Common perpendicular of two lines

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To find the common perpendicular between the two lines defined by the equations x / 2 = (y-1) / 2 = z and x + 1 = y – 2 = (z + 4) / 2, the direction vectors are identified as V1 = (2, 2, 1) and V2 = (1, 1, 2). The cross product of these direction vectors, calculated as V1 X V2, results in (3, -3, 0). This confirms the correctness of the calculation, indicating that the common perpendicular can be determined using this vector. The initial confusion regarding the distance being zero is clarified through this vector analysis. The discussion concludes with the affirmation of the cross product's accuracy.
ydan87
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Hi there,
The two given lines:
1) x / 2 = (y-1) / 2 = z
2) x + 1 = y – 2 = (z + 4) / 2
I need to find the common perpendicular between them.
Though when I tried to calculate the distance between them, I get 0.

Can someone please help?

Thanks in advance
 
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Think about the cross product of their direction vectors.
 
Easier than I thought! Thanks :)
 
Direction vectors of lines 1 and 2: V1 = (2, 2, 1), V2 = (1, 1, 2)
V1 X V2 = (3, -3, 0)

Is it correct?
 
Yes, it is.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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