Find an equation of the line in three space?

Question

Find an equation of the line through the point A (4,5,5) which meets the line (x-11)/3 = (y+8)/-1 = z-4 at right angles.



Any vector, , perpendicular to <3, -1, 1> must satisfy 3a-b+ c= 0 or c= b- 3a. (I did this through dot product)


Such a line passing through (4, 5, 5) must be of the form (x-4)/a= (y- 5)/b= (z-5)/(b- 3a).

Now, what are the conditions on a and b, in order to ensure that (x-4)/a= (y- 5)/b= (z-5)/(b- 3a) intersects (x-11)/3 = (y+8)/-1 = z-4 ?

This is where i have a hard time. help

Answer 1

Notice that a line has the general vector form:

r = r_0 + v * t where r_0 is the vector formed by a point on the line and the origin, v is the direction vector of the line, and t is a non-zero parameter.

At this moment, you have the following information about the two lines:

r_1 = <4+at, 5+bt, 5+ct>
r_2 = <11+3t, -8-t, 4+t>

You want to determine a, b, and c.

You want to make sure these two lines intersect and are perpendicular to each other.

Note that they intersect when 4+at=11+3t, 5+bt=-8-t and 5+ct=4+t. Solving for a, b, and c, we get that:

a = (7+3t)/t
b = (-13-t)/t
c = (-1+t)/t

From your work, we find that we get that is perpendicular to <3, -1, 1> if 3a - b + c = 0.

Plugging in our a, b, and c values, we have that:

3a - b + c = 3[(7+3t)/t] - (-13-t)/t + (-1+t)/t = 0

Multiplying both sides by t (note that t is assumed to be non-zero), we get that:

21+9t + 13 + t - 1 + t = 0
33 + 11t = 0
11t = -33t
t = -3

So solving for a, b, and c we get:

a = (7+3t)/t = (7+3(-3))/(-3) = 2/3
b = (-13-t)/t = (-13-(-3))/(-3) = 10/3
c = (-1+t)/t = (-1+(-3))/(-3) = 4/3

This gives the vector <2/3, 10/3, 4/3>.
We can use <2, 10, 4> as our vector (since it is in the same direction).

Thus, an equation of the line through the point (4,5,5) which meets the line (x-11)/3 = (y+8)/-1 = z-4 at right angles is the line (x-4)/2 = (b-5)/10 = (c-5)/4.

Hope that helped.

Answer 2

Any point on the line L:(x-11)/3 = (y+8)/-1 = z-4 =t (say) can be considered as P(3t+11,-t -8 ,t + 4).

Now direction ratios of AP is < 3t +11-4, -t-8-5, t+4-5 > =
<3t +7,-t-13,t-1>

Direction Ratios of L is <3, -1, 1>

Now L is perpendicular to AP. So, 3(3t+7)-1(-t-13)+1(t-1) = 0
This gives t = -3.

The point P is (2,-5,1).

The equation of line AP is (x-4)/2 =(y-5)/10 =(z-5)/4

A small tip:Whenever there are 2 information about a line its equation can always be determined. All we need to do is try in a different angle, when we get struck in one method. Enjoy solving problems.

Answer 3

You need to find the point on the given line closest to A, then draw the line through those two points: X+t*(A-X). But I forgot how to get the closest point....you could use calculus.......

Like (11,-8,4) is on the given line and it goes in direction (3,-1,1) so an equation would be (11+3t,-8-t,4+t). Then distance between point and line squared would be (11+3t-4)^2 +(-8-t-5)^2 +(4+t-5)^2 which expands to 219+66*t+11*t^2 which is minimzed at t=-3. So you need the line thru A(4,5,5) and X(2,-5,1) which would be given by (4+2t,5 +10t,5+4t) or w/o parameters (x-4)/2= (y-5)/10 = (z-4)/4

Using your method, solve given line for x in terms of y and substitute into the {x,y} equations with a.b in it then subs x in terms of z into the a-b-x-z equations to get 2 equations with a and b in them.

Answer 4

Assume the intersection point is (x₀, y₀, z₀).
(x₀-11)/3 = (y₀+8)/(-1) = z₀-4 = c......(1)

Also, the dot product of two perpendicular lines is zero
3(x₀-4)-(y₀-5)+(z₀-5) = 0......(2)

Plug in (1) into (2) with
x₀-4 = 3c+7
y₀-5 = -c-13
z₀-5 = c-1

3(3c+7)+(c+13)+(c-1) = 0......(3)

Solve (3) for c,
c = =-3

Therefore,
x₀ = 2
y₀ = -5
z₀ = 1

The direction of the line is,
(2, 10, 4)

The equation of the line is,
(x-4)/2 = (y-5)/10 = (z-5)/4