Evaluate: ∫_( 1) ^( 2) (A∙B × C) dt and ∫_( 1) ^( 2) A × (B × C) dt where, A = ti − 3j + 2tk, B = i − 2j + 2k, C = 3i + tj − k


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Question Number 56803 by Tawa1 last updated on 24/Mar/19

$Evaluate : \int_{1}^{2} (A \cdot B \times C) dt and \int_{1}^{2} A \times (B \times C) dt$ $where, A = ti - 3 j + 2 tk, B = i - 2 j + 2 k,$ $C = 3 i + tj - k$
Commented by tanmay.chaudhury50@gmail.com last updated on 24/Mar/19

$\int_{1}^{2} (A . B \times C) d t \leftarrow i s i t c o r r e c t q u e s t i o n$ $i t h i n k i t s h o u d b e \int_{1}^{2} A . (B \times C) d t$
Commented by Tawa1 last updated on 24/Mar/19

$Yes sir . Help .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 24/Mar/19
	$A×(B×C)=(A.C)B−(A.B)C =(3t−3t−2t)(i−2j+2k)−(t+6+4t)(3i+tj−k) =−2ti+4tj−4tk−(15ti+5t^2 j−5tk+18i+6tj−6k) =i(−2t−15t−18)+j(4t−5t^2 +6t)+k(−4t+5t+6) =i(−17t−18)+(10t−5t^2 )j+k(t+6) ∫_1 ^2 A×(B×C)dt ∣(((−17t^2 )/2)−18t)i+(((10t^2 )/2)−((5t^3 )/3))j+k((t^2 /2)+6t)∣_1 ^2 [i{((−17)/2)(2^2 −1^2 )−18(2−1)}+j{5(2^2 −1^2 )−(5/3)(2^3 −1^3 )}+k{(1/2)(2^2 −1^2 )+6(2−1)}] [i(((−51)/2)−18)+j(15−((35)/3))+k((3/2)+6)] =i(((−87)/2))+((10)/3)j+((15)/2)k$
	$A \times (B \times C) = (A . C) B - (A . B) C$ $= (3 t - 3 t - 2 t) (i - 2 j + 2 k) - (t + 6 + 4 t) (3 i + t j - k)$ $= - 2 t i + 4 t j - 4 t k - (15 t i + 5 t^{2} j - 5 t k + 18 i + 6 t j - 6 k)$ $= i (- 2 t - 15 t - 18) + j (4 t - 5 t^{2} + 6 t) + k (- 4 t + 5 t + 6)$ $= i (- 17 t - 18) + (10 t - 5 t^{2}) j + k (t + 6)$ $\int_{1}^{2} A \times (B \times C) d t$ $∣ (\frac{- 17 t^{2}}{2} - 18 t) i + (\frac{10 t^{2}}{2} - \frac{5 t^{3}}{3}) j + k (\frac{t^{2}}{2} + 6 t) ∣_{1}^{2}$ $[i {\frac{- 17}{2} (2^{2} - 1^{2}) - 18 (2 - 1)} + j {5 (2^{2} - 1^{2}) - \frac{5}{3} (2^{3} - 1^{3})} + k {\frac{1}{2} (2^{2} - 1^{2}) + 6 (2 - 1)}]$ $[i (\frac{- 51}{2} - 18) + j (15 - \frac{35}{3}) + k (\frac{3}{2} + 6)]$ $= i (\frac{- 87}{2}) + \frac{10}{3} j + \frac{15}{2} k$
	Commented by Tawa1 last updated on 24/Mar/19

	$God bless you sir$
	Answered by tanmay.chaudhury50@gmail.com last updated on 24/Mar/19
	$if the questionis →∫_1 ^2 (A.B)×Cdt then ∫_1 ^2 (t+6+4t)(3i+tj−k)dt ∫_1 ^2 [i(15t+18)+j(5t^2 +6t)+k(−5t−6)] dt ∣[i(((15t^2 )/2)+18t)+j(((5t^3 )/3)+3t^2 )+k(((−5t^2 )/2)−6t)]∣_1 ^2 =[i{((15)/2)(2^2 −1^2 )+18(2−1)}+j{(5/3)(2^3 −1^3 )+3(2^2 −1^2 )}−k{(5/2)(2^2 −1^2 )+6(2−1)}] =[i(((45)/2)+18)+j(((35)/3)+9)−k(((15)/2)+6)] =i(((81)/2))+j(((62)/3))−k(((27)/2)) now if the question →∫_1 ^2 A.(B×C)dt then B×C= ∣i j k∣ ∣1 −2 2∣ ∣3 t −1∣ =i(2−2t)−j(−1−6)+k(t+6) =i(2−2t)+7j+k(t+6) A.(B×C) =(ti−3j+2tk).{i(2−2t)+7j+k(t+6)} =(2t−2t^2 −21+2t^2 +12t) =14t−21 ∫_1 ^2 (14t−21)dt =∣((14t^2 )/2)−21t∣_1 ^2 =14(((2^2 −1^2 )/2))−21(2−1) =21−21=0$
	$i f t h e q u e s t i o n i s \to \int_{1}^{2} (A . B) \times C d t t h e n$ $\int_{1}^{2} (t + 6 + 4 t) (3 i + t j - k) d t$ $\int_{1}^{2} [i (15 t + 18) + j (5 t^{2} + 6 t) + k (- 5 t - 6)] d t$ $∣ [i (\frac{15 t^{2}}{2} + 18 t) + j (\frac{5 t^{3}}{3} + 3 t^{2}) + k (\frac{- 5 t^{2}}{2} - 6 t)] ∣_{1}^{2}$ $= [i {\frac{15}{2} (2^{2} - 1^{2}) + 18 (2 - 1)} + j {\frac{5}{3} (2^{3} - 1^{3}) + 3 (2^{2} - 1^{2})} - k {\frac{5}{2} (2^{2} - 1^{2}) + 6 (2 - 1)}]$ $= [i (\frac{45}{2} + 18) + j (\frac{35}{3} + 9) - k (\frac{15}{2} + 6)]$ $= i (\frac{81}{2}) + j (\frac{62}{3}) - k (\frac{27}{2})$ $n o w i f t h e q u e s t i o n \to \int_{1}^{2} A . (B \times C) d t t h e n$ $B \times C = ∣ i j k ∣$ $∣ 1 - 2 2 ∣$ $∣ 3 t - 1 ∣$ $= i (2 - 2 t) - j (- 1 - 6) + k (t + 6)$ $= i (2 - 2 t) + 7 j + k (t + 6)$ $A . (B \times C)$ $= (t i - 3 j + 2 t k) . {i (2 - 2 t) + 7 j + k (t + 6)}$ $= (2 t - 2 t^{2} - 21 + 2 t^{2} + 12 t)$ $= 14 t - 21$ $\int_{1}^{2} (14 t - 21) d t$ $=∣ \frac{14 t^{2}}{2} - 21 t ∣_{1}^{2}$ $= 14 (\frac{2^{2} - 1^{2}}{2}) - 21 (2 - 1)$ $= 21 - 21 = 0$
	Commented by Tawa1 last updated on 24/Mar/19

	$I really appreciate your effort sir . God bless you sir .$
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