Chapter 11: Problem 30
Let \(P\) be a point on a plane with normal vector \(\mathbf{n}\) and \(Q\) be a point off the plane. Show that the result of Example 10 of Section \(11.3\), the distance \(d\) between the point \(Q\) and the plane, can be expressed as $$ d=\frac{|\overrightarrow{P Q} \cdot \mathbf{n}|}{\|\mathbf{n}\|} $$ and use this result to find the distance from \((4,-2,3)\) to the plane \(4 x-4 y+2 z=2\).
Short Answer
Step by step solution
Understand the formula
Determine vector PQ
Find the normal vector
Calculate the dot product
Calculate the magnitude of n
Apply the distance formula
Verify your calculations
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Dot Product
- Helps determine the angle between vectors.
- Zero dot product indicates orthogonal vectors.
Normal Vector
- Gives direction to the plane.
- Helps in determining distances and angles related to the plane.
Plane Equation
- Defines the plane position and orientation.
- Integral for finding vectors and calculating distances.
Magnitude of Vector
- Essential for normalizing vectors.
- Used in various computations like distance and force.