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-

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 \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

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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