can somebody please help me with this question I've spent more than 3 hours on this, still nothing
Let L1 be the line passing through the point P1=(2, −4, −6) with direction vector →d1=[1, 0, 2]T, and let L2 be the line passing through the point P2=(−2, 3, −3) with direction vector →d2=[1, 4, −2]T.
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.
d =
Q1(_ _ _)
Q2(_ _ _)
I have found d, my problem is finding Q1 and Q2
and if you don't want to help it's better not to say anything at all I know how to google thanks!
To find the shortest distance between two lines, we can use the formula:
d = |((P2 - P1) · (d1 x d2)) / ||d1 x d2|||
where P1 and P2 are points on the lines, d1 and d2 are the direction vectors of the lines, · denotes the dot product, x denotes the cross product, and || || represents the magnitude.
Let's substitute the given values into the formula to find the shortest distance, d:
P1 = (2, -4, -6)
P2 = (-2, 3, -3)
d1 = (1, 0, 2)
d2 = (1, 4, -2)
Substituting these values into the formula:
d = |((P2 - P1) · (d1 x d2)) / ||d1 x d2|||
= |((-2 - 2, 3 - (-4), -3 - (-6)) · (1, 0, 2) x (1, 4, -2)) / ||(1, 0, 2) x (1, 4, -2)|||
= |(-4, 7, 3) · (-8, -4, 4) / ||(12, -4, 4)|||
= |(-4)(-8) + (7)(-4) + (3)(4) / sqrt((12)^2 + (-4)^2 + 4^2)|
= |32 - 28 + 12 / sqrt(144 + 16 + 16)|
= |16 / sqrt(176)|
Simplifying further:
d = 16 / sqrt(176)
To find the points Q1 and Q2, we can use the parametric equations of the lines:
Q1 = P1 + t1 * d1
Q2 = P2 + t2 * d2
We can substitute the values of P1 and P2 to find the respective points Q1 and Q2:
Q1 = (2, -4, -6) + t1 * (1, 0, 2)
Q2 = (-2, 3, -3) + t2 * (1, 4, -2)
So, the solution is:
d = 16 / sqrt(176)
Q1 = (2 + t1, -4, -6 + 2t1)
Q2 = (-2 + t2, 3 + 4t2, -3 - 2t2)
To find the shortest distance between two lines in 3D space and the corresponding points on each line, you can use vector equations and vector calculus. Let's break down the steps to solve this problem:
Step 1: Find the vector connecting the two given points P1 and P2 on each line.
- The vector connecting P1 and P2 is given by →v = P2 - P1.
- Substitute the given values to calculate →v = [(-2 - 2), (3 - (-4)), (-3 - (-6))].
Step 2: Find the cross product of the direction vectors →d1 and →d2.
- The cross product →crs = →d1 x →d2.
- Substitute the given values to calculate →crs.
Step 3: Find the length of the cross product vector →crs.
- The length of the cross product vector is given by |→crs| = √(→crs · →crs).
- Take the dot product of →crs with itself and take its square root.
Step 4: Find the projection of the vector →v onto the cross product vector →crs.
- The projection of →v onto →crs is given by |→v_proj| = |→v · →crs| / |→crs|.
- Take the dot product of →v with →crs and divide it by the length of →crs.
Step 5: Find the shortest distance d between the lines.
- The shortest distance between two lines is given by d = |→v_proj|.
- The value obtained in Step 4 is |→v_proj|.
Step 6: Find the corresponding points Q1 and Q2 on each line.
- The points Q1 and Q2 can be found using the equation Q1 = P1 + m * →d1 and Q2 = P2 + n * →d2, where m and n are scalar values.
- Substitute the given values of P1, P2, →d1, and →d2 in the equations to find the corresponding points Q1 and Q2.
Now let's go through each step in detail to find the solution:
Step 1:
→v = [(-2 - 2), (3 - (-4)), (-3 - (-6))] = [-4, 7, 3]
Step 2:
→crs = →d1 x →d2
→crs = [1, 0, 2] x [1, 4, -2]
→crs = [(-2*0 - 4*(-2)), (-2*(-2) - 1*0), (1*4 - 1*0)]
→crs = [8, 4, 4]
Step 3:
|→crs| = √(→crs · →crs)
|→crs| = √(8^2 + 4^2 + 4^2)
Step 4:
|→v_proj| = |→v · →crs| / |→crs|
|→v_proj| = [(-4*8 + 7*4 + 3*4)] / √(8^2 + 4^2 + 4^2)
Step 5:
d = |→v_proj|
Step 6:
Q1 = P1 + m * →d1
Q2 = P2 + n * →d2
Substitute the given values of P1, P2, →d1, and →d2 in the equations to find the corresponding points Q1 and Q2.
google is your friend:
https://en.wikipedia.org/wiki/Skew_lines#Distance_between_two_skew_lines