+1=0\) (2) \(x-4 y+3 z-1=0\) (3) \(x-4 y+3 z=0\) (4) \(x+4 y-3 z=0\)
Short Answer
Expert verified
(2) \(x - 4y + 3z - 1 = 0\)
Step by step solution
01
Write the general form of the plane equation
The equation of a plane passing through the intersection of the given planes \(x + 2y + z - 1 = 0\) and \(2x + y + 3z - 2 = 0\) can be written as: \( (x + 2y + z - 1) + \lambda (2x + y + 3z - 2) = 0\), where \(\lambda\) is a scalar.
02
Expand and simplify the equation
Expand and combine like terms: \(x + 2y + z - 1 + \lambda (2x + y + 3z - 2) = 0\)\(x + 2y + z - 1 + \lambda 2x + \lambda y + \lambda 3z - 2\lambda = 0\)\((x + 2\lambda x) + (2y + \lambda y) + (z + 3\lambda z) - (1 + 2\lambda) = 0\)\((1 + 2\lambda)x + (2 + \lambda)y + (1 + 3\lambda)z - (1 + 2\lambda) = 0\)
03
Determine perpendicular condition
For the resultant plane to be perpendicular to the plane \(x + y + z = 1\), their normal vectors must be orthogonal. The normal vector of the required plane is \((1 + 2\lambda, 2 + \lambda, 1 + 3\lambda)\). The normal vector of the given perpendicular plane is \((1, 1, 1)\). For orthogonality, their dot product must be zero: \((1 + 2\lambda) \cdot 1 + (2 + \lambda) \cdot 1 + (1 + 3\lambda) \cdot 1 = 0\).If we expand this, we get: \(1 + 2\lambda + 2 + \lambda + 1 + 3\lambda = 0\)\(4 + 6\lambda = 0\)\(\lambda = -\frac{2}{3}\)
04
Substitute \(\lambda\) back into the plane equation
Substitute \(\lambda = -\frac{2}{3}\) back into the generalized plane equation: \((1 + 2(-\frac{2}{3}))x + (2 - \frac{2}{3})y + (1 + 3(-\frac{2}{3}))z - (1 - 2 \cdot \frac{2}{3})= 0\)\((1 - \frac{4}{3})x + (2 - \frac{2}{3})y + (1 - 2)z - (1 - \frac{4}{3}) = 0\)\(-\frac{1}{3}x + \frac{4}{3}y - z + \frac{1}{3} = 0\)Multiplying through by 3 to clear the fractions: \(-x + 4y - 3z + 1 = 0\)
05
Express the final plane equation
Thus, the final equation of the plane is: \(-x + 4y - 3z + 1 = 0\)Which can be written as \(x - 4y + 3z - 1 = 0\).
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Equation of a Plane
A plane in three-dimensional geometry can be uniquely identified by its equation. This equation can take several forms, but one of the most common is the general form, given by: \( Ax + By + Cz + D = 0 \) where \(A\), \(B\), and \(C\) are the coefficients representing the normal vector to the plane, and \(D\) is the constant term.
In our problem, we need to find a plane that intersects two given planes, \(x + 2y + z - 1 = 0\) and \(2x + y + 3z - 2 = 0\).
The equation of a plane passing through the intersection of two planes can be represented using a parameter \(\backslash lambda\): \((x + 2y + z - 1) + \backslash lambda (2x + y + 3z - 2) = 0 \).
Expand and simplify this to combine like terms and express it in the general form. This will give you a linear combination of the two original planes.
Perpendicular Planes
Two planes are perpendicular if their normal vectors are orthogonal to each other. This means that the dot product of their normal vectors must be zero.
The normal vector of the plane \(x + y + z = 1\) is \( (1, 1, 1) \).
The normal for our plane (from the intersection equation) is \ where the coefficients are: \((1 + 2\backslash lambda, 2 + \backslash lambda, 1 + 3\backslash lambda) \).
For the given plane to be perpendicular to \((1, 1, 1)\), we need: \( (1 + 2\backslash lambda) \cdot 1 + (2 + \backslash lambda) \cdot 1 + (1 + 3\backslash lambda) \cdot 1 = 0 \).
This condition will help us determine the value of \(\backslash lambda\), solving for which we get \(\backslash lambda = -\backslash frac{2}{3}\).
Intersection of Planes
At the intersection of two planes, a line is formed. When we add the condition that the resulting plane is perpendicular to a third plane, we can find a specific plane.
Using the value of \(\backslash lambda = -\backslash frac{2}{3}\), we substitute it back into our plane equation:
\[ (1 + 2(-\backslash frac{2}{3}))x + (2 - \backslash frac{2}{3})y + (1 + 3(-\backslash frac{2}{3}))z - (1 - 2 \cdot \frac{2}{3})= 0 \]
Multiplying through by 3 to clear the fractions, we get:
\[ -x + 4y - 3z + 1 = 0 \]
Therefore, the final equation of the plane is: \( x - 4y + 3z - 1 = 0 \), which confirms the correctness of the intersection condition with the given planes.
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