Let L1 be the line passing through the point P1=(−1, −1, 8) with direction vector →d1=[1, 0, −3],
and let L2 be the line passing through the point P2=(7, −11, 14) with direction vector →d2=[1, 2, −3].
Find the shortest distance d between these two lines, and find a point Q1 on L1 and a point Q2 on L2 so that d(Q1,Q2) = d. Use the square root symbol '√' where needed to give an exact value for your answer.
I know how to use the cross product to find the normal and project the points onto the normal to find the distance, but i'm having trouble figuring out Q1 and Q2. Can someone offer advice?
Of course, I'd be happy to help! Finding the points Q1 and Q2 is actually quite simple once you have the direction vectors of the two lines.
To find Q1, you can start with the given point P1 on L1. Since the direction vector →d1 is parallel to the line, you can simply use a parameter t to move along the line.
Q1 = P1 + t * →d1
To find Q2, you follow the same process using the given point P2 on L2 and the direction vector →d2.
Q2 = P2 + s * →d2
Now, the shortest distance d between the two lines can be found by calculating the distance between any point on L1 and L2. In this case, we can take Q1 as the point on L1 and Q2 as the point on L2.
Once you have Q1 and Q2, you can calculate the distance between them using the distance formula:
d(Q1,Q2) = √[(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2]
Substituting the coordinates of Q1 and Q2 into the formula will give you the exact value of d.
I hope this helps! If you have any more questions, feel free to ask.
To find points Q1 and Q2 on lines L1 and L2, respectively, such that the distance between them is d, you can follow these steps:
1. Find a vector connecting the two lines by taking the difference of any two points, for example, P1 and P2. Let's call this vector →v1 = P1 - P2.
→v1 = [-1, -1, 8] - [7, -11, 14] = [-8, 10, -6].
2. Find the unit vector →u1 in the direction of →d1.
To find →u1, normalize →d1 by dividing it by its magnitude.
→u1 = (1/√(1^2 + 0^2 + (-3)^2)) * [1, 0, -3] = [1/√10, 0, -3/√10].
3. Find the projection →p1 of →v1 onto the line L1 using the dot product.
→p1 = ( →v1 · →u1 ) * →u1.
→p1 = [1/√10, 0, -3/√10] · [-8, 10, -6] * [1/√10, 0, -3/√10]
= ( -8/√10 + (-6)(-3/√10) ) * [1/√10, 0, -3/√10]
= ( -8/√10 + 18/√10 ) * [1/√10, 0, -3/√10]
= 10/√10 * [1/√10, 0, -3/√10]
= [1, 0, -3/√10].
4. Find the point Q1 on line L1 by adding the projection →p1 to the initial point P1.
Q1 = P1 + →p1 = [-1, -1, 8] + [1, 0, -3/√10] = [0, -1, 8 - 3/√10].
5. Now, find the unit vector →u2 in the direction of →d2. Normalize →d2 by dividing it by its magnitude.
→u2 = (1/√(1^2 + 2^2 + (-3)^2)) * [1, 2, -3] = [1/√14, 2/√14, -3/√14].
6. Since L1 and L2 are parallel, the line connecting Q1 and Q2 can be represented by →v1, so the direction vector →d of this line is →v1.
7. Find the projection →p2 of →v1 onto the line L2 using the dot product.
→p2 = ( →v1 · →u2 ) * →u2.
→p2 = [1/√14, 2/√14, -3/√14] · [-8, 10, -6] * [1/√14, 2/√14, -3/√14]
= ( (-8)(1/√14) + 10(2/√14) + (-6)(-3/√14) ) * [1/√14, 2/√14, -3/√14]
= ( -8/√14 + 20/√14 + 18/√14 ) * [1/√14, 2/√14, -3/√14]
= 30/√14 * [1/√14, 2/√14, -3/√14]
= [30/14, 60/14, -90/14]
= [15/7, 30/7, -45/7].
8. Find the point Q2 on line L2 by adding the projection →p2 to the initial point P2.
Q2 = P2 + →p2 = [7, -11, 14] + [15/7, 30/7, -45/7]
= [7 + 15/7, -11 + 30/7, 14 - 45/7]
= [(49 + 15)/7, (-77 + 30)/7, (98 - 45)/7]
= [64/7, -47/7, 53/7].
Therefore, Q1 = [0, -1, 8 - 3/√10] and Q2 = [64/7, -47/7, 53/7]. These points lie on lines L1 and L2, respectively, and the distance between them is d.
To find the shortest distance between two lines, you can use the concept of a perpendicular distance. The idea is to find the point on each line that lies on the line perpendicular to both lines.
To find a point Q1 on line L1, we need to find a point that lies on L1 and is perpendicular to L2. Similarly, we need to find a point Q2 on line L2 that is perpendicular to L1.
Let's start by finding Q1 on L1:
1. Define the two lines:
- Line L1: P1 = (-1, -1, 8) + t[1, 0, -3]
- Line L2: P2 = (7, -11, 14) + s[1, 2, -3]
2. Set up a system of equations:
- The direction vector of line L2, [1, 2, -3], should be perpendicular to the vector connecting points P1 and Q1 on line L1.
So, ([1, 2, -3] dot ((-1, -1, 8) - Q1)) = 0.
3. Solve the system of equations to find Q1:
- Substitute P1 = (-1, -1, 8) + t[1, 0, -3] into the equation above:
([1, 2, -3] dot ((-1, -1, 8) + t[1, 0, -3] - Q1)) = 0.
- Simplify the equation and solve for t:
[1, 2, -3] dot ([-1-t, -1, 8-3t] - Q1) = 0.
4. Once you find t, substitute it back into the equation for line L1 to find point Q1.
5. Repeat the above steps to find Q2 on L2:
- Set up a system of equations:
([1, 0, -3] dot ((7, -11, 14) - Q2)) = 0.
- Solve the system of equations and substitute the obtained t into the equation for line L2 to find point Q2.
6. Finally, calculate the distance d(Q1, Q2) between Q1 and Q2 using the distance formula:
- d(Q1, Q2) = sqrt((Q2x - Q1x)^2 + (Q2y - Q1y)^2 + (Q2z - Q1z)^2).
By following these steps, you should be able to find the shortest distance d between the two lines, as well as the corresponding points Q1 and Q2 on the respective lines.