/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Let \(\mathbf{y}=\left[\begin{ar... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 15

Let \(\mathbf{y}=\left[\begin{array}{r}{5} \\ {-9} \\ {5}\end{array}\right], \mathbf{u}_{1}=\left[\begin{array}{r}{-3} \\ {-5} \\ {1}\end{array}\right], \mathbf{u}_{2}=\left[\begin{array}{r}{-3} \\ {2} \\ {1}\end{array}\right] .\) Find the distance from y to the plane in \(\mathbb{R}^{3}\) spanned by \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\)

Short Answer

Expert verified
The distance from the vector \(\mathbf{y}\) to the plane is \(\sqrt{91}\).

Step by step solution

01

Understand the Problem

We are asked to find the distance from a vector \(\mathbf{y}\) to a plane spanned by two other vectors \(\mathbf{u}_1\) and \(\mathbf{u}_2\) in \(\mathbb{R}^3\). The distance from a point to a plane can be thought of as the magnitude of the vector from the point to its projection onto the plane.
02

Find a Normal Vector to the Plane

To find the distance, we first need a normal vector to the plane spanned by \(\mathbf{u}_1\) and \(\mathbf{u}_2\). A normal vector can be found via the cross product: \(\mathbf{n} = \mathbf{u}_1 \times \mathbf{u}_2\). Calculate: \[\mathbf{n} = \left(\begin{array}{r}{-3} \ {-5} \ {1}\end{array}\right) \times \left(\begin{array}{r}{-3} \ {2} \ {1}\end{array}\right) = \left(\begin{array}{r}(-5)(1) - (1)(2) \ (1)(-3) - (-3)(1) \ (-3)(2) - (-5)(-3)\end{array}\right) = \left(\begin{array}{r}-7 \ 0 \ -21\end{array}\right)\]Thus, \(\mathbf{n} = \left(\begin{array}{r}{-7} \ {0} \ {-21}\end{array}\right)\).
03

Project \(\mathbf{y}\) onto the Normal Vector

The projection of \(\mathbf{y}\) onto the normal vector \(\mathbf{n}\) is calculated as follows:\[\text{proj}_{\mathbf{n}} \mathbf{y} = \frac{\mathbf{y} \cdot \mathbf{n}}{\|\mathbf{n}\|^2} \mathbf{n}\]Calculate \(\mathbf{y} \cdot \mathbf{n}\):\[\mathbf{y} \cdot \mathbf{n} = \begin{bmatrix} 5 \ -9 \ 5 \end{bmatrix} \cdot \begin{bmatrix} -7 \ 0 \ -21 \end{bmatrix} = 5(-7) + (-9)(0) + 5(-21) = -35 - 105 = -140\]Calculate \(\|\mathbf{n}\|^2\):\[\|\mathbf{n}\|^2 = (-7)^2 + 0^2 + (-21)^2 = 49 + 441 = 490\]Thus, the projection is:\[\text{proj}_{\mathbf{n}} \mathbf{y} = \frac{-140}{490} \begin{bmatrix} -7 \ 0 \ -21 \end{bmatrix} = \left(\frac{2}{7}\right) \begin{bmatrix} -7 \ 0 \ -21 \end{bmatrix} = \begin{bmatrix} 2 \ 0 \ 6 \end{bmatrix}\]
04

Calculate the Distance

The distance from \(\mathbf{y}\) to the plane is the magnitude of the vector created by subtracting the projection from \(\mathbf{y}\):\[\text{Distance} = \left\| \mathbf{y} - \text{proj}_{\mathbf{n}} \mathbf{y} \right\| = \left\| \begin{bmatrix} 5 \ -9 \ 5 \end{bmatrix} - \begin{bmatrix} 2 \ 0 \ 6 \end{bmatrix} \right\|\]Calculating:\[\begin{bmatrix} 5 \ -9 \ 5 \end{bmatrix} - \begin{bmatrix} 2 \ 0 \ 6 \end{bmatrix} = \begin{bmatrix} 3 \ -9 \ -1 \end{bmatrix}\]Then, compute the magnitude:\[\sqrt{3^2 + (-9)^2 + (-1)^2} = \sqrt{9 + 81 + 1} = \sqrt{91}\]
05

Conclude the Solution

Thus, the distance from \(\mathbf{y}\) to the plane is \(\sqrt{91}\). The projection of \(\mathbf{y}\) onto the plane's normal vector allows us to find this distance.

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Key Concepts

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

Cross Product
The cross product is a mathematical operation that takes two vectors in three-dimensional space and returns a new vector that is perpendicular to both of the original vectors. This orthogonal vector is often utilized to find a normal vector to a plane defined by the two original vectors. In our exercise, the plane is spanned by vectors \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \). The normal vector, \( \mathbf{n} \), is obtained using the cross product, defined as:
  • For vectors \( \mathbf{a} = [a_1, a_2, a_3] \) and \( \mathbf{b} = [b_1, b_2, b_3] \), their cross product is \( \mathbf{a} \times \mathbf{b} = [a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1] \).
  • The resulting vector is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \) and lies on a plane that passes through these vectors.
By applying this to \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \), we obtain the vector \( \mathbf{n} = [-7, 0, -21] \). This vector serves as the normal vector to our plane.
Vector Projection
Vector projection is the operation of projecting one vector onto another vector. It allows us to find the component of the first vector in the direction of the second vector. This is particularly useful for calculating the shortest distance from a point to a plane. In the current problem, we needed to project vector \( \mathbf{y} \) onto the normal vector \( \mathbf{n} \) of the plane.
  • The formula for the projection of vector \( \mathbf{y} \) onto a vector \( \mathbf{n} \) is given by: \[ \text{proj}_{\mathbf{n}} \mathbf{y} = \frac{\mathbf{y} \cdot \mathbf{n}}{\|\mathbf{n}\|^2} \mathbf{n} \]
  • This operation gives the shadow of \( \mathbf{y} \) on \( \mathbf{n} \), which represents how much of \( \mathbf{y} \) goes in the direction of \( \mathbf{n} \).
In our case, the result of the projection is \( \begin{bmatrix} 2 & 0 & 6 \end{bmatrix} \), a key factor determining the distance from \( \mathbf{y} \) to the plane.
Magnitude of a Vector
The magnitude of a vector is a measure of its length, calculated using the Pythagorean theorem in three-dimensional space. This concept is important for finding distances and lengths in vector space.
  • For a vector \( \mathbf{a} = [a_1, a_2, a_3] \), its magnitude is calculated as: \[ \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]
  • It represents the length of the vector from origin to the point \([a_1, a_2, a_3] \) in space.
  • Magnitude is always a non-negative quantity.
In our problem, the distance from \( \mathbf{y} \) to the plane is found by computing the magnitude of the vector \( [3, -9, -1] \), resulting in \( \sqrt{91} \). This reflects the shortest path along perpendicular to the plane.
Normal Vector to a Plane
A normal vector to a plane in three-dimensional space is a vector that is perpendicular to the plane. In essence, it defines the orientation of the plane. To find a normal vector, we frequently utilize the cross product of two non-parallel vectors that lie in the plane.
  • Given vectors \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \) within a plane, the cross product \( \mathbf{u}_1 \times \mathbf{u}_2 \) yields a vector normal to the plane.
  • This normal vector is crucial in defining the plane's equation and is used to calculate distances to the plane.
In the exercise, the normal vector to the plane spanned by \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \) is \( \mathbf{n} = [-7, 0, -21] \). This normal vector is essential for projecting \( \mathbf{y} \) and determining the exact distance from \( \mathbf{y} \) to the plane.

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Most popular questions from this chapter

In Exercises 11 and \(12,\) find the closest point to \(y\) in the subspace \(W\) spanned by \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2} .\) $$ \mathbf{y}=\left[\begin{array}{r}{3} \\ {-1} \\ {1} \\\ {13}\end{array}\right], \mathbf{v}_{1}=\left[\begin{array}{r}{1} \\ {-2} \\\ {-1} \\ {2}\end{array}\right], \mathbf{v}_{2}=\left[\begin{array}{r}{-4} \\\ {1} \\ {0} \\ {3}\end{array}\right] $$

In Exercises \(1-4,\) find a least-squares solution of \(A \mathbf{x}=\mathbf{b}\) by (a) constructing the normal equations for \(\hat{\mathbf{x}}\) and (b) solving for \(\hat{\mathbf{x}}\) . $$ A=\left[\begin{array}{rr}{1} & {-2} \\ {-1} & {2} \\ {0} & {3} \\ {2} & {5}\end{array}\right], \mathbf{b}=\left[\begin{array}{r}{3} \\ {1} \\ {-4} \\\ {2}\end{array}\right] $$

In Exercises 13 and \(14,\) find the best approximation to \(z\) by vectors of the form \(c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}\) $$ \mathbf{z}=\left[\begin{array}{r}{2} \\ {4} \\ {0} \\\ {-1}\end{array}\right], \mathbf{v}_{1}=\left[\begin{array}{r}{2} \\ {0} \\\ {-1} \\ {-3}\end{array}\right], \mathbf{v}_{2}=\left[\begin{array}{r}{5} \\\ {-2} \\ {4} \\ {2}\end{array}\right] $$

Let \(\mathbb{P}_{3}\) have the inner product given by evaluation at \(-3,-1\) \(1,\) and \(3 .\) Let \(p_{0}(t)=1, p_{1}(t)=t,\) and \(p_{2}(t)=t^{2}\) a. Compute the orthogonal projection of \(p_{2}\) onto the sub- space spanned by \(p_{0}\) and \(p_{1}\) . b. Find a polynomial \(q\) that is orthogonal to \(p_{0}\) and \(p_{1},\) such that \(\left\\{p_{0}, p_{1}, q\right\\}\) is an orthogonal basis for \(\operatorname{Span}\left\\{p_{0}, p_{1}, p_{2}\right\\} .\) Scale the polynomial \(q\) so that its vec- tor of values at \((-3,-1,1,3)\) is \((1,-1,-1,1)\)

In Exercises 13 and \(14,\) the columns of \(Q\) were obtained by applying the Gram-Schmidt process to the columns of \(A .\) Find an upper triangular matrix \(R\) such that \(A=Q R .\) Check your work. $$ A=\left[\begin{array}{rr}{5} & {9} \\ {1} & {7} \\ {-3} & {-5} \\ {1} & {5}\end{array}\right], Q=\left[\begin{array}{rr}{5 / 6} & {-1 / 6} \\ {1 / 6} & {5 / 6} \\ {-3 / 6} & {1 / 6} \\ {1 / 6} & {3 / 6}\end{array}\right] $$
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