Linear Algebra - Norms | Practice Hub

Find the norm of the vector $\vec{v} = (4, 3, 0)$

$\sqrt{7}$

Explanation

To find the norm, square each component, add, then take the square root:

$\sqrt{4^2 + 3^2 + 0^2 } = \sqrt{16 + 9 } = \sqrt{25 } = 5$

$\textbf{u} = \left (\frac{3}{5} \cos \theta ,\frac{4}{5} \sin \theta , -\frac{3}{5} \sin \theta, -\frac{4}{5} \cos \theta \right )$

True or false: $\textbf{u}$ is a unit vector regardless of the value of $\theta$ .

True

False

Explanation

$\textbf{u}$ is a unit vector if and only if $\left | \textbf{u} \right | = 1$

$\left | \textbf{u} \right |$ , the norm, or length, of $\textbf{u}$ can be found by adding the squares of the entries and taking the square root of the sum:

$\left | \textbf{u} \right | =\sqrt{ \left (\frac{3}{5} \cos \theta \right ) ^{2}+ \left (\frac{4}{5} \sin \theta\right )^{2} + \left ( -\frac{3}{5} \sin \theta\right )^{2}+ \left ( -\frac{4}{5} \cos \theta \right )^{2}}$

$\left | \textbf{u} \right | =\sqrt{ \frac{9}{25} \cos^{2} \theta + \frac{16}{25} \sin^{2} \theta+ \frac{9}{25} \sin^{2} \theta + \frac{16}{25} \cos^{2} \theta}$

$\left | \textbf{u} \right | =\sqrt{ \sin^{2} \theta + \cos^{2} \theta}$

Applying a trigonometric identity:

$\left | \textbf{u} \right | =\sqrt{1} = 1$ .

Therefore, $\textbf{u}$ is a unit vector regardless of the value of $\theta$ .

Which of these functions could be that of a Euclidean norm operator? You may assume each function is onto.

$f:\mathbb{R}^7 \rightarrow \mathbb{R}$

$g: \mathbb{Q}^2 \rightarrow \mathbb{N}$

$h: \mathbb{N}^3 \rightarrow \mathbb{Q}$

All of the other answers are norm operators

Explanation

This function's range is $\mathbb{R}$ , the set of all real numbers. In short, this is set of all possible "distances between two given numbers" in elementary linear algebra. would not be a norm. For example, $h(1,1,1) = \sqrt{1+1+1} =\sqrt{3}$ , which is not a rational number (part of $\mathbb{Q}$ ). Similarly, is also not a norm. We have $g(1,1) = \sqrt{1+1} = \sqrt{2}$ , which is not a natural number.

$\textbf{u} = \left ( \cos \theta , \frac{1}{6}, \sin \theta \right )$ .

To the nearest hundredth (radian), which of the following values of $\theta$ would make $\textbf{u}$ a unit vector?

$\textbf{u}$ cannot be a unit vector regardless of the value of $\theta$ .

Explanation

$\textbf{u}$ is a unit vector if and only if $\left | \textbf{u} \right | = 1$

$\left | \textbf{u} \right |$ , the norm, or length, of $\textbf{u}$ can be found by adding the squares of the entries and taking the square root of the sum:

$\left | \textbf{u} \right | =\sqrt{ \cos ^{2}\theta +\left ( \frac{1}{6} \right ) ^{2} + \sin ^{2}\theta}$

$\left | \textbf{u} \right | =\sqrt{ \sin ^{2}\theta+\cos ^{2}\theta + \frac{1}{36} }$

Since by a trigonometric identity,

$\sin ^{2}\theta+\cos ^{2}\theta =1$ for all $\theta$ ,

$\left | \textbf{u} \right | =\sqrt{ 1+ \frac{1}{36} }$

$\left | \textbf{u} \right | =\sqrt{ \frac{37}{36} } = \frac{\sqrt{ 37 } } {6}$ .

Therefore, for any value of $\theta$ , $\left | \textbf{u} \right | e 1$ . $\textbf{u}$ cannot be a unit vector.

$\textbf{u} = \left (\cos \theta , \frac{1}{2} , \frac{2}{3} \right )$

Evaluate $\theta \in [0, 2 \pi]$ (nearest hundredth of a radian) to make $\textbf{u}$ a unit vector.

$\textbf{u}$ cannot be a unit vector regardless of the value of $\theta$ .

Explanation

$\textbf{u}$ is a unit vector if and only if $\left | \textbf{u} \right | = 1$

$\left | \textbf{u} \right |$ , the norm, or length, of $\textbf{u}$ can be found by adding the squares of the entries and taking the square root of the sum:

$\left | \textbf{u} \right | = \sqrt{ \cos^{2} \theta+ \left ( \frac{1}{2} \right ) ^{2}+ \left ( \frac{2}{3} \right ) ^{2}}$

$\left | \textbf{u} \right | = \sqrt{ \cos^{2} \theta+ \frac{1}{4} + \frac{4}{9} }$

$\left | \textbf{u} \right | = \sqrt{ \cos^{2} \theta+ \frac{25}{36} }$

Set this value equal to 1:

$\sqrt{ \cos^{2} \theta+ \frac{25}{36} } = 1$

$\cos^{2} \theta+ \frac{25}{36} = 1$

$\cos^{2} \theta= \frac{11}{36}$

$\cos \theta=\sqrt{ \frac{11}{36}} \approx 0.5528$

We are looking for a value in radians $\theta \in [0, 2 \pi]$ , so

$\theta \approx \cos^{-1} 0.5528 \approx 0.99$ .

Find a unit vector in the same direction as $\vec{v} = (2, -3, \sqrt{3})$

$(\frac{1}{2} , -\frac{3}{4} , \frac{\sqrt{3}}{4})$

$(1, -1.5, \frac{\sqrt3}{2} )$

$(\frac{2}{\sqrt{2}}, -\frac{3}{\sqrt{2}}, \sqrt{\frac{3}{2}})$

Explanation

First, find the length of the vector: $\sqrt{ 2^2 + (-3)^2 + (\sqrt{3})^2 } = \sqrt{4+9+3} = \sqrt{16} = 4$

Because this vector has the length of 4 and a unit vector would have a length of 1, divide everything by 4:

$( \frac{ 2}{4} , \frac{-3}{4} , \frac{\sqrt{3}}{4}) = (\frac{1}{2} , -\frac{3}{4} , \frac{\sqrt{3}}{4})$

$\textbf{u} = \left ( \frac{1}{4}, \frac{1}{3} , \ln a \right )$ .

Evaluate to make $\textbf{u}$ a unit vector.

$\textbf{u}$ cannot be a unit vector regardless of the value of .

Explanation

$\textbf{u}$ is a unit vector if and only if $\left | \textbf{u} \right | = 1$

$\left | \textbf{u} \right |$ , the norm, or length, of $\textbf{u}$ can be found by adding the squares of the entries and taking the square root of the sum: