Norms - Linear Algebra

Card 0 of 212

Question

$\textbf{u} = (3, -1, 6)$

$\textbf{v}= (5, 2, a)$ , an integer.

For which values of does it hold that $\left | \textbf{u} \right |< \left | \textbf{v} \right |$ ?

Answer

$\left | \textbf{u} \right |$ , the norm, or length, of $\textbf{u}$ , can be calculated by adding the squares of the numbers and taking the square root of the sum; $\left | \textbf{v} \right |$ can be calculated similarly.

We are seeking the real values of so that $\left | \textbf{u} \right |< \left | \textbf{v} \right |$ ; since both norms must be nonnegative, it suffices to find so that $\left | \textbf{u} \right |^{2} < \left | \textbf{v} \right | ^{2}$ .

$\left | \textbf{u} \right |^{2} = 3^{2}+ (-1)^{2}+ 6^{2} = 9 + 1 + 36 = 46$

$\left | \textbf{v} \right | ^{2} = 5^{2}+ 2^{2}+ a^{2} = 25 + 4 + a^{2} = a^{2}+ 29$

For $\left | \textbf{u} \right |^{2} < \left | \textbf{v} \right | ^{2}$ to hold, it must hold that

$46 < a^{2}+ 29$ , or

$a^{2}+ 29> 46$

This is true if

$a^{2} > 17$ ,

which in turn holds if

$|a| > \sqrt{17}$ .

Since it is specified that is an integer, it holds that $a \in \left \{ ...-7, -6, -5, 5, 6, 7, ...\right \}$ .

Compare your answer with the correct one above

Question

$\textbf{u} = (3, -1, 6)$

$\textbf{v}= (5, 2, a)$ , an integer.

For which values of does it hold that $\left | \textbf{u} \right |< \left | \textbf{v} \right |$ ?

Answer

$\left | \textbf{u} \right |$ , the norm, or length, of $\textbf{u}$ , can be calculated by adding the squares of the numbers and taking the square root of the sum; $\left | \textbf{v} \right |$ can be calculated similarly.

We are seeking the real values of so that $\left | \textbf{u} \right |< \left | \textbf{v} \right |$ ; since both norms must be nonnegative, it suffices to find so that $\left | \textbf{u} \right |^{2} < \left | \textbf{v} \right | ^{2}$ .

$\left | \textbf{u} \right |^{2} = 3^{2}+ (-1)^{2}+ 6^{2} = 9 + 1 + 36 = 46$

$\left | \textbf{v} \right | ^{2} = 5^{2}+ 2^{2}+ a^{2} = 25 + 4 + a^{2} = a^{2}+ 29$

For $\left | \textbf{u} \right |^{2} < \left | \textbf{v} \right | ^{2}$ to hold, it must hold that

$46 < a^{2}+ 29$ , or

$a^{2}+ 29> 46$

This is true if

$a^{2} > 17$ ,

which in turn holds if

$|a| > \sqrt{17}$ .

Since it is specified that is an integer, it holds that $a \in \left \{ ...-7, -6, -5, 5, 6, 7, ...\right \}$ .

Compare your answer with the correct one above

Question

$\textbf{u} = (3, -1, 6)$

$\textbf{v}= (5, 2, a)$ , an integer.

For which values of does it hold that $\left | \textbf{u} \right |< \left | \textbf{v} \right |$ ?

Answer

$\left | \textbf{u} \right |$ , the norm, or length, of $\textbf{u}$ , can be calculated by adding the squares of the numbers and taking the square root of the sum; $\left | \textbf{v} \right |$ can be calculated similarly.

We are seeking the real values of so that $\left | \textbf{u} \right |< \left | \textbf{v} \right |$ ; since both norms must be nonnegative, it suffices to find so that $\left | \textbf{u} \right |^{2} < \left | \textbf{v} \right | ^{2}$ .

$\left | \textbf{u} \right |^{2} = 3^{2}+ (-1)^{2}+ 6^{2} = 9 + 1 + 36 = 46$

$\left | \textbf{v} \right | ^{2} = 5^{2}+ 2^{2}+ a^{2} = 25 + 4 + a^{2} = a^{2}+ 29$

For $\left | \textbf{u} \right |^{2} < \left | \textbf{v} \right | ^{2}$ to hold, it must hold that

$46 < a^{2}+ 29$ , or

$a^{2}+ 29> 46$

This is true if

$a^{2} > 17$ ,

which in turn holds if

$|a| > \sqrt{17}$ .

Since it is specified that is an integer, it holds that $a \in \left \{ ...-7, -6, -5, 5, 6, 7, ...\right \}$ .

Compare your answer with the correct one above

Question

$\textbf{u} = (3, -1, 6)$

$\textbf{v}= (5, 2, a)$ , an integer.

For which values of does it hold that $\left | \textbf{u} \right |< \left | \textbf{v} \right |$ ?

Answer

$\left | \textbf{u} \right |$ , the norm, or length, of $\textbf{u}$ , can be calculated by adding the squares of the numbers and taking the square root of the sum; $\left | \textbf{v} \right |$ can be calculated similarly.

We are seeking the real values of so that $\left | \textbf{u} \right |< \left | \textbf{v} \right |$ ; since both norms must be nonnegative, it suffices to find so that $\left | \textbf{u} \right |^{2} < \left | \textbf{v} \right | ^{2}$ .

$\left | \textbf{u} \right |^{2} = 3^{2}+ (-1)^{2}+ 6^{2} = 9 + 1 + 36 = 46$

$\left | \textbf{v} \right | ^{2} = 5^{2}+ 2^{2}+ a^{2} = 25 + 4 + a^{2} = a^{2}+ 29$

For $\left | \textbf{u} \right |^{2} < \left | \textbf{v} \right | ^{2}$ to hold, it must hold that

$46 < a^{2}+ 29$ , or

$a^{2}+ 29> 46$

This is true if

$a^{2} > 17$ ,

which in turn holds if

$|a| > \sqrt{17}$ .

Since it is specified that is an integer, it holds that $a \in \left \{ ...-7, -6, -5, 5, 6, 7, ...\right \}$ .

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

Find the norm of the following vector.

Answer

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

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

Compare your answer with the correct one above

Question

Find the norm of vector .

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

Answer

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

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

Compare your answer with the correct one above

Question

$\textbf{u} \in \mathbb{R}^{3} , \textbf{v} \in \mathbb{R}^{4}$

True or false: $\left | \textbf{u} \right |+ \left | \textbf{v} \right |$ is an undefined expression.

Answer

$\left | \textbf{u} \right |$ refers to the norm of a vector, which is always a scalar quantity regardless of what vector space the vector falls in. It follows that $\left | \textbf{u} \right |+ \left | \textbf{v} \right |$ , the sum of two scalars, itself a scalar - a defined expression.

Compare your answer with the correct one above

Question

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

Answer

By definition,

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

therefore,

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

Compare your answer with the correct one above

Question

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

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

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

Answer

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

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

Therefore,

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

Compare your answer with the correct one above

Question

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

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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:

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

Compare your answer with the correct one above

Question

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

Answer

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

This can be simplified:

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

Compare your answer with the correct one above

Question

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

Answer

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

This can be simplified:

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

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

$|| \vec{v} || = \sqrt{3^2 + (-2)^2 + (-2)^2 } = \sqrt{9+4+4} = \sqrt{17}$

Compare your answer with the correct one above

Question

Find the norm of the vector $\vec{v} = (4, -1, 3 )$

Answer

$|| \vec{v} || = \sqrt{ 4^2 + (-1)^2 + 3^2 } = \sqrt{16+1+9 } = \sqrt{26}$

Compare your answer with the correct one above

Question

Find the norm of the vector $\vec{v} = (3, 4, 5 )$

Answer

$|| \vec{v} || = \sqrt{ 3^2 + 4^2 + 5^2 } = \sqrt{9+16+25} = \sqrt{50 }$

This can be simplified:

$\sqrt{2 \cdot 5 \cdot 5 } = 5 \sqrt{2 }$

Compare your answer with the correct one above

Question

Let $\textbf{u} = (2, 3, a, 1 )$ for some real number .

Give such that $\left | \textbf{u} \right | = 4$ .

Answer

$\left | \textbf{u} \right |$ , the norm, or length, of vector $\textbf{u}$ , is equal to the square root of the sum of the squares of its elements. Therefore,

$\left | \textbf{u} \right | = \sqrt{2^{2}+3^{2}+a^{2}+1^{2}}$

$= \sqrt{4+9+a^{2}+1}$

$= \sqrt{ a^{2}+14}$

Set this equal to 4:

$\sqrt{ a^{2}+14} = 4$

$a^{2}+14 = 16$

$a^{2}= 2$

$a = \pm \sqrt{2}$

Compare your answer with the correct one above

Question

$\textbf{u}= (-3, 2, x, x)$ ,

where is a real number.

In terms of , give $\left | \textbf{u} \right |$ .

Answer

$\left | \textbf{u} \right |$ , the norm, or length, of vector $\textbf{u}$ , is equal to the square root of the sum of the squares of its elements. Therefore,

$\left | \textbf{u} \right |= \sqrt{(-3)^{2}+ 2 ^{2}+ x ^{2} +x ^{2}}$

$=\sqrt{9+4+ x ^{2} +x ^{2}}$

$=\sqrt{2 x ^{2} +13}$

Compare your answer with the correct one above

Tap the card to reveal the answer