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当前位置:首页 > 商业/管理/HR > 质量控制/管理 > 黄克智版张量分析第一章习题解析
1.1求证:u×(v×w)=(u·w)v-(u·v)w并问:u×(v×w)与(u×v)×w是否相等?u,v,w为矢量。证明:kjikjiwvxyyxzxxzyzzyzyxzyxwvwvwvwvwvwvkjikjiwvuyzzyyzxxzxxyyxxyzzyzzxxzzxyyxyxyyxzxxzyzzyzyxwvwvuwvwvuwvwvuwvwvuwvwvuwvwvuwvwvwvwvwvwvuuukjivwuzyxzzyyxxvvvwuwuwukjiwvuzyxzzyyxxkjikjikjiwvuvwuyzzyyzxxzxxyyxxyzzyzzxxzzxyyxyzyxzzyyxxzyxzzyyxxwvwvuwvwvuwvwvuwvwvuwvwvuwvwvuwvuvwuwvukjikjivuxyyxzxxzyzzyzyxzyxvuvuvuvuvuvuvvvuuukjikjiwvuzxxzxyzzyyyzzyzxyyxxxyyxyzxxzzzyxxyyxzxxzyzzyvuvuwvuvuwvuvuwvuvuwvuvuwvuvuwvuwvu1.2求证:(A×B)×(C×D)=B(A·C×D)-A(B·C×D)=C(A·B×D)-D(A·B×C)证明:kjikjiBAxyyxzxxzyzzyzyxzyxBABABABABABABBBAAAkjikjiDCxyyxzxxzyzzyzyxzyxDCDCDCDCDCDCDDDCCCkjikjiCBxyyxzxxzyzzyzyxzyxCBCBCBCBCBCBCCCBBBkjikjiDBxyyxzxxzyzzyzyxzyxDBDBDBDBDBDBDDDBBBxyyxzxxzyzzyxyyxzxxzyzzyDCDCDCDCDCDCBABABABABABAkjiDCBAxyyxzzxxzyyzzyxDCDCADCDCADCDCADCAxyyxzzxxzyyzzyxDCDCBDCDCBDCDCBDCBkjikjikjiDCBADCABzxxzyyzzyxzzxxzyyzzyxzxyyxzyzzyxyxyyxzyzzyxyxyyxzzxxzyxxyyxzzxxzyxzyxxyyxzzxxzyyzzyxzyxxyyxzzxxzyyzzyxDCDCBDCDCBADCDCADCDCABDCDCBDCDCBADCDCADCDCABDCDCBDCDCBADCDCADCDCABAAADCDCBDCDCBDCDCBBBBDCDCADCDCADCDCAxyyxzzxxzyyzzyxDCDBADBDBADBDBADBAxyyxzzxxzyyzzyxCBCBACBCBACBCBACBAkjikjikjiCBADDBACyxxyzxyyxzzyxxyzxyyxzzxzzxyzxxzyyxzzxyzxxzyyzyyzxyzzyxxzyyzxyzzyxxzyxxyyxzzxxzyyzzyxzyxxyyxzzxxzyyzzyxCACABCBCBADDADABDBDBACCACABCBCBADDADABDBDBACCACABCBCBADDADABDBDBACDDDCBCBACBCBACBCBACCCDBDBADBDBADBDBA1.3求证矢量的非退化性。即:若矢量v与它所属的矢量空间中的任意矢量u都正交,即:u·v=0,则矢量v=0。证明:因为u为任意,所以可取u1,u2,u3,使得0det333222111zyxzyxzyxuuuuuuuuuU由u·v=0得000333222111zzyyxxzzyyxxzzyyxxuvuvuvuvuvuvuvuvuv因为detU≠0,所以vx=vy=vz=0是唯一零解,即:v=0。1.4已知:矢量u,v,求证:vuvu证明:vuvuvuvu,sin1.5求证:a,b线性相关。0ba0kjikjibaxyyxzxxzyzzyzyxzyxbababababababbbaaa证明:000xyyxzxxzyzzybabababababa即或cbababazzyyxx故bac即,a,b线性相关。1.6求证:a,b,c线性相关。0cba证明:0cbacba即cba或a,b,c共面。三维空间中共面的三矢量线性相关。1.7已知:矢量b=2i+j-2k,c=i+2j+3k,i,j,k为笛卡儿基;若将c分解为与b平行的矢量及垂直于b的矢量a之和,即c=a+mb。求a;m(其中b·a=0)解:kjikjikjibcammmmm23221223209223222122322122mmmmmmmkjikjiab92mkjia9239209131.8利用证明gij是对称正定的。,iixddgr证明:0ddddddd2ijjijiijjjiixxgxxgxxggr即gij是对称正定的。1.9求证:对于一组非共面的gi,存在唯一的gj,gj也是非共面的。证明:参见:1.2.2.4由协变基矢量求逆变基矢量式(1.2.17)及式(1.2.25)。1.10已知:以i,j,k表示三维空间中笛卡坐标基矢量,jigkigkjg321,,(1)按公式(1.2.17),求g1,g2,g3以i,j,k表示的式子;(2)求grs。解:2011101110321gggkjikjigg01110132kjikjigg11001113kjig213kjig211kjig212kjikjigg10111021212112333223223113211211jijijikikikijikjkikjkjkjggggggggg1.11根据上题结果验算公式:gj=gjigi解:kjkjikjikjigggg21212123132121111gggkikjikjikjigggg21212213232221212gggjikjikjikjigggg21221213332321313ggg1.12已知:u=2g1+3g2-g3,v=g1-g2+g3,基矢量同上题。运用1.11题求得的grs计算:(1)u·v;(2)u,v的协变分量。解:2211121311223232333231232221131211321321gggggggggggggggvu376132211121112,321333231232221131211321uuuggggggggguuugvvguuijjiijji202111211121112321333231232221131211321vvvgggggggggvvv22021322111376iiiivuvuvu1.13已知:(1)圆柱坐标系如图(a),r=x1,=x2,z=x3。(2)球坐标系如图(b),r=x1,=x2,=x3。x3'Ox2‘x1'zrx3'Ox2‘x1'r求:两种坐标系中:(1)gi通过笛卡儿基i,j,k的表达式,画出简图。(2)求gi,说明gi和gi的大小与方向有何关系。(3)由gi求gij,gij,。2dr解:(1)kjigkjigkjig333231323222121312111xxxxxxxxxxxxxxxxxx圆柱坐标系:33212211sinsincoscosxzxxxrxxxrxkgjigjig321212221cossinsincosxxxxxxjigkjigkjig3213213213213212232321cossinsinsinsinsincoscoscoscossinsincossinxxxxxxxxxxxxxxxxxxx球坐标系:21332123211coscossinsinsinsincossincossinxxrxxxxrxxxxrxx3'Ox2‘x1'zrg1g2g3x3'Ox2‘x1'rg1g2g3(2)圆柱坐标系:321213321132321321,,ggggggggggggggggggrrrrrkkkkjijigggggg22321321sincoscossinsincos11cossincossin1gijkjigrrr2221sincos1sincos1gijjikgrrr3223sincoscossinsincos1gkkkjijigrrr球坐标系:sincossinsinsincossinsinsincossincossinsinsinsincoscoscossinsincossincossincossincossincossincossinsinsinsinsincoscoscoscossinsincossin222222222321321rrrrrrrrrrrrrrrrrrrrjijijiijikjkjikjikjigggggg1222221cossinsincossincossinsinsinsincossincoscoss
本文标题:黄克智版张量分析第一章习题解析
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