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Chapter 5: Problem 10

Find the distance from the point (1,1,1) to the plane \(2 x+2 y+z=0\)

Short Answer

Expert verified
The distance between the point (1,1,1) and the plane \(2x + 2y + z = 0\) is \(\frac{5}{3}\) units.

Step by step solution

01

Identify the equation of the plane

In this exercise, we have the equation of the plane: \[2x + 2y + z = 0\]
02

Rewrite the equation of the plane

To simplify the process of finding the distance, rewrite the equation of the plane in the form ax + by + cz = d, where a, b, and c are the coefficients of x, y, and z respectively. In our case, we already have the plane equation in the required form: \[2x + 2y + z = 0\] Here, a = 2, b = 2, c = 1, and d = 0.
03

Calculate the components of the point and the plane equation

We have the coefficient values a, b, and c from the plane equation, and we need to find the distance (D) between the point (x鈧�, y鈧�, z鈧�) and the plane. The given point is (1,1,1).
04

Use the point-to-plane distance formula

To find the distance between a point and a plane, we use the point-to-plane distance formula: \(D = \frac{|ax_{0}+by_{0}+cz_{0}+d|}{\sqrt{a^{2}+b^{2}+c^{2}}}\) In our case, we have: \(D = \frac{|2(1)+2(1)+1(1)+0|}{\sqrt{2^{2}+2^{2}+1^{2}}}\)
05

Solve for the distance

Plug the given values into the formula: \(D = \frac{|2+2+1+0|}{\sqrt{4+4+1}}\) \(D = \frac{5}{\sqrt{9}}\) \(D = \frac{5}{3}\) The distance between the point (1,1,1) and the plane \(2x + 2y + z = 0\) is \(\frac{5}{3}\) units.

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Most popular questions from this chapter

Compute \(\|\mathbf{x}\|_{1},\|\mathbf{x}\|_{2},\) and \(\|\mathbf{x}\|_{\infty}\) for each of the following vectors in \(\mathbb{R}^{3}\) : (a) \(\mathbf{x}=(-3,4,0)^{T}\) (b) \(\mathbf{x}=(-1,-1,2)^{T}\) (c) \(\mathbf{x}=(1,1,1)^{T}\)

Prove that, for any \(\mathbf{u}\) and \(\mathbf{v}\) in an inner product space \(V\) $$\|\mathbf{u}+\mathbf{v}\|^{2}+\|\mathbf{u}-\mathbf{v}\|^{2}=2\|\mathbf{u}\|^{2}+2\|\mathbf{v}\|^{2}$$ Give a geometric interpretation of this result for the vector space \(\mathbb{R}^{2}\)

Show that $$\|\mathbf{x}\|_{1}=\sum_{i=1}^{n}\left|x_{i}\right|$$ defines a norm on \(\mathbb{R}^{n}\)

Which of the following sets of vectors form an orthonormal basis for \(\mathbb{R}^{2} ?\) (a) \(\left\\{(1,0)^{T},(0,1)^{T}\right\\}\) (b) \(\left\\{\left(\frac{3}{5}, \frac{4}{5}\right)^{T},\left(\frac{5}{13}, \frac{12}{13}\right)^{T}\right\\}\) (c) \(\left\\{(1,-1)^{T},(1,1)^{T}\right\\}\) \((\mathbf{d})\left\\{\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)^{T},\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)^{T}\right\\}\)

Let \(A \in \mathbb{R}^{m \times n}\) and let \(\hat{\mathbf{x}}\) be a solution of the least squares problem \(A \mathbf{x}=\mathbf{b} .\) Show that a vector \(\mathbf{y} \in \mathbb{R}^{n}\) will also be a solution if and only if \(\mathbf{y}=\hat{\mathbf{x}}+\mathbf{z},\) for some vector \(\mathbf{z} \in N(A)\) \(\left[\text {Hint}: N\left(A^{T} A\right)=N(A) .\right]\)
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