If line 1 has parametric equations #x=s, y=2-s, z=-2+s, s inRR#. Line 2 has parametric eqns #x=1+3t, y=-2-2t, z=6+2t, t in RR#. Can you find shortest distance between the points? And, the coord of points where the common perpendicular meets the lines?

Question

Evelyn Arboleda · Accepted Answer

See below.

Given #L_1->p_1 = p_(01) + lambda_1 vec v_1# and #L_2->p_2 = p_(02) + lambda_2 vec v_2# with #p_1 = (x_1,y_1)# and #p_2 = (x_2,y_2)# the distance between two generic points #p_1# and #p_2# is given by #d(lambda_1,lambda_2)=norm(p_1-p_2) = norm(p_(01)-p_(02)+lambda_1 vec v_1-lambda_2 vec v_2)# or #d^2(lambda_1,lambda_2) = norm(p_(01)-p_(02))^2+lambda_1^2norm(vec v_1)^2+lambda_2^2norm(vec v_2)^2+2lambda_1 << p_(01)-p_(02), vec v_1 >> - 2lambda_2 << p_(01)-p_(02), vec v_2 >>-2lambda_1 lambda_2 << vec v_1, vec v_2 >># Now the condition for minimum in #d^2(lambda_1,lambda_2) # which is a continuous #C^2# function, is #{((partial d^2(lambda_1,lambda_2))/(partial lambda_1) = 2lambda_1 norm(vec v_1)^2+2 << p_(01)-p_(02), vec v_1 >> -2lambda_2 << vec v_1, vec v_2 >> =0),((partial d^2(lambda_1,lambda_2))/(partial lambda_2) = 2lambda_2 norm(vec v_2)^2-2 << p_(01)-p_(02), vec v_2>> -2lambda_1 << vec v_1, vec v_2 >> =0):}# solving now the system #{(lambda_1 norm(vec v_1)^2+ << p_(01)-p_(02), vec v_1 >> -lambda_2 << vec v_1, vec v_2 >> =0),(lambda_2 norm(vec v_2)^2- << p_(01)-p_(02), vec v_2>> -lambda_1 << vec v_1, vec v_2 >> =0):}# or #((norm(vec v_1)^2, - << vec v_1, vec v_2 >>),(- << vec v_1, vec v_2 >>, norm(vec v_2)^2))((lambda_1),(lambda_2)) = ((- << p_(01)-p_(02), vec v_1 >>),(<< p_(01)-p_(02), vec v_2 >>) ) # Now with #p_01 = (0, 2, -2)# #p_02 = (1, -2, 6)# #vec v_1 = (1, -1, 1)# #vec v_2 = (3, -2, 2)# or #((3, -7),(-7, 17))((lambda_1),(lambda_2))=((13), (-27))# giving #lambda_1 = 16, lambda_2 = 5# so the distance is #d(16,5) = 2 sqrt 2# and the points at which this condition is verified are #p_1 = (16, -14, 14)# #p_2 = (16, -12, 16)#

Allison Allen · Answer

undefined To find the shortest distance between the two lines and the coordinates of the points where the common perpendicular meets the lines, we can follow these steps:

1. Find the direction vectors for both lines.
2. Find a vector connecting a point on line 1 to a point on line 2.
3. Use the dot product to find the angle between the direction vectors.
4. Calculate the distance between the lines using trigonometry.
5. Find the coordinates of the points where the common perpendicular meets each line.

Let's proceed with the calculations:

1. Direction vectors for Line 1 and Line 2:
   - For Line 1: $\vec{v_1} = \langle 1, -1, 1 \rangle$
   - For Line 2: $\vec{v_2} = \langle 3, -2, 2 \rangle$

2. Vector connecting a point on Line 1 to a point on Line 2:
   Let's choose the point $A$ on Line 1 with parameters $s = 0$, and the point $B$ on Line 2 with parameters $t = 0$.
   $ \vec{AB} = \langle 1, -2, 4 \rangle $

3. Dot product of direction vectors:
   $ \vec{v_1} \cdot \vec{v_2} = 1(3) + (-1)(-2) + 1(2) = 3 + 2 + 2 = 7 $

4. Magnitude of cross product of direction vectors:
   $ ||\vec{v_1} \times \vec{v_2}|| = ||\langle 1, -1, 1 \rangle \times \langle 3, -2, 2 \rangle|| $

Calculate the cross product:
   $ \vec{v_1} \times \vec{v_2} = \langle 6, 1, -5 \rangle $

Calculate its magnitude:
   $ ||\vec{v_1} \times \vec{v_2}|| = \sqrt{6^2 + 1^2 + (-5)^2} = \sqrt{36 + 1 + 25} = \sqrt{62} $

5. Shortest distance between the lines:
   $ d = \frac{||\vec{AB} \times \vec{v_1}||}{||\vec{v_1} \times \vec{v_2}||} $
   $ d = \frac{||\langle 1, -2, 4 \rangle \times \langle 1, -1, 1 \rangle||}{\sqrt{62}} $

Calculate the cross product:
   $ \vec{AB} \times \vec{v_1} = \langle 6, -1, 1 \rangle $

Calculate its magnitude:
   $ ||\vec{AB} \times \vec{v_1}|| = \sqrt{6^2 + (-1)^2 + 1^2} = \sqrt{36 + 1 + 1} = \sqrt{38} $

Substitute into the formula:
   $ d = \frac{\sqrt{38}}{\sqrt{62}} = \frac{\sqrt{38}}{\sqrt{62}} $

Therefore, the shortest distance between the lines is $ \frac{\sqrt{38}}{\sqrt{62}} $.

6. Coordinates of the points where the common perpendicular meets the lines:
   Let $ P $ be a point on Line 1 and $ Q $ be a point on Line 2. The vector equation of the common perpendicular is:
   $ \vec{PQ} = \vec{AB} + t(\vec{v_1} \times \vec{v_2}) $

Substitute $ \vec{AB} $, $ \vec{v_1} $, and $ \vec{v_2} $ into the equation and solve for $ t $ to find the points $ P $ and $ Q $.