솔루션피아

$$\begin{cases} \vec p:(3.50,220^\circ)\\ \vec q:(6.30,75.0^\circ)\\ \end{cases} $$ (a)$\vec p \times \vec q=?$ $$ \begin{aligned} \Ans =& \vec p \times \vec q\\ =&pq\sin\phi\\ =&3.50\cdot6.30\sin(220^\circ-75.0^\circ)\\ \approx&12.7 \, ,-z \text{ Direction} \end{aligned} $$ (b)$\vec p \cdot \vec q=?$ $$ \begin{aligned} \Ans =& \vec p \cdot \vec q\\ =&pq\cos\phi\\ =&3.50\cdot6.30\cos(220^\circ-7..

$$\begin{cases} \vec a= 3.0\hat i + 5.0\hat j\\ \vec b = 2.0\hat i + 4.0\hat j\\ \end{cases} $$ (a)$\vec a \times \vec b=?$ $$ \begin{aligned} \Ans =& \vec a \times \vec b\\ =& \begin{vmatrix} \hat i & \hat j & \hat k\\ 3.0 & 5.0 & 0\\ 2.0 & 4.0 & 0\\ \end{vmatrix}\\ =& \begin{vmatrix} 5.0 & 0 \\ 4.0 & 0 \\ \end{vmatrix} \hat i + \begin{vmatrix} 0 & 3.0 \\ 0 & 2.0 \\ \end{vmatrix} \hat j + \begi..

(a)$\vec r_a=\overrightarrow{(0,0,0)(a,a,a)}=?$ $$\vec r_a = a\hat i + a\hat j + a\hat k$$ (b)$\vec r_b=\overrightarrow{(a,0,0)(0,a,a)}=?$ $$\vec r_b = (-a)\hat i + a\hat j + a\hat k$$ (c)$\vec r_c=\overrightarrow{(0,a,0)(a,0,a)}=?$ $$\vec r_c = a\hat i + (-a)\hat j + a\hat k$$ (d)$\vec r_d=\overrightarrow{(a,a,0)(0,0,a)}=?$ $$\vec r_d = (-a)\hat i + (-a)\hat j + a\hat k$$ (e)$\phi_{\vec r \to \..

(풀이자 주:x축이 아니라 z축을 나타내지 않았는데 오타로 추정됩니다. x축은 나타나있습니다.) $$\begin{cases} a:(4,0)\\ b:(3,\frac{\pi}{2})\\ c:(5,\pi+\tan^{-1}\(\frac{3}{4}\))\\ \end{cases}$$ (a)$\abs{\vec a \times \vec b}=?$ $$ \begin{aligned} \Ans=&\abs{\vec a \times \vec b}\\ =& ab\sin\phi_{ab}\\ =& 4\cdot3\cdot1\\ =&12\\ \end{aligned}$$ (b)$\text{Direction of }\vec a \times \vec b=?$ $$ + z $$ (c)$\abs{\vec a \times \vec c}=?$ $$..

$$\vec a = (15\cos56.0^\circ)\hat i + (15\sin56.0^\circ)\hat j$$ (a)$a_x=?$ $$15\cos56.0^\circ\approx8.39\ut{m}$$ (b)$a_y=?$ $$15\sin56.0^\circ\approx12.4\ut{m}$$ (c),(d)$\vec a'=?$ $$ \vec a':(15.0,56^\circ-18^\circ)=(15.0,38^\circ)$$ $$ \vec a'=\(15\cos38^\circ\)\hat i + \(15\sin38^\circ\)\hat j$$ (c)$a_x'=?$ $$15\cos38^\circ\approx11.8\ut{m}$$ (d)$a_y'=?$ $$15\sin38^\circ\approx9.24\ut{m}$$

$$\begin{cases} \vec a=(4.0 \ut{m})\hat i + (-3.0 \ut{m})\hat j\\ \vec b= (6.0\ut{m})\hat i + (8.0\ut{m})\hat j\\ \end{cases} $$ (a)$a=?$ $$ a=\sqrt{4^2+(-3)^2}=5.0\ut{m}$$ (b)$\theta_a=?$ $$ \tan^{-1}\(\frac{-3}{4}\)\approx-0.64\ut{rad}$$ (c)$b=?$ $$ b=\sqrt{6^2+8^2}=10.0\ut{m}$$ (d)$\theta_b=?$ $$ \tan^{-1}\(\frac{8}{6}\)\approx0.93\ut{rad}$$ (e),(f) $$\vec a +\vec b = (10.0 \ut{m})\hat i + (5..

(a),(b) $$\begin{cases} \vec w:(2.00\ut{cm},60^\circ)\\ \vec v:(2.00\ut{cm},90^\circ)\\ \vec i:(2.00\ut{cm},120^\circ)\\ \vec h:(2.00\ut{cm},90^\circ)\\ \end{cases} $$ $$ \begin{aligned} \vec w =& (2\cos60^\circ)\hat i + (2\sin60^\circ)\hat j\\ =&\hat i + \sqrt3\hat j \end{aligned} $$ $$ \begin{aligned} \vec v =& \vec h\\ =& (2\cos90^\circ)\hat i + (2\sin90^\circ)\hat j\\ =&2\hat j\\ \end{aligne..

$$\begin{cases} \vec a_1 = 0.50\ut{m}\hat i\\ \vec a_2:(0.70\ut{m},30^\circ)\\ \vec b_1:(1.6\ut{m},90-40^\circ)\\ \Sigma \vec a = \Sigma \vec b\\ \end{cases} $$ $$ \vec b_2 = \Sigma \vec a - \vec b_1$$ $$ \begin{aligned} \vec a_2 =& (0.7\cos30^\circ)\hat i + 0.7\sin30^\circ)\hat j\\ =&\(\frac{7 }{20}\sqrt3\)\hat i + \(\frac{7}{20}\)\hat j\\ \end{aligned} $$ $$ \begin{aligned} \vec b_1 =& (1.6\co..

$$\begin{cases} \vec d_1 + \vec d_2=5\vec d_3\\ \vec d_1 - \vec d_2=3\vec d_3\\ \end{cases} $$ $$\vec d_3=2\hat i+4\hat j$$ $$\begin{cases} \vec d_1=4\vec d_3\\ \vec d_2=\vec d_3\\ \end{cases} $$ (a)$\vec d_1=?$ $$ \begin{aligned} \vec d_1 =&4\vec d_3\\ =&4(2\hat i+4\hat j)\\ =&8\hat i+16\hat j \end{aligned} $$ (b)$\vec d_2=?$ $$ \vec d_2 = \vec d_3 = 2\hat i+4\hat j$$

$$ \begin{aligned} \vec A=&(2.0\ut{m})\hat i + (3.00\ut{m})\hat j\\ \vec B:&(4.00\ut{m}, +65.0^\circ)\\ \vec C =& ( - 4.00 \ut{m})\hat i - (6.00 \ut{m} )\hat j\\ \vec D:&(5.00\ut{m}, -235^\circ)\\ \end{aligned} $$ $$ \begin{aligned} B=\(4\cos65.0^\circ\)\hat i + \(4\sin65.0^\circ\)\hat j\\ \end{aligned} $$ $$ \begin{aligned} D=\bra{5\cos(-235^\circ)}\hat i + \bra{5\sin(-235^\circ)}\hat j\\ \end{..