솔루션피아

(풀이자 주:d2 변위가 xy평면위에 있다면 z성분은 반드시 0입니다. 주어진 모든 조건을 만족시킬 방법은 없고 다만 오타로 의심되어 y성분이 양의 값인것으로 풀었습니다.) $$\begin{cases} \vec d_{1x}=0\\ \vec d_{1y}=4.80\cos63^\circ\\ \vec d_{1z}=4.80\sin63^\circ\\ \end{cases}$$ $$\begin{cases} \vec d_{2x}=1.40\cos30^\circ=0.7\sqrt3\\ \vec d_{2y}=1.40\sin30^\circ=0.7\\ \vec d_{2z}=0\\ \end{cases}$$ $$\begin{cases} \vec d_1=(4.80\cos63^\circ)\j+(4.80\sin63^\circ)\k\\ \v..

$$\begin{cases} \vec A:(6.00,\theta_A)\\ \vec B:(7.00,\theta_B)\\ \vec A\cdot\vec B=14.0\\ \end{cases} $$ $$\phi_{AB}=?$$ $$ \begin{aligned} \phi_{AB}=&\cos^{-1}\(\frac{\vec A\cdot\vec B}{AB}\)\\ =&\cos^{-1}\(\frac{14.0}{42.0}\)\\ \approx&1.23\ut{rad} \end{aligned} $$

$$\begin{cases} \vec A = 2.00\i + 3.00\j - 4.00\k\\ \vec B = -3.00\i + 4.00\j+ 2 .00\k\\ \vec C= 7.00\i- 8.00 \j\\ \end{cases} $$ $$3\vec C\cdot (2\vec A \times \vec B)=?$$ $$ \begin{aligned} \Ans=&3\vec C\cdot (2\vec A \times \vec B)\\ =&3\vec C\cdot \begin{vmatrix} \i & \j & \k\\ 4 &6 &-8 \\ -3 &4 &2 \\ \end{vmatrix}\\ =&3\vec C\cdot \( \begin{vmatrix} 6 &-8 \\ 4 &2 \\ \end{vmatrix}\i+ \begin{..

$$\begin{cases} \vec a = 3.0\i + 3.0\j +(- 2.0)\k\\ \vec b =(-1.0)\i +(-4.0)\j +(-2.0)\k\\ \vec c = 2.0\i + 2.0\j + 1.0\k\\ \end{cases} $$ (a)$\vec a\cdot(\vec b \times \vec c)$ $$ \begin{aligned} \Ans=&\vec a\cdot(\vec b \times \vec c)\\ =&\vec a\cdot \begin{vmatrix} \i & \j & \k\\ -1&-4&-2\\ 2&2&1\\ \end{vmatrix}\\ =&\vec a\cdot \( \begin{vmatrix} -4&-2\\ 2&1\\ \end{vmatrix}\i+ \begin{vmatrix}..

$$\begin{cases} \vec p_1 = 4\hat i - 3 \hat j\\ \vec p_2= -6\hat i + 3\hat j - 2\hat k\\ \end{cases} $$ $$ \begin{aligned} \Ans =& (\vec p_1 + \vec p_2)\cdot(\vec p_1\times \vec p_2)\\ =&\bra{(-2)\i+(-2)\k}\cdot \begin{vmatrix} \i & \j & \k\\ 4&-3&0\\ -6&3&-2\\ \end{vmatrix}\\ =&\bra{(-2)\i+(-2)\k}\\ &\cdot\( \begin{vmatrix} -3&0\\ 3&-2\\ \end{vmatrix}\i+ \begin{vmatrix} 0&4\\ -2&-6\\ \end{vmatr..

$$\begin{cases} \vec p:(3.50,220^\circ)\\ \vec q:(6.30,75.0^\circ)\\ \end{cases} $$ (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}$$