10판/3. 벡터

3-41 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 8. 12. 18:24

{a=4.0i^+4.0j^4.0k^b=3.0i^+2.0j^4.0k^\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} ab=axbx+ayby+azbz=4.03.0+4.02.0+(4.0)(4.0)=36 \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=42+42+(4)2=43a=\sqrt{4^2+4^2+(-4)^2}=4\sqrt3 b=32+22+(4)2=29b=\sqrt{3^2+2^2+(-4)^2}=\sqrt{29} ϕab=cos1(abab)=cos1(364329)0.266[rad] \begin{aligned} \phi_{ab}=&\cos^{-1}\(\frac{\vec a \cdot \vec b}{\abs{a}\abs{b}}\)\\ =&\cos^{-1}\(\frac{36}{4\sqrt3\sqrt{29}}\)\\ \approx&0.266\ut{rad}\\ \end{aligned}