Find the Unit Vector in the Same Direction as a Given Vector - Pre-Calculus

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Question

Find the unit vector that is in the same direction as the vector $\vec{v}= [3,6,1]$

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Answer

To find the unit vector in the same direction as a vector, we divide it by its magnitude.

The magnitude of $\vec{v}$ is $\left | \vec{v} \right |= $$\sqrt{3^2$$+6^2$$+1^2$}$=$\sqrt{46}$$ .

We divide vector $\vec{v}$ by its magnitude to get the unit vector $\vec{u}_v$ :

$\vec{u}_v= $\frac{\vec{v}$}{\left| \vec{v} \right|}=\frac{1}{\sqrt{46}$}\cdot[3,6,1]$

or

$\vec{u}_v= \left [ {$\frac{3}{\sqrt{46}$}},$\frac{6}{\sqrt{46}$} , $\frac{1}{\sqrt{46}$} \right ]$

All unit vectors have a magnitude of , so to verify we are correct:

$\left | \vec{u}_v \right | = $\sqrt{ $(\frac{3}{\sqrt{46}$})^2$ + $($\frac{6}{\sqrt{46}$})^2$ + $($\frac{9}{\sqrt{46}$})^2$ } = $\sqrt{\frac{9}{46}$ + $\frac{36}{46}$ + $\frac{1}{46}$ } = 1$

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Question

A unit vector has length .

Given the vector

find the unit vector in the same direction.

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Answer

First, you must find the length of the vector. This is given by the equation:

Then, dividing the vector by its length gives the unit vector in the same direction.

$$\frac{v}{\left | v\right |}$=\frac{<3,6,44,-5,12>}{\sqrt{2150}$}\ \ \approx<0.065,0.129,0.949,-0.108,0.259>$

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Question

Put the vector $\small (1,3)$ in unit vector form.

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Answer

To get the unit vector that is in the same direction as the original vector $\small v$ , we divide the vector by the magnitude of the vector.

For $\small (1,3)$ , the magnitude is:

$\small $$\sqrt{1^2$$+3^2$}$=$\sqrt{10}$$ .

This means the unit vector in the same direction of $\small (1,3)$ is,

$\small $\frac{1}{\sqrt{10}$}(1,3)=\left($\frac{1}{\sqrt{10}$},$\frac{3}{\sqrt{10}$}\right)$ .

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Question

Find the unit vector of

$\left<3,,4\right>$ .

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Answer

In order to find the unit vector u of a given vector v, we follow the formula

$u=\frac{v}{\left | v\right |}$$

Let

$v=\left<a,b\right>$

The magnitude of v follows the formula

$\left| v\right $|=$\sqrt{a^{2}$$$+b^{2}$}$ .

For this vector in the problem

$\left| v\right $|=$\sqrt{3^{2}$$$+4^{2}$}$

$=$\sqrt{9+16}$$

$=$\sqrt{25}$$

$\left| v\right|=5$ .

Following the unit vector formula and substituting for the vector and magnitude

$u=\frac{\left< 3,4\right>}{5}$$ .

As such,

$u,=,\left<$\frac{3}{5}$,, $\frac{4}{5}\right>$ .

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Question

Find the unit vector of

$\left<5,,12\right>$

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Answer

In order to find the unit vector u of a given vector v, we follow the formula

$u=\frac{v}{\left | v\right |}$$

Let

$v=\left<a,b\right>$

The magnitude of v follows the formula

$\left| v\right $|=$\sqrt{a^{2}$$$+b^{2}$}$

For this vector in the problem

$\left| v\right $|=$\sqrt{5^{2}$$$+12^{2}$}$

$=$\sqrt{25+144}$$

$=$\sqrt{169}$$

$\left| v\right|=13$

Following the unit vector formula and substituting for the vector and magnitude

$u=\frac{\left< 5,12\right>}{13}$$

As such,

$u,=,\left<$\frac{5}{13}$,, $\frac{12}{13}\right>$

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Question

Find the unit vector that is in the same direction as the vector $\vec{v}= [3,6,1]$

Tap to reveal answer

Answer

To find the unit vector in the same direction as a vector, we divide it by its magnitude.

The magnitude of $\vec{v}$ is $\left | \vec{v} \right |= $$\sqrt{3^2$$+6^2$$+1^2$}$=$\sqrt{46}$$ .

We divide vector $\vec{v}$ by its magnitude to get the unit vector $\vec{u}_v$ :

$\vec{u}_v= $\frac{\vec{v}$}{\left| \vec{v} \right|}=\frac{1}{\sqrt{46}$}\cdot[3,6,1]$

or

$\vec{u}_v= \left [ {$\frac{3}{\sqrt{46}$}},$\frac{6}{\sqrt{46}$} , $\frac{1}{\sqrt{46}$} \right ]$

All unit vectors have a magnitude of , so to verify we are correct:

$\left | \vec{u}_v \right | = $\sqrt{ $(\frac{3}{\sqrt{46}$})^2$ + $($\frac{6}{\sqrt{46}$})^2$ + $($\frac{9}{\sqrt{46}$})^2$ } = $\sqrt{\frac{9}{46}$ + $\frac{36}{46}$ + $\frac{1}{46}$ } = 1$

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Question

A unit vector has length .

Given the vector

find the unit vector in the same direction.

Tap to reveal answer

Answer

First, you must find the length of the vector. This is given by the equation:

Then, dividing the vector by its length gives the unit vector in the same direction.

$$\frac{v}{\left | v\right |}$=\frac{<3,6,44,-5,12>}{\sqrt{2150}$}\ \ \approx<0.065,0.129,0.949,-0.108,0.259>$

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Question

Put the vector $\small (1,3)$ in unit vector form.

Tap to reveal answer

Answer

To get the unit vector that is in the same direction as the original vector $\small v$ , we divide the vector by the magnitude of the vector.

For $\small (1,3)$ , the magnitude is:

$\small $$\sqrt{1^2$$+3^2$}$=$\sqrt{10}$$ .

This means the unit vector in the same direction of $\small (1,3)$ is,

$\small $\frac{1}{\sqrt{10}$}(1,3)=\left($\frac{1}{\sqrt{10}$},$\frac{3}{\sqrt{10}$}\right)$ .

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Question

Find the unit vector of

$\left<3,,4\right>$ .

Tap to reveal answer

Answer

In order to find the unit vector u of a given vector v, we follow the formula

$u=\frac{v}{\left | v\right |}$$

Let

$v=\left<a,b\right>$

The magnitude of v follows the formula

$\left| v\right $|=$\sqrt{a^{2}$$$+b^{2}$}$ .

For this vector in the problem

$\left| v\right $|=$\sqrt{3^{2}$$$+4^{2}$}$

$=$\sqrt{9+16}$$

$=$\sqrt{25}$$

$\left| v\right|=5$ .

Following the unit vector formula and substituting for the vector and magnitude

$u=\frac{\left< 3,4\right>}{5}$$ .

As such,

$u,=,\left<$\frac{3}{5}$,, $\frac{4}{5}\right>$ .

← Didn't Know|Knew It →

Question

Find the unit vector of

$\left<5,,12\right>$

Tap to reveal answer

Answer

In order to find the unit vector u of a given vector v, we follow the formula

$u=\frac{v}{\left | v\right |}$$

Let

$v=\left<a,b\right>$

The magnitude of v follows the formula

$\left| v\right $|=$\sqrt{a^{2}$$$+b^{2}$}$

For this vector in the problem

$\left| v\right $|=$\sqrt{5^{2}$$$+12^{2}$}$

$=$\sqrt{25+144}$$

$=$\sqrt{169}$$

$\left| v\right|=13$

Following the unit vector formula and substituting for the vector and magnitude

$u=\frac{\left< 5,12\right>}{13}$$

As such,

$u,=,\left<$\frac{5}{13}$,, $\frac{12}{13}\right>$

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Question

Find the unit vector that is in the same direction as the vector $\vec{v}= [3,6,1]$

Tap to reveal answer

Answer

To find the unit vector in the same direction as a vector, we divide it by its magnitude.

The magnitude of $\vec{v}$ is $\left | \vec{v} \right |= $$\sqrt{3^2$$+6^2$$+1^2$}$=$\sqrt{46}$$ .

We divide vector $\vec{v}$ by its magnitude to get the unit vector $\vec{u}_v$ :

$\vec{u}_v= $\frac{\vec{v}$}{\left| \vec{v} \right|}=\frac{1}{\sqrt{46}$}\cdot[3,6,1]$

or

$\vec{u}_v= \left [ {$\frac{3}{\sqrt{46}$}},$\frac{6}{\sqrt{46}$} , $\frac{1}{\sqrt{46}$} \right ]$

All unit vectors have a magnitude of , so to verify we are correct:

$\left | \vec{u}_v \right | = $\sqrt{ $(\frac{3}{\sqrt{46}$})^2$ + $($\frac{6}{\sqrt{46}$})^2$ + $($\frac{9}{\sqrt{46}$})^2$ } = $\sqrt{\frac{9}{46}$ + $\frac{36}{46}$ + $\frac{1}{46}$ } = 1$

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Question

A unit vector has length .

Given the vector

find the unit vector in the same direction.

Tap to reveal answer

Answer

First, you must find the length of the vector. This is given by the equation:

Then, dividing the vector by its length gives the unit vector in the same direction.

$$\frac{v}{\left | v\right |}$=\frac{<3,6,44,-5,12>}{\sqrt{2150}$}\ \ \approx<0.065,0.129,0.949,-0.108,0.259>$

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Question

Put the vector $\small (1,3)$ in unit vector form.

Tap to reveal answer

Answer

To get the unit vector that is in the same direction as the original vector $\small v$ , we divide the vector by the magnitude of the vector.

For $\small (1,3)$ , the magnitude is:

$\small $$\sqrt{1^2$$+3^2$}$=$\sqrt{10}$$ .

This means the unit vector in the same direction of $\small (1,3)$ is,

$\small $\frac{1}{\sqrt{10}$}(1,3)=\left($\frac{1}{\sqrt{10}$},$\frac{3}{\sqrt{10}$}\right)$ .

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Question

Find the unit vector of

$\left<3,,4\right>$ .

Tap to reveal answer

Answer

In order to find the unit vector u of a given vector v, we follow the formula

$u=\frac{v}{\left | v\right |}$$

Let

$v=\left<a,b\right>$

The magnitude of v follows the formula

$\left| v\right $|=$\sqrt{a^{2}$$$+b^{2}$}$ .

For this vector in the problem

$\left| v\right $|=$\sqrt{3^{2}$$$+4^{2}$}$

$=$\sqrt{9+16}$$

$=$\sqrt{25}$$

$\left| v\right|=5$ .

Following the unit vector formula and substituting for the vector and magnitude

$u=\frac{\left< 3,4\right>}{5}$$ .

As such,

$u,=,\left<$\frac{3}{5}$,, $\frac{4}{5}\right>$ .

← Didn't Know|Knew It →

Question

Find the unit vector of

$\left<5,,12\right>$

Tap to reveal answer

Answer

In order to find the unit vector u of a given vector v, we follow the formula

$u=\frac{v}{\left | v\right |}$$

Let

$v=\left<a,b\right>$

The magnitude of v follows the formula

$\left| v\right $|=$\sqrt{a^{2}$$$+b^{2}$}$

For this vector in the problem

$\left| v\right $|=$\sqrt{5^{2}$$$+12^{2}$}$

$=$\sqrt{25+144}$$

$=$\sqrt{169}$$

$\left| v\right|=13$

Following the unit vector formula and substituting for the vector and magnitude

$u=\frac{\left< 5,12\right>}{13}$$

As such,

$u,=,\left<$\frac{5}{13}$,, $\frac{12}{13}\right>$

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Question

Find the unit vector that is in the same direction as the vector $\vec{v}= [3,6,1]$

Tap to reveal answer

Answer

To find the unit vector in the same direction as a vector, we divide it by its magnitude.

The magnitude of $\vec{v}$ is $\left | \vec{v} \right |= $$\sqrt{3^2$$+6^2$$+1^2$}$=$\sqrt{46}$$ .

We divide vector $\vec{v}$ by its magnitude to get the unit vector $\vec{u}_v$ :

$\vec{u}_v= $\frac{\vec{v}$}{\left| \vec{v} \right|}=\frac{1}{\sqrt{46}$}\cdot[3,6,1]$

or

$\vec{u}_v= \left [ {$\frac{3}{\sqrt{46}$}},$\frac{6}{\sqrt{46}$} , $\frac{1}{\sqrt{46}$} \right ]$

All unit vectors have a magnitude of , so to verify we are correct:

$\left | \vec{u}_v \right | = $\sqrt{ $(\frac{3}{\sqrt{46}$})^2$ + $($\frac{6}{\sqrt{46}$})^2$ + $($\frac{9}{\sqrt{46}$})^2$ } = $\sqrt{\frac{9}{46}$ + $\frac{36}{46}$ + $\frac{1}{46}$ } = 1$

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Question

A unit vector has length .

Given the vector

find the unit vector in the same direction.

Tap to reveal answer

Answer

First, you must find the length of the vector. This is given by the equation:

Then, dividing the vector by its length gives the unit vector in the same direction.

$$\frac{v}{\left | v\right |}$=\frac{<3,6,44,-5,12>}{\sqrt{2150}$}\ \ \approx<0.065,0.129,0.949,-0.108,0.259>$

← Didn't Know|Knew It →

Question

Put the vector $\small (1,3)$ in unit vector form.

Tap to reveal answer

Answer

To get the unit vector that is in the same direction as the original vector $\small v$ , we divide the vector by the magnitude of the vector.

For $\small (1,3)$ , the magnitude is:

$\small $$\sqrt{1^2$$+3^2$}$=$\sqrt{10}$$ .

This means the unit vector in the same direction of $\small (1,3)$ is,

$\small $\frac{1}{\sqrt{10}$}(1,3)=\left($\frac{1}{\sqrt{10}$},$\frac{3}{\sqrt{10}$}\right)$ .

← Didn't Know|Knew It →

Question

Find the unit vector of

$\left<3,,4\right>$ .

Tap to reveal answer

Answer

In order to find the unit vector u of a given vector v, we follow the formula

$u=\frac{v}{\left | v\right |}$$

Let

$v=\left<a,b\right>$

The magnitude of v follows the formula

$\left| v\right $|=$\sqrt{a^{2}$$$+b^{2}$}$ .

For this vector in the problem

$\left| v\right $|=$\sqrt{3^{2}$$$+4^{2}$}$

$=$\sqrt{9+16}$$

$=$\sqrt{25}$$

$\left| v\right|=5$ .

Following the unit vector formula and substituting for the vector and magnitude

$u=\frac{\left< 3,4\right>}{5}$$ .

As such,

$u,=,\left<$\frac{3}{5}$,, $\frac{4}{5}\right>$ .

← Didn't Know|Knew It →

Question

Find the unit vector of

$\left<5,,12\right>$

Tap to reveal answer

Answer

In order to find the unit vector u of a given vector v, we follow the formula

$u=\frac{v}{\left | v\right |}$$

Let

$v=\left<a,b\right>$

The magnitude of v follows the formula

$\left| v\right $|=$\sqrt{a^{2}$$$+b^{2}$}$

For this vector in the problem

$\left| v\right $|=$\sqrt{5^{2}$$$+12^{2}$}$

$=$\sqrt{25+144}$$

$=$\sqrt{169}$$

$\left| v\right|=13$

Following the unit vector formula and substituting for the vector and magnitude

$u=\frac{\left< 5,12\right>}{13}$$

As such,

$u,=,\left<$\frac{5}{13}$,, $\frac{12}{13}\right>$

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