10판/3. 벡터

3-33 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 8. 10. 17:23

(a)ra=(0,0,0)(a,a,a)=?\vec r_a=\overrightarrow{(0,0,0)(a,a,a)}=? ra=ai^+aj^+ak^\vec r_a = a\hat i + a\hat j + a\hat k
(b)rb=(a,0,0)(0,a,a)=?\vec r_b=\overrightarrow{(a,0,0)(0,a,a)}=? rb=(a)i^+aj^+ak^\vec r_b = (-a)\hat i + a\hat j + a\hat k
(c)rc=(0,a,0)(a,0,a)=?\vec r_c=\overrightarrow{(0,a,0)(a,0,a)}=? rc=ai^+(a)j^+ak^\vec r_c = a\hat i + (-a)\hat j + a\hat k
(d)rd=(a,a,0)(0,0,a)=?\vec r_d=\overrightarrow{(a,a,0)(0,0,a)}=? rd=(a)i^+(a)j^+ak^\vec r_d = (-a)\hat i + (-a)\hat j + a\hat k
(e)ϕri^,j^,k^=?\phi_{\vec r \to \hat i,\hat j,\hat k}=? ϕri^=cos1(ri^ri)=cos1((ai^+aj^+ak^)(i^)a3×1)=cos1(a1+a0+a0a3)=cos1(aa3)=cos1(13)0.9553[rad] \begin{aligned} \phi_{\vec r \to \hat i} =& \cos^{-1}\(\frac{\vec r \cdot \hat i}{ri}\)\\ =&\cos^{-1}\(\frac{(a\hat i + a\hat j + a\hat k)\cdot(\hat i)}{a\sqrt3\times1}\)\\ =&\cos^{-1}\(\frac{a\cdot1+a\cdot0+a\cdot0}{a\sqrt3}\)\\ =&\cos^{-1}\(\frac{a}{a\sqrt3}\)\\ =&\cos^{-1}\(\frac{1}{\sqrt3}\)\\ \approx&0.9553\ut{rad}\\ \end{aligned} ϕrj^=cos1(rj^rj)=cos1((ai^+aj^+ak^)(j^)a3×1)=cos1(a0+a1+a0a3)=cos1(aa3)=cos1(13)0.9553[rad] \begin{aligned} \phi_{\vec r \to \hat j} =& \cos^{-1}\(\frac{\vec r \cdot \hat j}{rj}\)\\ =&\cos^{-1}\(\frac{(a\hat i + a\hat j + a\hat k)\cdot(\hat j)}{a\sqrt3\times1}\)\\ =&\cos^{-1}\(\frac{a\cdot0+a\cdot1+a\cdot0}{a\sqrt3}\)\\ =&\cos^{-1}\(\frac{a}{a\sqrt3}\)\\ =&\cos^{-1}\(\frac{1}{\sqrt3}\)\\ \approx&0.9553\ut{rad}\\ \end{aligned} ϕrk^=cos1(rk^rk)=cos1((ai^+aj^+ak^)(k^)a3×1)=cos1(a0+a0+a1a3)=cos1(aa3)=cos1(13)0.9553[rad] \begin{aligned} \phi_{\vec r \to \hat k} =& \cos^{-1}\(\frac{\vec r \cdot \hat k}{rk}\)\\ =&\cos^{-1}\(\frac{(a\hat i + a\hat j + a\hat k)\cdot(\hat k)}{a\sqrt3\times1}\)\\ =&\cos^{-1}\(\frac{a\cdot0+a\cdot0+a\cdot1}{a\sqrt3}\)\\ =&\cos^{-1}\(\frac{a}{a\sqrt3}\)\\ =&\cos^{-1}\(\frac{1}{\sqrt3}\)\\ \approx&0.9553\ut{rad}\\ \end{aligned}
(f)r=?r=? r=a2+a2+a2=3a2=a31.732a \begin{aligned} r=&\sqrt{a^2+a^2+a^2}\\ =&\sqrt{3a^2}\\ =&a\sqrt3\\ \approx&1.732a \end{aligned}