This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Hapa kuna suluhisho la swali la 1: a) Kutokana na vekta A = 2i + 3j - 2k na B = -3i - j + k. i) 2A + 3B; Step 1: Kokotoa 2A. 2A = 2(2i + 3j - 2k) = 4i + 6j - 4k Step 2: Kokotoa 3B. 3B = 3(-3i - j + k) = -9i - 3j + 3k Step 3: Jumlisha matokeo. 2A + 3B = (4i + 6j - 4k) + (-9i - 3j + 3k) 2A + 3B = (4-9)i + (6-3)j + (-4+3)k 2A + 3B = -5i + 3j - k ii) |4A - B|; Step 1: Kokotoa 4A. 4A = 4(2i + 3j - 2k) = 8i + 12j - 8k Step 2: Kokotoa 4A - B. 4A - B = (8i + 12j - 8k) - (-3i - j + k) 4A - B = (8 - (-3))i + (12 - (-1))j + (-8 - 1)k 4A - B = (8+3)i + (12+1)j + (-8-1)k 4A - B = 11i + 13j - 9k Step 3: Kokotoa ukubwa wa vekta. |4A - B| = sqrt((11)^2 + (13)^2 + (-9)^2) |4A - B| = sqrt(121 + 169 + 81) |4A - B| = sqrt(371) |4A - B| = sqrt(371) b) Kutokana na vekta U = 3i - 4j + 2k ni perpendicular kwa vekta V = 2i + Pj + 3k. Step 1: Tumia sheria ya dot product kwa vekta perpendicular. Ikiwa vekta mbili ni perpendicular, dot product yao ni sifuri (U · V = 0). (3)(2) + (-4)(P) + (2)(3) = 0 Step 2: Tatua kwa P. 6 - 4P + 6 = 0 12 - 4P = 0 4P = 12 P = (12)/(4) P = 3 c) Tumia uchambuzi wa vekta kubainisha eneo la pembetatu ambayo pande zake mbili ni vekta P = 2i + 2j + k na Q = i + 6j + 2k. Step 1: Kokotoa cross product ya vekta P na Q. Eneo la pembetatu iliyoundwa na vekta mbili kama pande zake ni (1)/(2) |P × Q|. P × Q = i & j & k \\ 2 & 2 & 1 \\ 1 & 6 & 2 P × Q = i((2)(2) - (1)(6)) - j((2)(2) - (1)(1)) + k((2)(6) - (2)(1)) P × Q = i(4 - 6) - j(4 - 1) + k(12 - 2) P × Q = -2i - 3j + 10k Step 2: Kokotoa ukubwa wa cross product. |P × Q| = sqrt((-2)^2 + (-3)^2 + (10)^2) |P × Q| = sqrt(4 + 9 + 100) |P × Q| = sqrt(113) Step 3: Kokotoa eneo la pembetatu. Eneo = (1)/(2) |P × Q| Eneo = (1)/(2)sqrt(113) units^2 Tuma swali linalofuata 📸