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Chapter 1: Problem 4

Find the equation of the plane which is perpendicular to the vector \((1,-1,0)\) and which contains the point \((1,2,1)\). Find also the perpendicular distance of this plane from the origin.

Short Answer

Expert verified
The plane's equation is \(x - y + 1 = 0\), and the distance from the origin is \(\frac{1}{\sqrt{2}}\).

Step by step solution

01

Determine Plane's Normal Vector

In the equation of a plane, the normal vector to the plane plays a crucial role. Given the vector \((1, -1, 0)\), this will serve as our normal vector \(\mathbf{n} = (1, -1, 0)\).
02

Use Point-Normal Form to Write Plane's Equation

The equation of a plane with a normal vector \(\mathbf{n} = (a, b, c)\) and a point \((x_0, y_0, z_0)\) on it is expressed as \(a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\). Substitute \(\mathbf{n} = (1, -1, 0)\) and the point \((1, 2, 1)\) into the equation to get: \(1(x - 1) - 1(y - 2) + 0(z - 1) = 0\).
03

Simplify the Plane's Equation

Simplify the equation from Step 2: \(x - 1 - y + 2 = 0\), which simplifies further to: \(x - y + 1 = 0\). This is the equation of the plane.
04

Find Distance from Origin to Plane

The distance \(d\) from a point \((x_1, y_1, z_1)\) to the plane \(ax + by + cz + d = 0\) is calculated using: \(d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}\). For the origin \((0, 0, 0)\) and plane \(x - y + 1 = 0\), substitute into the formula: \(d = \frac{|1(0) - 1(0) + 1|}{\sqrt{1^2 + (-1)^2 + 0^2}} = \frac{|1|}{\sqrt{2}} = \frac{1}{\sqrt{2}}\).

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Key Concepts

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

Normal Vector
In geometry, a normal vector is an essential tool when discussing planes. Simply put, it is a vector that is perpendicular to a surface or a plane. For a plane, this means that the normal vector points straight out from the surface.
  • Key role: The normal vector helps define the orientation of the plane. In this case, the normal vector is given as \((1, -1, 0)\).
  • Finding the normal vector: When the problem states that the plane is perpendicular to a vector, that vector is the normal vector of the plane.
  • Importance: Knowing the normal vector allows us to form equations for the plane since it directly influences the coefficients in the equation.
Whether you are finding a plane's equation or determining distances, the normal vector is the anchor of the plane's orientation.
Point-Normal Form
The point-normal form of a plane is a powerful way to write its equation, especially useful when provided with a point on the plane and the plane's normal vector. It involves straightforward substitution into a formula that relates these components.
  • Formula: The equation is \(a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\), where \((a, b, c)\) is the normal vector and \((x_0, y_0, z_0)\) is a point on the plane.
  • Application: By substituting the normal vector \((1, -1, 0)\) and the point \((1, 2, 1)\), you derive the plane's equation as \(x - y + 1 = 0\).
  • Simplification: After applying the point-normal form, simplify the equation to make it more manageable or easier to understand, resulting in its simplest terms.
This form is particularly valuable because it directly connects geometric terms of the plane and provides quick access to the equation necessary for further calculations.
Distance from Point to Plane
Calculating the distance from a point to a plane is a vital concept in understanding spatial relationships. It involves projecting the point onto the plane along the direction of the normal vector.
  • General formula: The distance \(d\) from a point \((x_1, y_1, z_1)\) to a plane \(ax + by + cz + d = 0\) is \(d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}\).
  • Example: For the origin \((0, 0, 0)\) and the plane equation \(x - y + 1 = 0\), plug the values into our formula to find distance \(d = \frac{1}{\sqrt{2}}\).
  • Interpreting results: A smaller distance value indicates that the point is closer to the plane, while a greater distance indicates the opposite.
This technique is invaluable in applications requiring measurement, spatial analysis, or determining proximity between geometrical figures.

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Most popular questions from this chapter

Show that, if a, b, \(\mathbf{c} \in \mathbb{R}^{3}\), then $$ (\mathbf{a} \times \mathbf{b}) \times \mathbf{c}+(\mathbf{b} \times \mathbf{c}) \times \mathbf{a}+(\mathbf{c} \times \mathbf{a}) \times \mathbf{b}=\mathbf{0} $$

Find a vector \(\mathbf{c}\) which is orthogonal to \((1,3,1)=\mathbf{a}\) and to \((2,1,1)=\mathbf{b}\), and verify that \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) is a basis for \(\mathbb{R}^{3}\).

Show that the vectors \((1,1,1),(1,2,3)\) and \((1,-1,-3)\) are linearly dependent.

Prove that the diagonals of a parallelogram bisect each other.

Let \(O A B C\) be a parallelogram, \(P\) be the mid-point of \(O A, S\) be the midpoint of \(O C\) and \(R\) be the point of intersection of \(P B\) and \(A S\). Find \(P R: R B\).
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