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Determine whether the given orthogonal set of vectors is orthonormal. If it is not, normalize the vectors to form an orthonormal set. $$\left[\begin{array}{l}\frac{3}{5} \\\\\frac{4}{5}\end{array}\right],\left[\begin{array}{r}-\frac{4}{5} \\\\\frac{3}{5}\end{array}\right]$$

Short Answer

Expert verified
The set of vectors is already orthonormal.

Step by step solution

01

Verify Orthogonality

To check if the set of vectors is orthogonal, we need to compute the dot product of the two vectors.Let \( \mathbf{u} = \begin{bmatrix} \frac{3}{5} \ \frac{4}{5} \end{bmatrix} \) and \( \mathbf{v} = \begin{bmatrix} -\frac{4}{5} \ \frac{3}{5} \end{bmatrix} \).The dot product \( \mathbf{u} \cdot \mathbf{v} = \frac{3}{5} \times \left(-\frac{4}{5}\right) + \frac{4}{5} \times \frac{3}{5} = -\frac{12}{25} + \frac{12}{25} = 0 \).Since the dot product is zero, the vectors are orthogonal.
02

Check for Normalization

Vectors are orthonormal if they are both orthogonal and each vector has a magnitude (or length) of 1. The magnitude of \( \mathbf{u} \) is computed as \[ \| \mathbf{u} \| = \sqrt{ \left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2 } = \sqrt{\frac{9}{25} + \frac{16}{25}} = \sqrt{\frac{25}{25}} = 1. \]The magnitude of \( \mathbf{v} \) is \[ \| \mathbf{v} \| = \sqrt{ \left(-\frac{4}{5}\right)^2 + \left(\frac{3}{5}\right)^2 } = \sqrt{\frac{16}{25} + \frac{9}{25}} = \sqrt{\frac{25}{25}} = 1. \]Since both vectors have a magnitude of 1, they are already normalized.

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Key Concepts

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

Orthogonal Vectors
Orthogonal vectors are vectors that are at right angles to each other in a given space. To determine if two vectors are orthogonal, we look at their dot product. If the dot product of two vectors is zero, they are orthogonal. This is because a zero dot product signifies that the vectors are perpendicular, sharing no direction in common.
For example, consider vectors \( \mathbf{u} = \begin{bmatrix} \frac{3}{5} \ \frac{4}{5} \end{bmatrix} \) and \( \mathbf{v} = \begin{bmatrix} -\frac{4}{5} \ \frac{3}{5} \end{bmatrix} \). Calculating the dot product, we have:
  • \( \mathbf{u} \cdot \mathbf{v} = \frac{3}{5} \times \left(-\frac{4}{5}\right) + \frac{4}{5} \times \frac{3}{5} \) = \( -\frac{12}{25} + \frac{12}{25} = 0 \)
Since the result is zero, \( \mathbf{u} \) and \( \mathbf{v} \) are indeed orthogonal vectors.
This property is fundamental in mathematical computations, especially in simplifying problems involving multiple vectors.
Vector Normalization
Vector normalization is the process of adjusting the vector's length, or magnitude, so that it becomes 1, creating a unit vector. This is significant because unit vectors retain the direction of the original vector but adhere to a standardized length.
Great for usage in numerous fields like computer graphics and physics. The formula for normalizing a vector \( \mathbf{w} \) is:
  • \( \mathbf{w}_{norm} = \frac{\mathbf{w}}{\|\mathbf{w}\|} \)
Where \( \|\mathbf{w}\| \) is the magnitude of the vector. For vectors \( \mathbf{u} \) and \( \mathbf{v} \) from our example:
  • \( \| \mathbf{u} \| = \sqrt{ \left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2 } = 1 \)
  • \( \| \mathbf{v} \| = \sqrt{ \left(-\frac{4}{5}\right)^2 + \left(\frac{3}{5}\right)^2 } = 1 \)
Both magnitudes equal 1, so \( \mathbf{u} \) and \( \mathbf{v} \) are already unit vectors and thus normalized.
Dot Product Calculation
Calculating the dot product of two vectors helps in determining their relationship regarding direction. It's a fundamental operation in vector algebra and is used in computing angles between vectors, projections, and more.

The formula to compute the dot product for two vectors \( \mathbf{a} = \begin{bmatrix} a_1 \ a_2 \end{bmatrix} \) and \( \mathbf{b} = \begin{bmatrix} b_1 \ b_2 \end{bmatrix} \) is:
  • \( \mathbf{a} \cdot \mathbf{b} = a_1 \times b_1 + a_2 \times b_2 \)
As shown in our problem, the vectors
  • \( \mathbf{u} = \begin{bmatrix} \frac{3}{5} \ \frac{4}{5} \end{bmatrix} \) and \( \mathbf{v} = \begin{bmatrix} -\frac{4}{5} \ \frac{3}{5} \end{bmatrix} \)
produced a dot product of zero,
indicating orthogonal vectors. Understanding the dot product is crucial for analyzing vector sets and their dependencies or independencies.

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Most popular questions from this chapter

Diagonalize the quadratic forms \(\mathrm{by}\) finding an orthogonal matrix \(Q\) such that the change of variable \(\mathbf{x}=Q \mathbf{y}\) transforms the given form into one with no cross-product terms. Give Q and the new quadratic form. $$x^{2}+8 x y+y^{2}$$

Classify each of the quadratic forms as positive definite, positive semidefinite, negative definite negative semidefinite, or indefinite. $$x_{1}^{2}+2 x_{2}^{2}$$

G is a generator matrix for a code C. Bring \(G\) into standard form and determine whether the corresponding code is equal to \(C\). $$G=\left[\begin{array}{lll} 0 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right]$$

Let \(\mathcal{B}=\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{n}\right\\}\) be an orthonormal basis for \(\mathbb{R}^{n}\) (a) Prove that, for any x and y in \(\mathbb{R}^{n}\) \\[\begin{array}{r}\mathbf{x} \cdot \mathbf{y}=\left(\mathbf{x} \cdot \mathbf{v}_{1}\right)\left(\mathbf{y} \cdot \mathbf{v}_{1}\right)+\left(\mathbf{x} \cdot \mathbf{v}_{2}\right)\left(\mathbf{y} \cdot \mathbf{v}_{2}\right)+\cdots \\ +\left(\mathbf{x} \cdot \mathbf{v}_{n}\right)\left(\mathbf{y} \cdot \mathbf{v}_{n}\right)\end{array}\\] (b) What does Parseval's Identity imply about the relationship between the dot products \(x \cdot y\) and \([\mathbf{x}]_{B} \cdot[\mathbf{y}]_{B} ?\) (This identity is called Parseval's Identity.)

Identify the quadric with the given equation and give its equation in standard form. $$-x^{2}-y^{2}-z^{2}+4 x y+4 x z+4 y z=12$$

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