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Consider the planes \(P_{1}: 2 x+y+z+4=0\) \(P_{2}: y-z+4=0\) and \(P_{3}: 3 x+2 y+z+8=0\) Let \(L_{1}, L_{2}, L_{3}\) be the lines of intersection of the planes \(P_{2}\) and \(P_{3}, P_{3}\) and \(P_{1}\), and \(P_{1}\) and \(P_{2}\) respectively. Then, (a) at least two of the lines \(L_{1}, L_{2}\) and \(L_{3}\) are non-parallel (b) at least two of the lines \(L_{1}, L_{2}\) and \(L_{3}\) are parallel (c) the three planes intersect in a line (d) the three planes form a triangular prism

Short Answer

Expert verified
(b): At least two of the lines are parallel.

Step by step solution

01

Determine Line of Intersection L1

Identify line of intersection of planes \( P_2 \) and \( P_3 \). First, rewrite the planes:\[ P_2: y-z+4=0 \rightarrow y = z - 4 \]\[ P_3: 3x + 2y + z + 8 = 0 \rightarrow z = -3x - 2y - 8 \]Substitute \( y = z - 4 \) into \( z = -3x - 2y - 8 \):\[ z = -3x - 2(z-4) - 8 \]Solve this equation for \( z \):\[ z = -3x - 2z + 8 - 8 \]\[ 3z = -3x + 8 \rightarrow z = -x + \frac{8}{3} \]For \( y = z - 4 \), we have \( y = -x + \frac{8}{3} - 4 \).Thus, \( L_1 \) parameterizes to : \[ x = t, \quad y = -t + \frac{8}{3} - 4, \quad z = -t + \frac{8}{3} \].
02

Determine Line of Intersection L2

Identify line of intersection of planes \( P_3 \) and \( P_1 \). Starting with plane equations:\[ P_3: 3x + 2y + z + 8 = 0 \]\[ P_1: 2x + y + z + 4 = 0 \]Subtract \( P_1 \) from \( P_3 \):\[ (3x+2y+z+8) - (2x+y+z+4) = x + y + 4 \]Then solve the system:\[ y = -4 - x \]Control with \( P_1 \):\[ 2x + (-4-x) + z + 4 = 0 \rightarrow x + z = -6 \rightarrow z = -x - 6 \]Thus, \( L_2 \) can be parameterized by: \[ x = t, \quad y = -4 - t, \quad z = -t - 6 \].
03

Determine Line of Intersection L3

Investigate line of intersection of planes \( P_1 \) and \( P_2 \):\[ P_1: 2x + y + z + 4 = 0 \rightarrow z = -2x - y - 4 \]\[ P_2: y - z + 4 = 0 \rightarrow z = y + 4 \]Set \( z = y + 4 \) equal to \( z = -2x - y - 4 \):\[ y + 4 = -2x - y - 4 \rightarrow 2y = -2x - 8 \rightarrow y = -x - 4 \]Therefore, \( z = y + 4 = -x \) leads:\[ x = t, \quad y = -t - 4, \quad z = -t \].
04

Analyze Parallellism of Lines

Compare direction vectors of lines \( L_1: (1, -1, -1) \), \( L_2: (1, -1, -1) \), and \( L_3: (1, -1, 0) \).- \( L_1 \) and \( L_2 \) have identical directions, thus they are parallel.- \( L_3 \) has a different direction due to the component 0 in place of -1 for \( z \), hence not parallel to either \( L_1 \) or \( L_2 \).
05

Conclusion on Plane Intersection

Since two lines \( L_1 \) and \( L_2 \) are parallel and \( L_3 \) is not parallel to any, the correct answer is (b): At least two of the lines are parallel. The planes do not converge into a single line or triangular prism, therefore options (c) and (d) are incorrect.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Line of Intersection
When two planes intersect, the result is often a line. This is known as a line of intersection. To find this line, we need to solve the system of equations representing the two planes. For instance, consider planes \( P_2 \) and \( P_3 \) from the exercise. By manipulating and equating their equations, \( y = z - 4 \) for \( P_2 \) and \( z = -3x - 2y - 8 \) for \( P_3 \), substituting one into the other helps us isolate and find the coordinates as parametric equations. You might encounter a general form like \( x = t, y = -t + \frac{8}{3} - 4, z = -t + \frac{8}{3} \), representing a line's path. This approach is efficient for defining each point on the line as \( t \) varies. Lines of intersection are crucial in determining the spatial relationships between the planes.
Plane Equations
Plane equations are fundamental in vector algebra, usually written in the form \( ax + by + cz + d = 0 \). The coefficients \( a, b, \) and \( c \) represent the plane's normal vector, which is perpendicular to the plane itself. This normal vector determines the plane's orientation in space. Each plane equation in the exercise, such as \( P_1: 2x + y + z + 4 = 0 \), represents an infinite set of points lying on its surface.
Solving these plane equations by plug and substitute methods, particularly while finding intersections, is pivotal. One plane's equation can be rearranged to express one variable in terms of others, helping uncover intersection lines. Understanding plane equations allows for the exploration of their intersections, providing insights into the geometry of the planes' relationships.
Parallel Lines
In geometry, parallel lines are lines in the same plane that never meet. They have the same direction vectors or are scalar multiples of each other. For the given exercise, the lines \( L_1 \) and \( L_2 \) are parallel, having direction vectors of \( (1, -1, -1) \).
To ascertain this, compare the vector components: if they are proportional, the lines are parallel. Having parallel lines indicates that the intersection lines of different plane pairs have the same orientation but do not converge to a single point. This understanding helps to simplify the exhaustive analysis of spatial configurations where the combination of these planes gets analyzed.
Plane Intersections
The intersection of planes can vary dramatically based on plane orientation. Three planes can intersect:
  • At a single point.
  • Along a line if two planes intersect in a line, and the third plane intersects this line.
  • At two parallel lines if each pair shares a line that doesn't coincide with the third plane.
In the exercise example, \( L_1 \) and \( L_2 \) were parallel, hence no single line formed by all planes. Instead, the correct interpretation was that planes do not meet entirely in a line but create individual intersection lines. Plane intersections serve as geometric pieces assembled to understand how three-dimensional spaces and forms interact
and relate to each other effectively.

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Most popular questions from this chapter

The line \(\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-1}{-1}\) intersects the curve \(x^{2}+y^{2}=r^{2}, z=0\) then (a) Equation of the plane through \((0,0,0)\) perpendicular to the given line is \(3 x+2 y-z=0\) (b) \(r=\sqrt{26}\) (c) \(r=6\) (d) \(r=7\)

Assuming the plane \(4 x-3 y+7 z=0\) to be horizontal, the direction cosines of the line of greatest slope in the plane \(2 x+y-5 z=0\) are \(\begin{array}{ll}\text { (a) }\left(\frac{3}{\sqrt{11}}, \frac{-1}{\sqrt{11}}, \frac{1}{\sqrt{11}}\right) & \text { (b) }\left(\frac{3}{\sqrt{11}}, \frac{1}{\sqrt{11}}, \frac{-1}{\sqrt{11}}\right)\end{array}\) (c) \(\left(\frac{-3}{\sqrt{11}}, \frac{1}{\sqrt{11}}, \frac{1}{\sqrt{11}}\right)\) (d) \(\left(\frac{1}{\sqrt{11}}, \frac{3}{\sqrt{11}}, \frac{-1}{\sqrt{11}}\right)\)

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If the straight lines $$ \frac{\dot{x}-1}{k}=\frac{y-2}{2}=\frac{z-3}{3} \text { and } \frac{x-2}{3}=\frac{y-3}{k}=\frac{z-1}{2} $$ intersect at a point, then the integer \(k\) is equal to [AIEEE 2008] (a) \(-2\) (b) \(-5\) (c) 5 (d) 2

Statement I The point \(A(1,0,7)\) is the mirror image of the point \(B(1,6,3)\) in the line \(\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}\). Statement II The line \(\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}\) bisects the line segment joining \(A(1,0,7)\) and \(B(1,6,3)\). [AIEEE 2011] (a) Statement \(I\) is true, Statement II is true; Statement II is not a correct explanation for Statement I (b) Statement I is true, Statement II is false (c) Statement I is false, Statement II is true (d) Statement \(\mathrm{I}\) is true, Statement II is true; Statement II is a correct explanation for Statement I

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