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Chapter 11: Problem 31

How far is the plane \(x+y-z=1\) from (0,0,0) and also from (1,1,-1)\(?\) Find the nearest points.

Short Answer

Expert verified
The distances are \(\frac{1}{\sqrt{3}}\) from (0,0,0) and \(\frac{2}{\sqrt{3}}\) from (1,1,-1); nearest point: \(\left(\frac{1}{3}, \frac{1}{3}, -\frac{1}{3}\right)\).

Step by step solution

01

Determine the normal vector

For the plane given by the equation \(x+y-z=1\), the normal vector can be directly obtained from the coefficients of \(x\), \(y\), and \(z\). The normal vector \(\mathbf{n}\) is \([1, 1, -1]\).
02

Compute distance from origin

The distance \(d\) from a point \((x_0, y_0, z_0)\) to the plane \(ax+by+cz=d\) is given by:\[d = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}\]Substituting the values for the origin \((0,0,0)\) into the formula:\[d = \frac{|0 + 0 + 0 - 1|}{\sqrt{1^2 + 1^2 + (-1)^2}} = \frac{1}{\sqrt{3}}\]
03

Compute distance from point (1,1,-1)

Applying the distance formula to the point \((1,1,-1)\):\[d = \frac{|1 \cdot 1 + 1 \cdot 1 - 1 \cdot (-1) - 1|}{\sqrt{1^2 + 1^2 + (-1)^2}}\]Simplifying inside the absolute value gives:\[1 + 1 + 1 - 1 = 2\]Thus, the distance is:\[d = \frac{2}{\sqrt{3}}\]
04

Find nearest point on plane from origin

The point on the plane nearest to \((0,0,0)\) can be found by moving in the direction of the normal vector. If a point on the plane is \((x, y, z)\), then \(x+y-z=1\). The nearest point is along the normal from the origin and is obtained by solving:\[(x, y, z) = t \cdot [1, 1, -1]\]Substitute in plane equation:\[t \cdot 1 + t \cdot 1 - t \cdot (-1) = 1 \implies 3t = 1 \implies t=\frac{1}{3}\]So, the point is \(\left(\frac{1}{3}, \frac{1}{3}, -\frac{1}{3}\right)\).
05

Find nearest point on plane from (1,1,-1)

The direction vector from \((1, 1, -1)\) along the normal is:\[(1+t, 1+t, -1-t) \]Substitute into the plane's equation:\[ (1+t) + (1+t) - (-1-t) = 1 \]This simplifies to:\[3 + 3t = 1 \implies 3t = -2 \implies t = -\frac{2}{3}\]So, the nearest point on the plane is:\[\left(\frac{1}{3}, \frac{1}{3}, -\frac{1}{3}\right)\].

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Key Concepts

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

Normal Vector
In the context of planes in three-dimensional space, understanding the normal vector is crucial. The normal vector is a vector that is perpendicular to the plane. It is essential because it helps in defining the orientation of the plane.

Given a plane equation like \[ ax + by + cz = d \]The coefficients \(a\), \(b\), and \(c\) are components of the normal vector \( \mathbf{n} \).
  • These coefficients directly inform us about the direction the plane faces.
  • A change in any of these coefficients tilts the plane in space.
  • For the plane \(x + y - z = 1\), the normal vector is simply \([1, 1, -1]\).
It serves as the main tool to gauge distances and nearest points related to the plane.
Distance Formula
The distance formula measures how far a given point in space is from a defined plane. It is integral in many applications, such as physics and engineering.

The formula used is:\[ d = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}} \]Here's a breakdown of what each part means:
  • \((x_0, y_0, z_0)\) is the point for which you want to compute the distance.
  • \(a, b, c\) are from the normal vector of the plane.
  • \(d\) represents the constant on the right of the plane equation.

For example, using this formula, we found the distances from (0,0,0) and (1,1,-1) to the plane were \(\frac{1}{\sqrt{3}}\) and \(\frac{2}{\sqrt{3}}\) respectively.
Point on Plane
Finding a point on a plane from a specific point is about determining where the perpendicular from the given point punctures the plane.

If you have a plane described by the equation \(x+y-z=1\), you need to ensure any point \((x, y, z)\) you derive satisfies this equation. The approach often involves starting from a baseline point:
  • Identify any given on- or off-plane point.
  • Move along the direction of the normal vector.
By applying these steps, you determine the nearest point from any point in space to the plane by altering the distance along the normal vector.
Nearest Point Calculation
When determining the nearest point along a plane from a given point, the normal vector is your best friend. It serves as the guide to trace the shortest route. This calculation is a geometric filtering process:
  • Start from your point, say the origin.
  • Move along the vector perpendicular to the plane, the normal vector.
  • Apply it directionally: if a point on the plane is \((x, y, z)\) and the equation is \(x+y-z=1\), substitute in \(t \cdot \mathbf{n}\).

For example, using this method, from \((0, 0, 0)\) the nearest point is \(\left(\frac{1}{3}, \frac{1}{3}, -\frac{1}{3}\right)\). By leveraging the normal vector, you can efficiently determine proximity from any point to a plane.

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Most popular questions from this chapter

Use the matrices \(A, B, C\). $$A=\left[\begin{array}{lll}1 & 4 & 0 \\ 0 & 2 & 6 \\ 0 & 0 & 3\end{array}\right] \quad B=\left[\begin{array}{lll}0 & 0 & 1 \\ 2 & 1 & 0 \\\ 6 & 4 & 0\end{array}\right] \quad C=\left[\begin{array}{rrr}1 & -1 & -3 \\\ -1 & 2 & 0 \\ 0 & -1 & 3\end{array}\right]$$ Which four entries of \(A\) give the upper left corner entry \(p\) of \(A^{-1}\), aftcr dividing by \(D=\operatorname{det} A\) ? Which four entries of \(A\) give the entry \(q\) in row \(1,\) column 2 of \(A^{-1}\) ? Find \(p\) and \(q\).

The equations \(a x+y=0, x+a y=0\) have the solution \(x=y=0 .\) For which two values of \(a\) are there other solutions (and what are the other solutions)?

True or false: (a) \(A \times B\) never equals \(A \cdot B\). (b) If \(\mathbf{A} \times \mathbf{B}=\mathbf{0}\) and \(\mathbf{A} \cdot \mathbf{B}=0,\) then either \(\mathbf{A}=\mathbf{0}\) or \(\mathbf{B}=\mathbf{0}\) (c) If \(\mathbf{A} \times \mathbf{B}=\mathbf{A} \times \mathbf{C}\) and \(\mathbf{A} \neq 0\), then \(\mathbf{B}=\mathbf{C}\)

If \((1,2,3),(3,4,7),\) and (2,1,2) are corners of a parallelogram, find all possible fourth corners.

The plane containing \(P=(0,1,1)\) and \(Q=(1,0,1)\) and \(R=(1,1,0)\) is perpendicular to the cross product \(N=\)_____. Find the equation of the plane and the area of triangle \(P Q R\).
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