The shortest distance between the lines x-51= y-22= z-4-3 and x+31= y+44= z-1-5

Question

Accepted Answer

Correct answer: C. (c) solution . Shortest distance between two lines x-x_1a_1= y-y_1a_2= z-z_1b_2 x-x_2b_1= y-y_2b_2= z-z_2b_3 (x_1-x_2y_1-y_2z_1-z_2a_1a_2a_3b_1b_2b_3) (a_1b_3 )^2+(a_1b_3-a_3b_1)^2+(a_1b_2-a_2b_1)^2 (5- (3 )2-(-5)4-112-314-5) (-10+12)^2+(-5+3)^2+(4-2)^2 (87312-314-5) (2 )^2+ (2 )^2+ (2 )^2 = |9(-10+12)-7(-5+3)+3(4-1) 4+4+4 = 16+14+6 12= 36 12= 362 3 = 183=6 3

$The shortest distance between the lines \frac{x - 5}{1} = \frac{y - 2}{2} = \frac{z - 4}{- 3} and \frac{x + 3}{1} = \frac{y + 4}{4} = \frac{z - 1}{- 5}$

Solution

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