Unit Vector in the Direction of a Given Vector

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\tdplotsinandcos }[3]{\pgfmathsetmacro {#1}{sin(#3)}\pgfmathsetmacro {#2}{cos(#3)}} \newcommand {\tdplotmult }[3]{\pgfmathsetmacro {#1}{#2*#3}} \newcommand {\tdplotdiv }[3]{\pgfmathsetmacro {#1}{#2/#3}} \newcommand {\tdplotcheckdiff }[5]{\par \par \pgfmathparse { abs(#2 -#1)<#3 } \par \ifthenelse {\equal {\pgfmathresult }{1}}{#4}{#5} } \newcommand {\tdplotsetmaincoords }[2]{\pgfmathsetmacro {\tdplotmaintheta }{#1} \pgfmathsetmacro {\tdplotmainphi }{#2} \tdplotcalctransformmainscreen \tikzset {tdplot_main_coords/.style={x={(\raarot cm,\rbarot cm)},y={(\rabrot cm,\rbbrot cm)},z={(\racrot cm,\rbcrot cm)}}}} \newcommand {\tdplotcalctransformmainscreen }[0]{\tdplotsinandcos {\sintheta }{\costheta }{\tdplotmaintheta }\tdplotsinandcos {\sinphi }{\cosphi 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{fill,stroke} }{} } } \newcommand {\tdplotshowargcolorguide }[4]{ \par \pgfmathsetmacro {\tdplotx }{#1} \pgfmathsetmacro {\tdploty }{#2} \pgfmathsetmacro {\tdplothuestep }{5} \pgfmathsetmacro {\tdplotxsize }{#3} \pgfmathsetmacro {\tdplotysize }{#4} \par \pgfmathsetmacro {\tdplotyscale }{\tdplotysize /360} \par \foreach \tdplotphi in {0,\tdplothuestep ,...,360} { \pgfmathdivide {\tdplotphi }{360} \definecolor {tdplotfillcolor}{hsb}{\pgfmathresult ,1,1} \color {tdplotfillcolor} \par \pgfmathsetmacro {\tdplotstarty }{\tdploty + \tdplotphi * \tdplotyscale } \pgfmathsetmacro {\tdplotstopy }{\tdplotstarty + \tdplothuestep * \tdplotyscale } \pgfmathsetmacro {\tdplotstartx }{\tdplotx } \pgfmathsetmacro {\tdplotstopx }{\tdplotx + \tdplotxsize } \filldraw [tdplot_screen_coords] (\tdplotstartx ,\tdplotstarty ) rectangle (\tdplotstopx ,\tdplotstopy ); } \par \pgfmathsetmacro {\tdplotstopy }{\tdploty + (360+\tdplothuestep )*\tdplotyscale } \pgfmathsetmacro {\tdplotstopx }{\tdplotx + \tdplotxsize } 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Unit Vector in the Direction of a Given Vector

Recall that a unit vector is a vector of length 1. Given a non-zero vector $\vec {v}$ , we can find a unit vector in the same direction by multiplying $\vec {v}$ by an appropriate scalar. For example, if $\vec {v}=\begin {bmatrix}a\\b\end {bmatrix}$ and $\norm {\vec {v}}=3$ , then a unit vector $\vec {u}$ in the same direction is given by $\vec {u}=\begin {bmatrix}a/3\\b/3\end {bmatrix}=\begin {bmatrix}a/\norm {\vec {v}}\\b/\norm {\vec {v}}\end {bmatrix}$ .

In general, dividing a non-zero vector by its own magnitude produces a unit vector in the same direction. We summarize this observation in a theorem.

Let $\vec {v}=\begin {bmatrix}v_1\\v_2\\\vdots \\v_n\end {bmatrix}$ be a non-zero vector in $\mathbb {R}^n$ . Vector $\vec {u}$ given by $\begin{equation*} \vec{u}=\frac{1}{\norm{\vec{v}}}\vec{v}=\begin{bmatrix}v_1/\norm{\vec{v}}\\v_2/\norm{\vec{v}}\\\vdots \\v_n/\norm{\vec{v}}\end{bmatrix} \end{equation*}$ is a unit vector in the direction of $\vec {v}$ .

Proof: Because $\vec {u}$ is a positive scalar multiple of $\vec {v}$ , $\vec {u}$ points in the direction of $\vec {v}$ . We now show that $\norm {\vec {u}}=1$ .
$\begin{eqnarray*} \norm{\vec{u}}&=&\sqrt{ \Big (\frac{v_1}{\norm{\vec{v}}}\Big )^2+\Big (\frac{v_2}{\norm{\vec{v}}}\Big )^2+\ldots +\Big (\frac{v_n}{\norm{\vec{v}}}\Big )^2}\\ &=&\frac{1}{\norm{\vec{v}}}\sqrt{v_1^2+v_2^2+\ldots +v_n^2}\\ &=&\frac{\norm{\vec{v}}}{\norm{\vec{v}}}=1 \end{eqnarray*}$
$\blacksquare$

Find a unit vector in the direction of $\vec {v}=\begin {bmatrix}2\\-3\\1\\0\\1\end {bmatrix}$ .

We first compute $\norm {\vec {v}}$ . $\norm {\vec {v}}=\sqrt {4+9+1+1}=\sqrt {15}$ By Theorem th:unit, $\vec {u}=\begin {bmatrix}2/\sqrt {15}\\-3/\sqrt {15}\\1/\sqrt {15}\\0\\1/\sqrt {15}\end {bmatrix}=\frac {1}{\sqrt {15}}\begin {bmatrix}2\\-3\\1\\0\\1\end {bmatrix}$

Practice Problems

Problems prob:00361-prob:00363

Find a unit vector in the direction of the given vector $\vec {v}$ .

$\vec {v}=\begin {bmatrix}-3\\4\end {bmatrix}$

Answer: $\vec {u}=\begin {bmatrix}\answer {-0.6}\\\answer {0.8}\end {bmatrix}$

$\vec {v}=\begin {bmatrix}2\\-3\\6\end {bmatrix}$

Answer: $\vec {u}=\begin {bmatrix}\answer {2/7}\\\answer {-3/7}\\\answer {6/7}\end {bmatrix}$

$\vec {v}=\begin {bmatrix}1\\3\\-2\end {bmatrix}$

Answer: $\vec {u}=\begin {bmatrix}\answer {1/\sqrt {14}}\\\answer {3/\sqrt {14}}\\\answer {-2/\sqrt {14}}\end {bmatrix}$

Let $\vec {v}=\begin {bmatrix}-1\\1\\\sqrt {7}\end {bmatrix}$ . Apply the concepts from this section to find a vector $\vec {w}$ that points in the same direction as $\vec {v}$ and whose length is 5.

Answer: $\vec {w}=\begin {bmatrix} \answer {-5/3}\\\answer {5/3}\\\answer {5\sqrt {7}/3} \end {bmatrix}$

2024-09-11 17:59:28

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defi	$\displaystyle\int_{\blue{[?]}}^{\blue{[?]}}\blue{[?]}d\blue{[?]}$
deriv	$\displaystyle\frac{d}{d\blue{[?]}}\blue{[?]}$
sum	$\displaystyle\sum_{\blue{[?]}}^{\blue{[?]}}\blue{[?]}$
prod	$\displaystyle\prod_{\blue{[?]}}^{\blue{[?]}}\blue{[?]}$
root	$\sqrt[\blue{[?]}]{\blue{[?]}}$
vec	$\left\langle \blue{[?]} \right\rangle$
mat	$\left(\begin{matrix} \blue{[?]} \end{matrix}\right)$
*	$\cdot$
infinity	$\infty$
arcsin	$\arcsin\left(\blue{[?]}\right)$
arccos	$\arccos\left(\blue{[?]}\right)$
arctan	$\arctan\left(\blue{[?]}\right)$
sin	$\sin\left(\blue{[?]}\right)$
cos	$\cos\left(\blue{[?]}\right)$
tan	$\tan\left(\blue{[?]}\right)$
sec	$\sec\left(\blue{[?]}\right)$
csc	$\csc\left(\blue{[?]}\right)$
cot	$\cot\left(\blue{[?]}\right)$
log	$\log\left(\blue{[?]}\right)$
ln	$\ln\left(\blue{[?]}\right)$
alpha	$\alpha$
beta	$\beta$
gamma	$\gamma$
delta	$\delta$
epsilon	$\epsilon$
zeta	$\zeta$
eta	$\eta$
theta	$\theta$
iota	$\iota$
kappa	$\kappa$
lambda	$\lambda$
mu	$\mu$
nu	$\nu$
xi	$\xi$
omicron	$\omicron$
pi	$\pi$
rho	$\rho$
sigma	$\sigma$
tau	$\tau$
upsilon	$\upsilon$
phi	$\phi$
chi	$\chi$
psi	$\psi$
omega	$\omega$
Gamma	$\Gamma$
Delta	$\Delta$
Theta	$\Theta$
Lambda	$\Lambda$
Xi	$\Xi$
Pi	$\Pi$
Sigma	$\Sigma$
Phi	$\Phi$
Psi	$\Psi$
Omega	$\Omega$

Unit Vector in the Direction of a Given Vector

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