솔루션피아

$$\begin{cases} \vec c = \(7.4 \)\hat i + \(-3.8 \)\hat j + \(-6.1 \)\hat k\\ \vec d = \(4.4 \)\hat i + \(-2.0 \)\hat j + \(3.3 \)\hat k\\ \end{cases} $$ $$ \begin{aligned} \vec r &= \(7.4+4.4 \)\hat i + \(-3.8-2.0 \)\hat j + \(-6.1+3.3 \)\hat k\\ &=\(11.8 \)\hat i + \(-5.8 \)\hat j + \(-2.8 \)\hat k \end{aligned} $$ (a)$r_x=11.8$ (b)$r_y=-5.8$ (c)$r_z=-2.8$

$$\begin{cases} \vec a = (5.0)\hat i - (4.0)\hat j + (2 . 0)\hat k\\ \vec b = (-2.0m)\hat i +(2 .0m)\hat j + (5.0m)\hat k\\ \end{cases} $$ (a)$\vec a + \vec b$ $$\vec a + \vec b = \(5.0-2.0m \)\hat i + \(-4.0+2.0m \)\hat j + \(2.0+5.0m \)\hat k$$ (b)$\vec a- \vec b$ $$\vec a - \vec b = \(5.0+2.0m \)\hat i + \(-4.0-2.0m \)\hat j + \(2.0-5.0m \)\hat k$$ (c)$\vec a - \vec b + \vec c = 0, \vec c=?$ ..

$$\begin{cases} \vec a:(0,2.5)\\ \vec b:(-3.0,0)\\ \vec c:(0,-4.2)\\ \end{cases} $$ (a) $$\vec x = \vec a+\vec b+\vec c = (-3.0,-1.7)$$ (b)$\abs{\vec x}=?$ $$ \begin{aligned} \abs{\vec x} &= \sqrt{(-3.0)^2+(-1.7)^2}\\ &=\frac{\sqrt{1189}}{10}\ut{m}\\ &\approx 3.5\ut{m}\\ \end{aligned} $$ (c)$\theta_x=?$ $$ \begin{aligned} \theta_x&=\tan^{-1}\(\frac{-1.7}{-3.0}\)\\ &\approx -2.6\ut{rad} \end{alig..

$$\begin{cases} \vec a:(3\ut{m},\theta_a)\\ \vec b:(4\ut{m},\theta_b)\\ \theta_a-\theta_b=\phi \end{cases} $$ $$\begin{cases} \vec a = 3\cos\theta_a\hat i+3\sin\theta_a\hat j\\ \vec b = 4\cos\theta_b\hat i+4\sin\theta_b\hat j\\ \end{cases} $$ $$\vec a+\vec b = \(3\cos\theta_a+4\cos\theta_b\)\hat i+\(3\sin\theta_a+4\sin\theta_b\)\hat j$$ $$ \begin{aligned} \abs{\vec a+\vec b} &= \sqrt{\(3\cos\the..

$$ \vec a:(10.5\ut{m},20.0^\circ) $$ $$\vec a = 10.5\cos\(20.0^\circ\)\hat i+10.5\sin\(20.0^\circ\)\hat j$$ (a)$a_y=?$ $$ \begin{aligned} a_x&=10.5\sin\(20.0^\circ\)\\ &\approx 3.59\ut{m}\\ \end{aligned} $$ (b)$a_x=?$ $$ \begin{aligned} a_x&=10.5\cos\(20.0^\circ\)\\ &\approx 9.87\ut{m}\\ \end{aligned} $$

$$\begin{cases} \vec a_1 = 3.66\hat j\ut{m}\\ \vec a_2:(1.83\ut{m},-\frac{\pi}{4})\\ \vec a_3:(0.91\ut{m},-\frac{3}{4}\pi)\\ \end{cases} $$ $$ \begin{aligned} \vec a_2 &= 1.83\cos\(-\frac{\pi}{4}\)\hat i+1.83\sin\(-\frac{\pi}{4}\)\hat j\\ &= \frac{1.83}{\sqrt{2}}\hat i-\frac{1.83}{\sqrt{2}}\hat j\\ \end{aligned} $$ $$ \begin{aligned} \vec a_3 &= 0.91\cos\(-\frac{3}{4}\pi\)\hat i+0.91\sin\(-\frac..

(a) 20.0° $$ 20.0^\circ =20.0^\circ \cdot \frac{\pi\ut{rad}}{180^\circ}=\frac{\pi}{9}\ut{rad}\approx0.349\ut{rad}$$ (b) 50.0° $$ 50.0^\circ =50.0^\circ \cdot \frac{\pi\ut{rad}}{180^\circ}=\frac{5}{18}\pi\ut{rad}\approx0.873\ut{rad}$$ (c) 100° $$ 100^\circ =100^\circ \cdot \frac{\pi\ut{rad}}{180^\circ}=\frac{5}{9}\pi\ut{rad}\approx1.75\ut{rad}$$ (d) 0.330 rad $$ 0.330\ut{rad} = 0.330\ut{rad} \cdo..

$$\vec x = 15\ut{m}\hat i + 15\ut{m}\hat j + 10\ut{m}\hat k$$ $$x=?$$ $$ \begin{aligned} x&=\abs{\vec x}\\ &=\sqrt{15^2+15^2+10^2}\ut{m}\\ &=5\sqrt{22}\ut{m}\\ &\approx 23.45207879911715\ut{m}\\ &\approx 23\ut{m}\\ \end{aligned} $$

$$\begin{cases} \vec r:(12\ut{m},30^\circ)\\ \vec r =r_x \hat i+ r_y \hat j\\ \end{cases} $$ (a)$r_x=?$ $$ \begin{aligned} r_x&=r\cos\theta\\ &=12\cos30^\circ\ut{m}\\ &=6\sqrt{3}\ut{m} \end{aligned} $$ (a)$r_y=?$ $$ \begin{aligned} r_x&=r\sin\theta\\ &=12\sin30^\circ\ut{m}\\ &=6\ut{m} \end{aligned} $$

$$\begin{cases} \vec a =a_x \hat i+ a_y \hat j\\ \vec a:(a,\theta)\\ a_x=\frac{1}{2}a\\ \end{cases} $$ $$ \tan\theta = ?$$ $$a_x=a\cos\theta=\frac{1}{2}a$$ $$ \cos\theta=\frac{1}{2}$$ $$ \theta = \frac{\pi}{3} $$ $$ \therefore \tan\theta = \tan\frac{\pi}{3}=\sqrt{3}$$