Space Geometry+Locus of a point

[Perlita]

Hello everyone,
Here's an exercise I have to solve:
Given an orthonormal basis (O;i,j,k) and two lines m and n whose respective parametric equations are:
x = 2+t x = 4-3t'
y = 3-2t (t belongs to R) and y = 5-8t' (t' belongs to R)
z = 5-t z = 7-t'

1) Are lines m and n coplanar? Justify your answer.
2) Show that point A(2,3,5) belongs to line m.

Let B be a point on line n.
Find the locus of point I, midpoint of segment AB, while B moves along line n.

3) Let M be a point of line m, and B be a point of line n.
Find the locus of the midpoint of segment MB while M and B move along lines m and n respectively.

The answers I gave:
1) We try to find out if the lines intersect, by solving the system:
2+t = 4-3t
3-2t = 5-8t
5-t = 7-t

We get t= 1/2, t= 1/3, 5=7. which is impossible, hence the lines do not intersect, so they are not coplanar.
2) By substitution in the equation of line m, we see that A belongs to m.
Considering a point B on line n, and taking the midpoint I of segment AB, we get I(3-(3/2)t , 4-4t , 6-(1/2)t). So I belongs to a line.

3)the locus is a line equidistant from m and n.

My questions:
-Did I answer correctly on parts 2 and 3?
-May you tell me how I can prove that the locus in 3) is a line? (if it's correct!)

Thank you

 

[DrPhil]

Hello everyone,
Here's an exercise I have to solve:
Given an orthonormal basis (O;i,j,k) and two lines m and n whose respective parametric equations are:
x = 2+t............................. x = 4-3t'
y = 3-2t (t belongs to R) and y = 5-8t' (t' belongs to R)
z = 5-t.............................. z = 7-t'

1) Are lines m and n coplanar? Justify your answer.
2) Show that point A(2,3,5) belongs to line m.

Let B be a point on line n.
Find the locus of point I, midpoint of segment AB, while B moves along line n.

3) Let M be a point of line m, and B be a point of line n.
Find the locus of the midpoint of segment MB while M and B move along lines m and n respectively.

The answers I gave:
1) We try to find out if the lines intersect, by solving the system:
2+t = 4-3t'
3-2t = 5-8t'
5-t = 7-t' The parameters t and t' are not necessarily equal

We get t= 1/2, t= 1/3, 5=7. which is impossible, hence the lines do not intersect, so they are not coplanar.
2) By substitution in the equation of line m, we see that A belongs to m.
Considering a point B on line n, and taking the midpoint I of segment AB, we get I(3-(3/2)t , 4-4t , 6-(1/2)t). So I belongs to a line.

3)the locus is a line equidistant from m and n.

My questions:
-Did I answer correctly on parts 2 and 3?
-May you tell me how I can prove that the locus in 3) is a line? (if it's correct!)

Thank you

1) If the lines intersect, there exist valuers of (t, t') such that the (x,y,z) of the two lines are equal
t + 3t' = 2
2t - 8t' = -2
t - t' = -2
Failure of THIS system of equations is what shows there is no intersection.

2) point A = (2,3,5) corresponds to t=0
....point B has the form (4,5,7) + (-3,-8,-1)t'
To find the midpoint, add the two points and divide by 2. Write what you got the form of a constant + a vector times t':
....I = (3,4,6) + (-3/2,-4,-1/2)t'
Does that look like the parametric form of a line?
The result shows that the locus of points halfway between a given POINT and a line is a line

3) point M = (2,3,5) + (+1,-2,-1)t
The midpoint is (M + B)/2
Is that a straight line?
[Remember that t and t' are independent of each other.]

 

[Perlita]

Thanks a lot Dr Phil, your answers are very helpful. I understood that if the coordinates of a point can be written as a constant+a vector * parameter, then we can say that the point belongs to a line...

But I got stuck again in part 3)!!!
Doing (M+B)/2 will give us an expression with 2 different parameters t and t' that we can't combine. What to do in this case? May you give me the correct answer?