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Chapter 5: Problem 11

Find the distance from the point (2,1,-2) to the plane $$6(x-1)+2(y-3)+3(z+4)=0$$

Short Answer

Expert verified
The distance from the point (2, 1, -2) to the plane \(6x + 2y + 3z = 0\) is \(\frac{8}{7}\).

Step by step solution

01

Convert the plane equation to standard form

Firstly, we need to expand and simplify the given equation. Expand the given plane equation: $$6x - 6 + 2y - 6 + 3z + 12 = 0$$ Now, we group the variables: $$6x + 2y + 3z - 6 - 6 + 12 = 0$$ Our simplified equation becomes: $$6x + 2y + 3z = 0$$ Now, the equation is in the standard form: $$Ax + By + Cz + D = 0$$ So, A = 6, B = 2, C = 3, and D = 0.
02

Apply the point-to-plane distance formula

Now that we have the coefficients and the constant term from the plane equation, we can plug them into the point-to-plane distance formula along with the given point (2,1,-2), to find the distance: $$Distance = \frac{|6(2) + 2(1) + 3(-2) + 0|}{\sqrt{6^2 + 2^2 + 3^2}}$$
03

Calculate the distance

Carry out the calculations in the numerator: $$|12 + 2 - 6| = |8| = 8$$ And the denominator: $$\sqrt{6^2 + 2^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$$ Now, divide the absolute value in the numerator by the square root in the denominator: $$Distance = \frac{8}{7}$$ The distance from the point (2,1,-2) to the given plane is: $$\frac{8}{7}$$

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Most popular questions from this chapter

Let $$A=\left(\begin{array}{ll} 2 & 1 \\ 1 & 1 \\ 2 & 1 \end{array}\right) \quad \text { and } \quad \mathbf{b}=\left(\begin{array}{r} 12 \\ 6 \\ 18 \end{array}\right)$$ (a) Use the Gram-Schmidt process to find an orthonormal basis for the column space of \(A\) (b) Factor \(A\) into a product \(Q R,\) where \(Q\) has an orthonormal set of column vectors and \(R\) is upper triangular. (c) Solve the least squares problem \(A \mathbf{x}=\mathbf{b}\)

Let $$\mathbf{u}_{1}=\left(\begin{array}{r} \frac{1}{3 \sqrt{2}} \\ \frac{1}{3 \sqrt{2}} \\ -\frac{4}{3 \sqrt{2}} \end{array}\right), \mathbf{u}_{2}=\left(\begin{array}{c} \frac{2}{3} \\ \frac{2}{3} \\ \frac{1}{3} \end{array}\right), \mathbf{u}_{3}=\left(\begin{array}{r} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \\ 0 \end{array}\right)$$ (a) Show that \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\right\\}\) is an orthonormal basis for \(\mathbb{R}^{3}\) (b) Let \(\mathbf{x}=(1,1,1)^{T}\). Write \(\mathbf{x}\) as a linear combination of \(\mathbf{u}_{1}, \mathbf{u}_{2},\) and \(\mathbf{u}_{3}\) using Theorem 5.5 .2 and use Parseval's formula to compute \(\|\mathbf{x}\|\)

Let \(A\) be an \(m \times n\) matrix whose column vectors are mutually orthogonal, and let \(\mathbf{b} \in \mathbb{R}^{m}\). Show that if \(\mathbf{y}\) is the least squares solution of the system \(A \mathbf{x}=\mathbf{b},\) then $$y_{i}=\frac{\mathbf{b}^{T} \mathbf{a}_{i}}{\mathbf{a}_{i}^{T} \mathbf{a}_{i}} \quad i=1, \ldots, n$$

The functions \(\cos x\) and \(\sin x\) form an orthonormal \(\operatorname{set} \operatorname{in} C[-\pi, \pi] .\) If \(f(x)=3 \cos x+2 \sin x\) and \(g(x)=\cos x-\sin x\) use Corollary 5.5 .3 to determine the value of $$\langle f, g\rangle=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) g(x) d x$$

If \(A\) is an \(m \times n\) matrix of rank \(r,\) what are the dimensions of \(N(A)\) and \(N\left(A^{T}\right) ?\) Explain.
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