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Linear Algebra : Norms

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Example Questions

Example Question #1 : Norms

Find the norm of the following vector.

$\displaystyle A=[3,4,5]$

Possible Answers:

$\left \| A\right \|=\sqrt{2}$ $\displaystyle \left \| A\right \|=\sqrt{2}$

$\left \| A\right \|=5$ $\displaystyle \left \| A\right \|=5$

$\left \| A\right \|=5\sqrt{2}$ $\displaystyle \left \| A\right \|=5\sqrt{2}$

$\left \| A\right \|=2\sqrt{5}$ $\displaystyle \left \| A\right \|=2\sqrt{5}$

$\left \| A\right \|=\sqrt{52}$ $\displaystyle \left \| A\right \|=\sqrt{52}$

Correct answer:

$\left \| A\right \|=5\sqrt{2}$ $\displaystyle \left \| A\right \|=5\sqrt{2}$

Explanation:

The norm of a vector is simply the square root of the sum of each component squared.

$\left \| A\right \|=\sqrt{3^2+4^2+5^2}=\sqrt{9+16+25}=\sqrt{50}=5\sqrt{2}$ $\displaystyle \left \| A\right \|=\sqrt{3^2+4^2+5^2}=\sqrt{9+16+25}=\sqrt{50}=5\sqrt{2}$

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Example Question #1 : Norms

Find the norm of vector $\displaystyle A$.

$A=(\cos(x), \sin(x))$ $\displaystyle A=(\cos(x), \sin(x))$

Possible Answers:

$\left \| A\right \|=1$ $\displaystyle \left \| A\right \|=1$

$\left \| A\right \|=1+\sin^2(x)$ $\displaystyle \left \| A\right \|=1+\sin^2(x)$

$\left \| A\right \|=1+\cos^2(x)$ $\displaystyle \left \| A\right \|=1+\cos^2(x)$

$\left \| A\right \|=\cos(x)+\sin(x)$ $\displaystyle \left \| A\right \|=\cos(x)+\sin(x)$

$\left \| A\right \|=0$ $\displaystyle \left \| A\right \|=0$

Correct answer:

$\left \| A\right \|=1$ $\displaystyle \left \| A\right \|=1$

Explanation:

In order to find the norm, we need to square each component, sum them up, and then take the square root.

$\left \| A\right \|=\sqrt{\cos^2(x)+\sin^2(x)}=\sqrt{1}=1$ $\displaystyle \left \| A\right \|=\sqrt{\cos^2(x)+\sin^2(x)}=\sqrt{1}=1$

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Example Question #1 : Norms

Find the norm, $\begin{Vmatrix} \bf{a} \end{Vmatrix}$ $\displaystyle \begin{Vmatrix} \bf{a} \end{Vmatrix}$, given $\bf{a}= \begin{bmatrix} 2\\ 1\\ 7\\ -2 \end{bmatrix}$ $\displaystyle \bf{a}= \begin{bmatrix} 2\\ 1\\ 7\\ -2 \end{bmatrix}$

Possible Answers:

$\begin{Vmatrix} \bf{a} \end{Vmatrix}=\sqrt{8}$ $\displaystyle \begin{Vmatrix} \bf{a} \end{Vmatrix}=\sqrt{8}$

$\begin{Vmatrix} \bf{a} \end{Vmatrix}=-28$ $\displaystyle \begin{Vmatrix} \bf{a} \end{Vmatrix}=-28$

$\begin{Vmatrix} \bf{a} \end{Vmatrix}=58$ $\displaystyle \begin{Vmatrix} \bf{a} \end{Vmatrix}=58$

$\begin{Vmatrix} \bf{a} \end{Vmatrix}=\sqrt{58}$ $\displaystyle \begin{Vmatrix} \bf{a} \end{Vmatrix}=\sqrt{58}$

Correct answer:

$\begin{Vmatrix} \bf{a} \end{Vmatrix}=\sqrt{58}$ $\displaystyle \begin{Vmatrix} \bf{a} \end{Vmatrix}=\sqrt{58}$

Explanation:

By definition,

$\begin{Vmatrix} \bf{a} \end{Vmatrix}= \sqrt{a_1^2+a_2^2+a_3^2+...+a_n^2}$ $\displaystyle \begin{Vmatrix} \bf{a} \end{Vmatrix}= \sqrt{a_1^2+a_2^2+a_3^2+...+a_n^2}$,

therefore,

$\begin{Vmatrix} \bf{a} \end{Vmatrix}= \sqrt{2^2+1^2+7^2+(-2)^2}= \sqrt{58}$ $\displaystyle \begin{Vmatrix} \bf{a} \end{Vmatrix}= \sqrt{2^2+1^2+7^2+(-2)^2}= \sqrt{58}$.

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Example Question #1 : Norms

Calculate the norm of $\bf{A}\times \bf{B}$ $\displaystyle \bf{A}\times \bf{B}$ , or $\begin{Vmatrix} \bf{A}\times \bf{B} \end{Vmatrix}$ $\displaystyle \begin{Vmatrix} \bf{A}\times \bf{B} \end{Vmatrix}$, given

$\bf{A} = 3 \hat{i} - 1 \hat{j}$ $\displaystyle \bf{A} = 3 \hat{i} - 1 \hat{j}$,

$\bf{B} = -1 \hat{k}$ $\displaystyle \bf{B} = -1 \hat{k}$.

Possible Answers:

$\begin{Vmatrix} \bf{A}\times \bf{B} \end{Vmatrix}= 10$ $\displaystyle \begin{Vmatrix} \bf{A}\times \bf{B} \end{Vmatrix}= 10$

$\begin{Vmatrix} \bf{A}\times \bf{B} \end{Vmatrix}=\sqrt{10}$ $\displaystyle \begin{Vmatrix} \bf{A}\times \bf{B} \end{Vmatrix}=\sqrt{10}$

$\begin{Vmatrix} \bf{A}\times \bf{B} \end{Vmatrix}= 4$ $\displaystyle \begin{Vmatrix} \bf{A}\times \bf{B} \end{Vmatrix}= 4$

Can not be determined.

Correct answer:

$\begin{Vmatrix} \bf{A}\times \bf{B} \end{Vmatrix}=\sqrt{10}$ $\displaystyle \begin{Vmatrix} \bf{A}\times \bf{B} \end{Vmatrix}=\sqrt{10}$

Explanation:

First, we need to find $\bf{A}\times \bf{B}$ $\displaystyle \bf{A}\times \bf{B}$. This is, by definition,

$\bf{A \times \bf{B}}= \begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ 3& -1& 0\\ 0&0 &-1 \end{vmatrix}= 1\hat{i}+3\hat{j}$ $\displaystyle \bf{A \times \bf{B}}= \begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ 3& -1& 0\\ 0&0 &-1 \end{vmatrix}= 1\hat{i}+3\hat{j}$.

Therefore,

$\begin{Vmatrix} \bf{A}\times \bf{B} \end{Vmatrix}=\sqrt{1^2+3^2}=\sqrt{10}$ $\displaystyle \begin{Vmatrix} \bf{A}\times \bf{B} \end{Vmatrix}=\sqrt{1^2+3^2}=\sqrt{10}$.

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Example Question #1 : Norms

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

Possible Answers:

$\displaystyle 7$

$\displaystyle 1$

$\sqrt{7}$ $\displaystyle \sqrt{7}$

$\displaystyle 5$

Correct answer:

$\displaystyle 5$

Explanation:

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

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

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Example Question #1 : Norms

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

Possible Answers:

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

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

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

(4,4,4) $\displaystyle (4,4,4)$

Correct answer:

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

Explanation:

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

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

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

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Example Question #1 : Norms

Find the norm of the vector $\vec{v} = (-1, 3, 2,2)$ $\displaystyle \vec{v} = (-1, 3, 2,2)$

Possible Answers:

$2 \sqrt{3}$ $\displaystyle 2 \sqrt{3}$

$\sqrt{6}$ $\displaystyle \sqrt{6}$

$\displaystyle 3$

$3 \sqrt{2}$ $\displaystyle 3 \sqrt{2}$

Correct answer:

$3 \sqrt{2}$ $\displaystyle 3 \sqrt{2}$

Explanation:

$|| \vec{v} || = \sqrt{ (-1)^2 + 3^2 + 2^2 + 2^2 } = \sqrt{ 1 + 9 + 4 + 4 } = \sqrt{18}$ $\displaystyle || \vec{v} || = \sqrt{ (-1)^2 + 3^2 + 2^2 + 2^2 } = \sqrt{ 1 + 9 + 4 + 4 } = \sqrt{18}$

This can be simplified:

$\sqrt{18 } = \sqrt{3\cdot3\cdot2 } = 3 \sqrt{2 }$ $\displaystyle \sqrt{18 } = \sqrt{3\cdot3\cdot2 } = 3 \sqrt{2 }$

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Example Question #1 : Norms

Find the norm of the vector $\vec{v} = (2, -3, 1 )$ $\displaystyle \vec{v} = (2, -3, 1 )$

Possible Answers:

$\sqrt{4}$ $\displaystyle \sqrt{4}$

$\displaystyle 0$

$\sqrt{14 }$ $\displaystyle \sqrt{14 }$

$\sqrt{10}$ $\displaystyle \sqrt{10}$

Correct answer:

$\sqrt{14 }$ $\displaystyle \sqrt{14 }$

Explanation:

$|| \vec{v} || = \sqrt{2^2 + (-3)^2 + 1^2 } = \sqrt{4+9+1} = \sqrt{14}$ $\displaystyle || \vec{v} || = \sqrt{2^2 + (-3)^2 + 1^2 } = \sqrt{4+9+1} = \sqrt{14}$

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Example Question #2 : Norms

Find the norm of the vector $\vec{v}= (4, 2, -5 )$ $\displaystyle \vec{v}= (4, 2, -5 )$

Possible Answers:

$5 \sqrt{2}$ $\displaystyle 5 \sqrt{2}$

$\displaystyle 5$

$\displaystyle 9$

$3 \sqrt{5}$ $\displaystyle 3 \sqrt{5}$

$\displaystyle 3$

Correct answer:

$3 \sqrt{5}$ $\displaystyle 3 \sqrt{5}$

Explanation:

$|| \vec{v} || = \sqrt{ 4^2 + 2^2 + (-5)^2 } = \sqrt{16 + 4 + 25 } = \sqrt{45}$ $\displaystyle || \vec{v} || = \sqrt{ 4^2 + 2^2 + (-5)^2 } = \sqrt{16 + 4 + 25 } = \sqrt{45}$

This can be simplified:

$\sqrt{3 \cdot 3 \cdot 5 } = 3 \sqrt {5}$ $\displaystyle \sqrt{3 \cdot 3 \cdot 5 } = 3 \sqrt {5}$

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Example Question #1 : Norms

Find the norm of the vector $\vec{v} = (2, -3, -6 )$ $\displaystyle \vec{v} = (2, -3, -6 )$

Possible Answers:

$\sqrt{23}$ $\displaystyle \sqrt{23}$

$\sqrt{41 }$ $\displaystyle \sqrt{41 }$

$\displaystyle 7$

$\displaystyle 6$

Correct answer:

$\displaystyle 7$

Explanation:

$|| \vec{v} || = \sqrt{ 2^2 + (-3)^2 + (-6)^2 } = \sqrt{4 + 9 + 36 } = \sqrt{49} = 7$ $\displaystyle || \vec{v} || = \sqrt{ 2^2 + (-3)^2 + (-6)^2 } = \sqrt{4 + 9 + 36 } = \sqrt{49} = 7$

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Linear Algebra : Norms

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #1 : Norms

Example Question #1 : Norms

Example Question #1 : Norms

Example Question #1 : Norms

Example Question #1 : Norms

Example Question #1 : Norms

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