What is the distance between the line #(x+1)/7=(y+1)/6=(z+1)/1# and the line #(x-3)/1=(y-5)/(-2)=(z-7)/1# ?

3 Answers
Mar 8, 2017

Answer:

Shortest distance: # 2 sqrt 2#

Explanation:

Your 2 lines are:

#mathbf l_1: (x+1)/7=(y+1)/6=z+1 qquad [ = p]#

#mathbf l_2: x-3=(y-5)/-2=z-7 qquad [ = q]#

We can write these in parameterised form:

#mathbf l_1: ((x),(y),(z)) = ((-1),(-1),(-1)) + p ((7),(6),(1))#

#mathbf l_2: ((x),(y),(z)) = ((3),(5),(7)) + q ((1),(-2),(1))#

If the line connecting the points on the shortest distance touches #mathbf l_1# and #mathbf l_2#, respectively at points #A# and #B#, then we can say, using the distance formula, that the square of the distance between these points #A# and #B# is:

#s^2 = S = (-1 + 7p - 3- q)^2 + (- 1 + 6p - 5 + 2 q)^2 + (- 1 + 7p - 7 - q)^2#

# = (-4 + 7p - q)^2 + (- 6 + 6p + 2 q)^2 + (- 8 + 7p - q)^2#

Assuming you can use calculus, we can take partial differentials in order to looks for any critical points:

#(partial S)/(partial p) = 2(-4 + 7p - q) cdot 7 + 2(- 6 + 6p + 2 q) cdot 6 + 2(- 8 + 7p - q) cdot 7#

#= 4 (67 p - q - 60) = 0 qquad triangle#

# (partial S)/(partial q) = 2(-4 + 7p - q) cdot (-1) + 2 (- 6 + 6p + 2 q) cdot 2 + 2 (- 8 + 7p - q) cdot (-1) #

# = -4(p - 3q) = 0 qquad square#

#square# is fortutious because we see that for a critical point #p = 3q#. We can sub that straight into #triangle# to get:

#4 (67 (3q) - q - 60) = 0#

#implies q = 3/10, implies p = 9/10, implies S = 8#

The shortest distance is therefore: #s = 2 sqrt 2#

We could now check the nature of the critical point via the second derivative but, but from a practical perspective, these are skew lines. The CP must be a min.

I might post a non-calculus approach later if i get time. There, we would use the cross product and algebra.

Mar 9, 2017

Answer:

I make it #82/25sqrt(5)#.

Explanation:

Line 1:

#t_1 = (x+1)/7=(y+1)/6=(z+1)/1#

That is:

#(x, y, z) = (7t_1-1, 6t_1-1, t_1-1)#

#color(white)((x, y, z)) = (-1, -1, -1) + t_1(7, 6, 1)#

Line 2:

#t_2 = (x-3)/1=(y-5)/(-2)=(z-7)/1#

That is:

#(x, y, z) = (t_2+3, -2t_2+5, t_2+7)#

#color(white)((x, y, z)) = (3,5,7) + t_2(1, -2, 1)#

A line through the points of closest approach of these two lines is perpendicular to both of these lines, so has the same direction as the cross product of #(7, 6, 1)# and #(1, -2, 1)#:

#(7, 6, 1) xx (1, -2, 1) = abs((i, j, k),(7, 6, 1),(1, -2, 1))#

#color(white)((7, 6, 1) xx (1, -2, 1)) = (abs((6, 1),(-2, 1)), abs((1, 7),(1, 1)), abs((7, 6),(1, -2)))#

#color(white)((7, 6, 1) xx (1, -2, 1)) = (8, -6, -20)#

So we want to find #t_1, t_2, t_3# such that:

#(-1, -1, -1) + t_1(7, 6, 1) + t_3(8, -6, -20) = (3, 5, 7) + t_2(1, -2, 1)#

Adding #(1, 1, 1) + t_2(-1, 2, -1)# to both sides, this becomes:

#t_1(7, 6, 1) + t_2(-1, 2, -1) + t_3(8, -6, -20) = (4, 6, 8)#

Which we can rewrite as:

#((7, -1, 8),(6, 2, -6),(1,-1,-20))((t_1),(t_2),(t_3)) = ((4),(6),(8))#

Rewrite as an augmented matrix:

#((7, -1, 8, |, 4),(6, 2, -6, |, 6),(1,-1,-20, |, 8))#

Perform a sequence of row operations to make the left hand side of this matrix into a #3xx3# identity matrix.

#((7, -1, 8, |, 4),(6, 2, -6, |, 6),(1,-1,-20, |, 8)) ->#

#((1, -3, 14, |, -2),(6, 2, -6, |, 6),(1,-1,-20, |, 8)) ->#

#((1, -3, 14, |, -2),(3, 1, -3, |, 3),(1,-1,-20, |, 8)) ->#

#((1, -3, 14, |, -2),(0, 10, -45, |, 9),(0,2,-34, |, 10)) ->#

#((1, -3, 14, |, -2),(0, 1, -9/2, |, 9/10),(0,1,-17, |, 5)) ->#

#((1, -3, 14, |, -2),(0, 1, -9/2, |, 9/10),(0,0, -25/2, |, 41/10)) ->#

#((1, -3, 14, |, -2),(0, 1, -9/2, |, 9/10),(0,0, 1, |, -41/125)) ->#

#((1, 0, 1/2, |, 7/10),(0, 1, -9/2, |, 9/10),(0,0, 1, |, -41/125)) ->#

#((1, 0, 0, |, 108/125),(0, 1, 0, |, -72/125),(0,0, 1, |, -41/125))#

Hence:

#{ (t_1 = 108/125), (t_2 = -72/125), (t_3 = -41/125) :}#

So the two points of closest approach are:

#(-1, -1, -1) + t_1(7, 6, 1) = (-1, -1, -1) + 108/125(7, 6, 1)#

#color(white)((-1, -1, -1) + t_1(7, 6, 1)) = (631/125, 523/125, -17/125)#

#(3,5,7) + t_2(1, -2, 1) = (3,5,7) - 72/125(1, -2, 1)#

#color(white)((3,5,7) + t_2(1, -2, 1)) = (303/125, 769/125, 803/125)#

The distance between these two points is:

#1/125 sqrt((303-631)^2 + (769-523)^2+(803+17)^2)#

#= 1/125sqrt(107584+60516+672400)#

#= 1/125sqrt(840500)#

#= 1/125*410sqrt(5)#

#= 82/25sqrt(5)#

Mar 9, 2017

Answer:

#82/25 sqrt(5)#

Explanation:

#lambda_1=(x+1)/7=(y+1)/6=(z+1)/1#

#lambda_2=(x-3)/1=(y-5)/(-2)=(z-7)/1#

#{(x=-1+7lambda_1),(y=-1+6lambda_1),(z=-1+lambda_1):}->p=p_1+lambda_1 v_1#

#{(x=3+lambda_2),(y=5-2lambda_2),(z=7+lambda_2):}->p = p_2+lambda_2 v_2#

Now the distance between the two lines is

#d(lambda_1,lambda_2)=norm(p_1+lambda_1 v_1-p_2-lambda_2 v_2)#

but

#d^2(lambda_1,lambda_2)=norm(p_1-p_2)^2+lambda_1^2norm (v_1)^2+lambda_2^2norm(v_2)^2+2lambda_1 << p_1-p_2,v_1 >>-2lambda_2 << p_1-p_2,v_2 >>-2lambda_1lambda_2 << v_1, v_2 >>#

The minimum of #d^2(lambda_1,lambda_2)# is located at the solution of

#{(partial/(partial lambda_1) d^2 = 0 ),(partial/(partial lambda_2) d^2 = 0):}#

which is obtained solving the linear system

#((-norm(v_1)^2, << v_1,v_2>>),(- << v_1,v_2>>, norm(v_2)^2))((lambda_1),(lambda_2)) = ((<< p_1-p_2,v_1 >>),(<< p_1-p_2, v_2 >>))#

Solving for #lambda_1,lambda_2# we obtain

#lambda_1 = 108/125, lambda_2 = -72/125#

which substituted into #d(lambda_1,lambda_2) = 82/25 sqrt(5)#