 # Vector Dot Product Learning Goal: To understand the rules for Cheexorgeny 2021-12-16 Answered
Vector Dot Product Learning Goal: To understand the rules for computing dot products. Let vectors $A=\left(2,1,-4\right),B=\left(-3,0,1\right)$
Angle between the vectors A and B What is the angle $\theta AB$ between A and B? Express your answer numerically to three significant figures in radians. $\theta AB=$ radians
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$\stackrel{\to }{A}\cdot \stackrel{\to }{B}=|\stackrel{\to }{A}||\stackrel{\to }{B}|\mathrm{cos}\left(\theta \right)$
$\mathrm{cos}\left(\theta \right)=\frac{\stackrel{\to }{A}\cdot \stackrel{\to }{B}}{|\stackrel{\to }{A}||\stackrel{\to }{B}|}$ We have $\stackrel{\to }{A}\cdot \stackrel{\to }{B}=-10$
$|\stackrel{\to }{A}|=\sqrt{{2}^{2}+{1}^{2}+{\left(-4\right)}^{2}}$
$⇒|\stackrel{\to }{A}|=\sqrt{4+1+16}$
$⇒|\stackrel{\to }{A}|=\sqrt{21}$
$⇒|\stackrel{\to }{B}|=\sqrt{{\left(-3\right)}^{2}+{0}^{2}+{1}^{2}}$
$⇒|\stackrel{\to }{B}|=\sqrt{9+0+1}$
$⇒|\stackrel{\to }{B}|=\sqrt{10}$
$\therefore \mathrm{cos}\left(\theta \right)=\frac{-10}{\sqrt{21}\sqrt{10}}$
$⇒\mathrm{cos}\left(\theta \right)=\frac{\sqrt{-10}}{\sqrt{21}}=-0.69006$
$⇒\theta ={\mathrm{cos}}^{-1}=-0.69006$
$\therefore \theta =2.33$ radians.
###### Not exactly what you’re looking for? Paineow
The magnitude of vector $\stackrel{\to }{A}$ is,
$\stackrel{\to }{A}=\sqrt{{A}_{x}^{2}+{A}_{y}^{2}+{A}_{z}^{2}}$
Substitute 2 for ${A}_{x}$, 1 for ${A}_{y}$ and -4 for ${A}_{z}$, The magnitude is,
$\stackrel{\to }{A}=\sqrt{{\left(2\right)}^{2}+{\left(1\right)}^{2}+{\left(-4\right)}^{2}}$
$=4.58$
The magnitude of vector $\stackrel{\to }{B}$ is
$\stackrel{\to }{B}=\sqrt{{B}_{x}^{2}+{B}_{y}^{2}+{B}_{z}^{2}}$ Substitute -3 for ${B}_{x}$, 0 for ${B}_{y}$ and 1 for ${B}_{z}$, The magnitude is, $\stackrel{\to }{A}=\sqrt{{\left(-3\right)}^{2}+{\left(0\right)}^{2}+{\left(1\right)}^{2}}$
$=3.16$
The angle between vectors is,
${\theta }_{AB}={\mathrm{cos}}^{-1}\left(\frac{\stackrel{―}{A}\cdot \stackrel{―}{B}}{|\stackrel{―}{A}||\stackrel{―}{B}|}\right)$
${\theta }_{AB}={\mathrm{cos}}^{-1}\left(\frac{-10}{\left(4.58\right)\left(3.16\right)}\right)$
$={134}^{\circ }$
###### Not exactly what you’re looking for? nick1337