Chapter 26: Problem 74
The plane containing the line \(\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z-1}{3}\) and also containing its projection on the plane \(2 \mathrm{x}+3 \mathrm{y}-\mathrm{z}=5\), contains which one of the following points? [Jan. 11, 2019 (I)] (a) \((2,2,0)\) (b) \((-2,2,2)\) (c) \((0,-2,2)\) (d) \((2,0,-2)\)
Short Answer
Step by step solution
Identify Directional Vector and Point of the Line
Find Intersection with the Plane
Calculate the Projection Point
Equation of Plane Containing the Line
Checking Options
Re-Evaluation upon Error Discovery
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
3D Geometry
Coordinate geometry in three dimensions extends the concept of points, lines, and planes from a flat surface to the entire space. This allows for calculating distances, finding midpoints between two points, and understanding the spatial relationship of figures.
In problems involving planes and lines in space like the given exercise, grasping these fundamental principles aids in visually imagining and solving complex geometrical scenarios.
Line and Plane Intersection
To find the intersection, we substitute the parametric equations of the line into the plane's equation. This process determines whether a particular 't' value makes the equations hold true for both line and plane.
In practical terms, this means solving a system of equations where the line's coordinates fulfill the plane's equation. It's crucial for students to closely follow substitution steps to ensure they don't miss out on finding that specific intersection point.
Parametric Equations
For any point on a line, parametric equations will give values for x, y, and z depending on t. In essence, these equations offer a way to describe an entire line elegantly through the variation of a single variable.
In the exercise, parametric equations of the line were given by:
- \( x = 2t + 3 \)
- \( y = -t - 2 \)
- \( z = 3t + 1 \)
Vector Algebra
The direction vector from parametric equations, \( (2, -1, 3) \) , establishes a line's orientation in space. Meanwhile, differences between points can be expressed as vectors to aid in defining planes or finding a normal (perpendicular) vector.
Essentially, vector algebra simplifies computations involving distances, directions, and planes, playing a crucial role in any 3D geometrical analysis.