Find the points P and Q which are as close as possible.
P lies on (-3t, 2 - 4t, -3 + 6t).
Q lies on (-21, 29 - 3t, 27 + 6t).
Please use vectors, dot products, cross products, and the like for this problem (not calculus to minimize distance).
I've been trying to figure this out for a while, but I just can't get it...
P lies on (-3t, 2 - 4t, -3 + 6t).
Q lies on (-21, 29 - 3t, 27 + 6t).
Please use vectors, dot products, cross products, and the like for this problem (not calculus to minimize distance).
I've been trying to figure this out for a while, but I just can't get it...
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edit : this is a quicker way of doing it
Line #1 : (x,y,z) = ( −3t, 2−4t , −3+6t ) … (i)
Line #2 : (x,y,z) = ( -21, 29−3s, 27+6s ) … (ii)
The min distance is along the common perp. So we first find its direction d.
d is in direction perp to both lines : ( -3, -4, 6 ) X ( 0, -3, 6 ) = ( -6,18, 9 )
If P(t) & Q(s) are any two points on the lines then the direction of PQ is
( −3t+21, 2−4t−29+3s, −3+6t−27−6s )
If this is in direction of common perp then we must have
(−3t+21)/(−6) = ( −27−4t+3s)/18 = (−30+6t−6s)/9
This gives 2 equations which can be solved for s and t → s=1 and t=3
From (i) P is ( -9, −10, 15 ) and from (ii) Q is ( -21, 26, 33 )
Line #1 : (x,y,z) = ( −3t, 2−4t , −3+6t ) … (i)
Line #2 : (x,y,z) = ( -21, 29−3s, 27+6s ) … (ii)
The min distance is along the common perp. So we first find its direction d.
d is in direction perp to both lines : ( -3, -4, 6 ) X ( 0, -3, 6 ) = ( -6,18, 9 )
If P(t) & Q(s) are any two points on the lines then the direction of PQ is
( −3t+21, 2−4t−29+3s, −3+6t−27−6s )
If this is in direction of common perp then we must have
(−3t+21)/(−6) = ( −27−4t+3s)/18 = (−30+6t−6s)/9
This gives 2 equations which can be solved for s and t → s=1 and t=3
From (i) P is ( -9, −10, 15 ) and from (ii) Q is ( -21, 26, 33 )