솔루션피아

$$\begin{cases} \vec F = q\vec v \times \vec B\\ q = 3\\ \vec v = 2.0\i + 4.0\j +6. 0\k\\ \vec F = 4.0\i - 20\j + 1 2\k\\ \vec B = B_x\i+B_y\j+B_z\k\\ B_x=B_y\\ \end{cases} $$ $$\vec F = q\vec v \times \vec B$$ $$ \begin{aligned} \vec F=&q(v_x\i+v_y\j+v_z\k)\times (B_x\i+B_y\j+B_z\k)\\ =&q\begin{vmatrix} \i & \j & \k\\ v_x&v_y&v_z\\ B_x&B_y&B_z\\ \end{vmatrix}\\ =&q\begin{vmatrix} v_y&v_z\\ B_y&..

$$\begin{cases} \vec a:(3.00\ut{m},0)\\ \vec b:(4.00\ut{m},30.0^\circ)\\ \vec c:(10.0\ut{m},\frac{\pi}{2}+30.0^\circ)\\ \end{cases} $$ (a)(b) $$ \vec a = 3\i,$$ (a)$a_x=?$ $$a_x = 3.00$$ (b)$a_y=?$ $$a_y = 0$$ (c)(d) $$ \begin{aligned} \vec b =& 4\cos30.0^\circ\i+4\sin30.0^\circ\j\\ =&2\sqrt3\i+2\j\\ \end{aligned} $$ (c)$b_x=?$ $$ b_x=2\sqrt3 \approx3.46$$ (d)$b_y=?$ $$ b_y=2.00 $$ (e)(f) $$ \be..

$$\begin{cases} \vec d_1 = (4.0 \ut{m})\i + (5.0 \ut{m})\j\\ \vec d_2= (-3.0 \ut{m})\i + (4.0 \ut{m})\j\\ \end{cases} $$ (a) $\vec d_1 \times \vec d_2=?$ $$ \begin{aligned} \vec d_1 \times \vec d_2=&\begin{vmatrix} \i & \j & \k\\ 4&5&0\\ -3&4&0\\ \end{vmatrix}\\ =&\begin{vmatrix} 5&0\\ 4&0\\ \end{vmatrix}\i+ \begin{vmatrix} 0&4\\ 0&-3\\ \end{vmatrix}\j+ \begin{vmatrix} 4&5\\ -3&4\\ \end{vmatrix}..

$$\begin{cases} \vec a =4.0\i+4.0\j-4.0\k\\ \vec b=3.0\i+2.0\j-4.0\k\\ \end{cases} $$ $$ \begin{aligned} \vec a \cdot \vec b=&a_xb_x+a_yb_y+a_zb_z\\ =&4.0\cdot3.0+4.0\cdot2.0+(-4.0)\cdot(-4.0)\\ =&36 \end{aligned} $$ $$a=\sqrt{4^2+4^2+(-4)^2}=4\sqrt3$$ $$b=\sqrt{3^2+2^2+(-4)^2}=\sqrt{29}$$ $$ \begin{aligned} \phi_{ab}=&\cos^{-1}\(\frac{\vec a \cdot \vec b}{\abs{a}\abs{b}}\)\\ =&\cos^{-1}\(\frac{..

(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..