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Find the integral value of 'a' such that the shortest distance between the lines r = (-i + 3k) + lambda*(i - a*j) and r = (-j + 2k) + mu*(i - j + k) is sqrt(2/3).

1. 1
2. -1
3. 2
4. 4

Correct answer: 2

Solution

Direction vectors: d1 = i - a*j + 0*k, d2 = i - j + k. d1 x d2 = |i j k; 1 -a 0; 1 -1 1| = i*(-a - 0) - j*(1 - 0) + k*(-1 - (-a)) = -a*i - j + (a-1)*k. |d1 x d2|² = a² + 1 + (a-1)² = 2a² - 2a + 2. Connecting vector: A1A2 = (-j+2k) - (-i+3k) = i - j - k. (i-j-k).(-a*i - j + (a-1)*k) = (-a)(1) + (-1)(-1) + (-1)(a-1) = -a + 1 - a + 1 = 2 - 2a. SD = |2 - 2a| / sqrt(2a² - 2a + 2) = sqrt(2/3). (2-2a)² / (2a²-2a+2) = 2/3. 4(1-a)² / (2(a²-a+1)) = 2/3. 2(1-a)² / (a²-a+1) = 2/3. 3(1-a)² = a²-a+1. Let u = a-1: 3u² = (u+1)² - (u+1) + 1 = u² + 2u + 1 - u - 1 + 1 = u² + u + 1. 2u² - u - 1 = 0. (2u+1)(u-1) = 0. u = -1/2 or u = 1. So a = 1 + 1 = 2 or a = 1 - 1/2 = 1/2. Integer value: a = 2.

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