10판/3. 벡터

3-34 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 8. 10. 19:11

{a=3.0i^+5.0j^b=2.0i^+4.0j^\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)a×b=?\vec a \times \vec b=? Ans=a×b=i^j^k^3.05.002.04.00=5.004.00i^+03.002.0j^+3.05.02.04.0k^=(5.0004.0)i^+(02.03.00)j^+(3.04.05.02.0)k^=2.0k^ \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 + \begin{vmatrix} 3.0 & 5.0 \\ 2.0 & 4.0 \\ \end{vmatrix} \hat k\\ =&(5.0\cdot0-0\cdot4.0)\hat i+(0\cdot2.0-3.0\cdot0)\hat j \\ &+(3.0\cdot4.0-5.0\cdot2.0)\hat k\\ =&2.0\hat k\\ \end{aligned}
(b)ab=?\vec a \cdot \vec b=? Ans=ab=3.02.0+5.04.0=26 \begin{aligned} \Ans =& \vec a \cdot \vec b\\ =&3.0\cdot2.0+5.0\cdot4.0\\ =&26\\ \end{aligned}
(c)(a+b)b=?(\vec a+ \vec b) \cdot \vec b=? Ans=(a+b)b=(5.0i^+9j^)(2.0i^+4.0j^)=46 \begin{aligned} \Ans =& (\vec a+ \vec b) \cdot \vec b\\ =&(5.0\hat i+9\hat j)\cdot(2.0\hat i+4.0\hat j)\\ =&46\\ \end{aligned}
(d)acosϕ=?a\cos\phi=? Ans=acosϕ=abb=2622+42=1355.8 \begin{aligned} \Ans=&|\vec a|\cos\phi\\ =&\frac{\vec a \cdot \vec b}{|\vec b|}\\ =&\frac{26}{\sqrt{2^2+4^2}}\\ =&\frac{13}{\sqrt5}\\ \approx&5.8 \end{aligned}