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Linear Algebra : Norms

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Example Questions

Example Question #261 : Linear Algebra

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

In terms of , find the unit vector in the same direction as $\textbf{u}$ .

Possible Answers:

$\left (-\frac{3}{\sqrt{n^{2}-8}}, \frac{1}{\sqrt{n^{2}-8}} ,\frac{n}{\sqrt{n^{2}-8}} \right )$

$\left (-\frac{3}{\sqrt{n^{2}+ 10}}, \frac{1}{\sqrt{n^{2}+ 10}} ,\frac{n}{\sqrt{n^{2}+ 10}} \right )$

$\left (-\frac{3}{\sqrt{3n}}, \frac{1}{\sqrt{3n}} ,\frac{n}{\sqrt{3n}} \right )$

$\left (-\frac{3}{\sqrt{n-2}}, \frac{1}{\sqrt{n-2}} ,\frac{n}{\sqrt{n-2}} \right )$

$\left (-\frac{3}{\sqrt{n+4}}, \frac{1}{\sqrt{n+4}} ,\frac{n}{\sqrt{n+4}} \right )$

Correct answer:

$\left (-\frac{3}{\sqrt{n^{2}+ 10}}, \frac{1}{\sqrt{n^{2}+ 10}} ,\frac{n}{\sqrt{n^{2}+ 10}} \right )$

Explanation:

The unit vector in the same direction as $\textbf{u}$ is

$\frac{ \textbf{u} }{\left \| \textbf{u} \right \|}$

$\left \| \textbf{u} \right \|$ is the norm of $\textbf{u}$ , which can be calculated by finding the square root of the sum of the squares of its entries:

$\left \| \textbf{u} \right \| = \sqrt{(-3)^{2}+ 1^{2}+ n^{2}} = \sqrt{9+1+n^{2}} = \sqrt{n^{2}+ 10}$ .

Therefore, the unit vector is

$\frac{ \textbf{u} }{\left \| \textbf{u} \right \|} =\frac{(-3, 1,n)}{\sqrt{n^{2}+ 10}}$

$= \left (-\frac{3}{\sqrt{n^{2}+ 10}}, \frac{1}{\sqrt{n^{2}+ 10}} ,\frac{n}{\sqrt{n^{2}+ 10}} \right )$ .

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Example Question #262 : Linear Algebra

$\textbf{u} = (2, -1 , 4, -6)$ .

Find the unit vector in the same direction as $\textbf{u}$ .

Possible Answers:

$\left ( \frac{2}{\sqrt{57}}, -\frac{1 }{\sqrt{57}}, \frac{4}{\sqrt{57}}, -\frac{6}{\sqrt{57}} \right )$

$\textbf{u}$ is itself a unit vector

$\left ( \frac{2}{\sqrt{13}}, -\frac{1 }{\sqrt{13}}, \frac{4}{\sqrt{13}}, -\frac{6}{\sqrt{13}} \right )$

$\left ( \frac{2}{\sqrt{23}}, -\frac{1 }{\sqrt{23}}, \frac{4}{\sqrt{23}}, -\frac{6}{\sqrt{23}} \right )$

$\left ( \frac{2}{\sqrt{48}}, -\frac{1 }{\sqrt{48}}, \frac{4}{\sqrt{48}}, -\frac{6}{\sqrt{48}} \right )$

Correct answer:

$\left ( \frac{2}{\sqrt{57}}, -\frac{1 }{\sqrt{57}}, \frac{4}{\sqrt{57}}, -\frac{6}{\sqrt{57}} \right )$

Explanation:

The unit vector in the same direction as $\textbf{u}$ is $\frac{ \textbf{u} }{\left \| \textbf{u} \right \|}$ .

$\left \| \textbf{u} \right \|$ is the norm of $\textbf{u}$ , which can be calculated by finding the square root of the sum of the squares of its entries:

$\left \| \textbf{u} \right \| = \sqrt{2^{2}+(-1)^{2}+4^{2}+(-6)^{2}}$

$= \sqrt{4+1+16+36}$

$=\sqrt{57}$

Thus, the unit vector is

$\frac{ \textbf{u} }{\left \| \textbf{u} \right \|} =\frac{ (2, -1 , 4, -6)}{\sqrt{57}}$

$= \left ( \frac{2}{\sqrt{57}}, -\frac{1 }{\sqrt{57}}, \frac{4}{\sqrt{57}}, -\frac{6}{\sqrt{57}} \right )$

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Example Question #51 : Norms

$\textbf{u}=\left ( \frac{1}{5}, -\frac{1}{5}, \frac{4}{5} \right )$

Find the unit vector in the same direction as $\textbf{u}$ .

Possible Answers:

$\left ( \frac{1}{ 4}, -\frac{1}{4},1 \right )$

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

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

$\left ( \frac{1}{5}, -\frac{1}{ 5}, -\frac{4}{ 5} \right )$

$\textbf{u}$ itself is a unit vector.

Correct answer:

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

Explanation:

The unit vector in the same direction as $\textbf{u}$ is

$\frac{ \textbf{u} }{\left \| \textbf{u} \right \|}$

$\left \| \textbf{u} \right \|$ is the norm of $\textbf{u}$ , which can be calculated by finding the square root of the sum of the squares of its entries:

$\left \| \textbf{u} \right \|=\sqrt{\left ( \frac{1}{5} \right )^{2} + \left ( -\frac{1}{5} \right )^{2} +\left ( \frac{4}{5} \right )^{2} }$

$=\sqrt{\frac{1}{25}+ \frac{1}{25}+ \frac{16}{25} }$

$=\sqrt{\frac{18}{25} }$

$= \frac{\sqrt{18}}{5 }$

Thus, the unit vector is

$\frac{ \textbf{u} }{\left \| \textbf{u} \right \|}=\frac{\left ( \frac{1}{5}, -\frac{1}{5}, \frac{4}{5} \right ) }{ \frac{\sqrt{18}}{5 }}$

$=\left ( \frac{1}{ \sqrt{18}}, -\frac{1}{ \sqrt{18}}, \frac{4}{ \sqrt{18}} \right )$

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

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Linear Algebra : Norms

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #261 : Linear Algebra

Example Question #262 : Linear Algebra

Example Question #51 : Norms

All Linear Algebra Resources

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