Question Number 56803 by Tawa1 last updated on 24/Mar/19
$$\mathrm{Evaluate}:\:\:\:\int_{\:\mathrm{1}} ^{\:\mathrm{2}} \:\left(\mathrm{A}\centerdot\mathrm{B}\:×\:\mathrm{C}\right)\:\mathrm{dt}\:\:\:\:\mathrm{and}\:\:\:\int_{\:\mathrm{1}} ^{\:\mathrm{2}} \:\mathrm{A}\:×\:\left(\mathrm{B}\:×\:\mathrm{C}\right)\:\:\:\mathrm{dt} \\ $$$$\mathrm{where},\:\:\:\:\:\:\:\:\:\mathrm{A}\:\:=\:\:\mathrm{ti}\:−\:\mathrm{3j}\:+\:\mathrm{2tk},\:\:\:\:\:\:\:\mathrm{B}\:\:=\:\:\mathrm{i}\:−\:\mathrm{2j}\:+\:\mathrm{2k}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{C}\:\:=\:\:\mathrm{3i}\:+\:\mathrm{tj}\:−\:\mathrm{k} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 24/Mar/19
$$\int_{\mathrm{1}} ^{\mathrm{2}} \left({A}.{B}×{C}\right){dt}\:\leftarrow{is}\:{it}\:{correct}\:{question} \\ $$$${i}\:{think}\:{it}\:{shoud}\:{be}\:\int_{\mathrm{1}} ^{\mathrm{2}} {A}.\left({B}×{C}\right){dt} \\ $$
Commented by Tawa1 last updated on 24/Mar/19
$$\mathrm{Yes}\:\mathrm{sir}.\:\mathrm{Help}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 24/Mar/19
![A×(B×C)=(A.C)B−(A.B)C =(3t−3t−2t)(i−2j+2k)−(t+6+4t)(3i+tj−k) =−2ti+4tj−4tk−(15ti+5t^2 j−5tk+18i+6tj−6k) =i(−2t−15t−18)+j(4t−5t^2 +6t)+k(−4t+5t+6) =i(−17t−18)+(10t−5t^2 )j+k(t+6) ∫_1 ^2 A×(B×C)dt ∣(((−17t^2 )/2)−18t)i+(((10t^2 )/2)−((5t^3 )/3))j+k((t^2 /2)+6t)∣_1 ^2 [i{((−17)/2)(2^2 −1^2 )−18(2−1)}+j{5(2^2 −1^2 )−(5/3)(2^3 −1^3 )}+k{(1/2)(2^2 −1^2 )+6(2−1)}] [i(((−51)/2)−18)+j(15−((35)/3))+k((3/2)+6)] =i(((−87)/2))+((10)/3)j+((15)/2)k](Q56809.png)
$${A}×\left({B}×{C}\right)=\left({A}.{C}\right){B}−\left({A}.{B}\right){C} \\ $$$$\:\:=\left(\mathrm{3}{t}−\mathrm{3}{t}−\mathrm{2}{t}\right)\left({i}−\mathrm{2}{j}+\mathrm{2}{k}\right)−\left({t}+\mathrm{6}+\mathrm{4}{t}\right)\left(\mathrm{3}{i}+{tj}−{k}\right) \\ $$$$=−\mathrm{2}{ti}+\mathrm{4}{tj}−\mathrm{4}{tk}−\left(\mathrm{15}{ti}+\mathrm{5}{t}^{\mathrm{2}} {j}−\mathrm{5}{tk}+\mathrm{18}{i}+\mathrm{6}{tj}−\mathrm{6}{k}\right) \\ $$$$={i}\left(−\mathrm{2}{t}−\mathrm{15}{t}−\mathrm{18}\right)+{j}\left(\mathrm{4}{t}−\mathrm{5}{t}^{\mathrm{2}} +\mathrm{6}{t}\right)+{k}\left(−\mathrm{4}{t}+\mathrm{5}{t}+\mathrm{6}\right) \\ $$$$={i}\left(−\mathrm{17}{t}−\mathrm{18}\right)+\left(\mathrm{10}{t}−\mathrm{5}{t}^{\mathrm{2}} \right){j}+{k}\left({t}+\mathrm{6}\right) \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} {A}×\left({B}×{C}\right){dt} \\ $$$$\mid\left(\frac{−\mathrm{17}{t}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{18}{t}\right){i}+\left(\frac{\mathrm{10}{t}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{5}{t}^{\mathrm{3}} }{\mathrm{3}}\right){j}+{k}\left(\frac{{t}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{6}{t}\right)\mid_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\left[{i}\left\{\frac{−\mathrm{17}}{\mathrm{2}}\left(\mathrm{2}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} \right)−\mathrm{18}\left(\mathrm{2}−\mathrm{1}\right)\right\}+{j}\left\{\mathrm{5}\left(\mathrm{2}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} \right)−\frac{\mathrm{5}}{\mathrm{3}}\left(\mathrm{2}^{\mathrm{3}} −\mathrm{1}^{\mathrm{3}} \right)\right\}+{k}\left\{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} \right)+\mathrm{6}\left(\mathrm{2}−\mathrm{1}\right)\right\}\right] \\ $$$$\left[{i}\left(\frac{−\mathrm{51}}{\mathrm{2}}−\mathrm{18}\right)+{j}\left(\mathrm{15}−\frac{\mathrm{35}}{\mathrm{3}}\right)+{k}\left(\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{6}\right)\right] \\ $$$$={i}\left(\frac{−\mathrm{87}}{\mathrm{2}}\right)+\frac{\mathrm{10}}{\mathrm{3}}{j}+\frac{\mathrm{15}}{\mathrm{2}}{k} \\ $$
Commented by Tawa1 last updated on 24/Mar/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 24/Mar/19
![if the questionis →∫_1 ^2 (A.B)×Cdt then ∫_1 ^2 (t+6+4t)(3i+tj−k)dt ∫_1 ^2 [i(15t+18)+j(5t^2 +6t)+k(−5t−6)] dt ∣[i(((15t^2 )/2)+18t)+j(((5t^3 )/3)+3t^2 )+k(((−5t^2 )/2)−6t)]∣_1 ^2 =[i{((15)/2)(2^2 −1^2 )+18(2−1)}+j{(5/3)(2^3 −1^3 )+3(2^2 −1^2 )}−k{(5/2)(2^2 −1^2 )+6(2−1)}] =[i(((45)/2)+18)+j(((35)/3)+9)−k(((15)/2)+6)] =i(((81)/2))+j(((62)/3))−k(((27)/2)) now if the question →∫_1 ^2 A.(B×C)dt then B×C= ∣i j k∣ ∣1 −2 2∣ ∣3 t −1∣ =i(2−2t)−j(−1−6)+k(t+6) =i(2−2t)+7j+k(t+6) A.(B×C) =(ti−3j+2tk).{i(2−2t)+7j+k(t+6)} =(2t−2t^2 −21+2t^2 +12t) =14t−21 ∫_1 ^2 (14t−21)dt =∣((14t^2 )/2)−21t∣_1 ^2 =14(((2^2 −1^2 )/2))−21(2−1) =21−21=0](Q56815.png)
$${if}\:{the}\:{questionis}\:\rightarrow\int_{\mathrm{1}} ^{\mathrm{2}} \left({A}.{B}\right)×{Cdt}\:{then} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \left({t}+\mathrm{6}+\mathrm{4}{t}\right)\left(\mathrm{3}{i}+{tj}−{k}\right){dt} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \left[{i}\left(\mathrm{15}{t}+\mathrm{18}\right)+{j}\left(\mathrm{5}{t}^{\mathrm{2}} +\mathrm{6}{t}\right)+{k}\left(−\mathrm{5}{t}−\mathrm{6}\right)\right]\:\:{dt} \\ $$$$\mid\left[{i}\left(\frac{\mathrm{15}{t}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{18}{t}\right)+{j}\left(\frac{\mathrm{5}{t}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{3}{t}^{\mathrm{2}} \right)+{k}\left(\frac{−\mathrm{5}{t}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{6}{t}\right)\right]\mid_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$=\left[{i}\left\{\frac{\mathrm{15}}{\mathrm{2}}\left(\mathrm{2}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} \right)+\mathrm{18}\left(\mathrm{2}−\mathrm{1}\right)\right\}+{j}\left\{\frac{\mathrm{5}}{\mathrm{3}}\left(\mathrm{2}^{\mathrm{3}} −\mathrm{1}^{\mathrm{3}} \right)+\mathrm{3}\left(\mathrm{2}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} \right)\right\}−{k}\left\{\frac{\mathrm{5}}{\mathrm{2}}\left(\mathrm{2}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} \right)+\mathrm{6}\left(\mathrm{2}−\mathrm{1}\right)\right\}\right] \\ $$$$=\left[{i}\left(\frac{\mathrm{45}}{\mathrm{2}}+\mathrm{18}\right)+{j}\left(\frac{\mathrm{35}}{\mathrm{3}}+\mathrm{9}\right)−{k}\left(\frac{\mathrm{15}}{\mathrm{2}}+\mathrm{6}\right)\right] \\ $$$$={i}\left(\frac{\mathrm{81}}{\mathrm{2}}\right)+{j}\left(\frac{\mathrm{62}}{\mathrm{3}}\right)−{k}\left(\frac{\mathrm{27}}{\mathrm{2}}\right) \\ $$$${now}\:{if}\:{the}\:{question}\:\rightarrow\int_{\mathrm{1}} ^{\mathrm{2}} {A}.\left({B}×{C}\right){dt}\:{then} \\ $$$$\:\:\:\boldsymbol{{B}}×{C}=\:\:\:\:\:\mid{i}\:\:\:\:\:\:\:\:\:\:\:\:{j}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{k}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\mathrm{2}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:{t}\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={i}\left(\mathrm{2}−\mathrm{2}{t}\right)−{j}\left(−\mathrm{1}−\mathrm{6}\right)+{k}\left({t}+\mathrm{6}\right) \\ $$$$\:\:={i}\left(\mathrm{2}−\mathrm{2}{t}\right)+\mathrm{7}{j}+{k}\left({t}+\mathrm{6}\right) \\ $$$${A}.\left({B}×{C}\right) \\ $$$$=\left({ti}−\mathrm{3}{j}+\mathrm{2}{tk}\right).\left\{{i}\left(\mathrm{2}−\mathrm{2}{t}\right)+\mathrm{7}{j}+{k}\left({t}+\mathrm{6}\right)\right\} \\ $$$$=\left(\mathrm{2}{t}−\mathrm{2}{t}^{\mathrm{2}} −\mathrm{21}+\mathrm{2}{t}^{\mathrm{2}} +\mathrm{12}{t}\right) \\ $$$$=\mathrm{14}{t}−\mathrm{21} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \left(\mathrm{14}{t}−\mathrm{21}\right){dt} \\ $$$$=\mid\frac{\mathrm{14}{t}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{21}{t}\mid_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$=\mathrm{14}\left(\frac{\mathrm{2}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }{\mathrm{2}}\right)−\mathrm{21}\left(\mathrm{2}−\mathrm{1}\right) \\ $$$$=\mathrm{21}−\mathrm{21}=\mathrm{0} \\ $$$$\:\:\: \\ $$
Commented by Tawa1 last updated on 24/Mar/19
$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$