Find the Unit Vector in the Same Direction as a Given Vector

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

Questions 1 - 5

1

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

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

$\left (\frac{1}{10},\frac{3}{10}\right)$

$\small (3,1)$

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

$\left (\frac{3}{10},\frac{1}{10}\right)$

Explanation

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:

$\sqrt{x^2+y^2}$

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

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

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

2

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

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

$\vec{u}_v= \left [ 9,36,1 \right ]$

$\vec{u}_v= \left [ {\frac{9}{\sqrt{3}}},\frac{36}{\sqrt{6}} ,1 \right ]$

$\vec{u}_v= \left [ 9,3,1 \right ]$

Explanation

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:

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

3

A unit vector has length .

Given the vector

find the unit vector in the same direction.

$<1,2,\frac{44}{3},\frac{-5}{3},4>$

$<\frac{1}{2},1,\frac{22}{3},\frac{-5}{6},2>$

$<\frac{3}{44},\frac{3}{22},1,\frac{-5}{44},\frac{3}{11}>$

Explanation

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

$\sqrt{3^2+6^2+44^2+(-5)^2+12^2}=\left | v\right |$

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

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

4

Find the unit vector of

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

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

$\left<1,3\right>$

$\left<1,1\right>$

$\left<1,0\right>$

Explanation

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,

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

5

Find the unit vector of

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

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

$\left<1, \frac{12}{5}\right>$

$\left<2,1\right>$

$\left<\frac{1}{2},\frac{1}{2}\right>$

Explanation

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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