Thread: parametric equations

Let P be the point (3,1,-2) and L be the line given by x=-4+2t, y=2+2t, z=1+t.
Find parametric equations for the line passing through P and intersecting L at a right angle.

Could you please show me how to solve this question? Thank you very much.

Originally Posted by Jenny20

Let P be the point (3,1,-2) and L be the line given by x=-4+2t, y=2+2t, z=1+t.
Find parametric equations for the line passing through P and intersecting L at a right angle.

Could you please show me how to solve this question? Thank you very much.

There are probably neater ways of doing this, but you could find the value
of t which minimises the distance from a point on L to P, then you have a pair
of points on on the desired line from which the required parametric equation
can be derived.

If the distance between points is minimised, so is the square distance.

The square distance from P to L(t) is:

D(t)=(7-2t)^2+(-1-2t)^2+(-3-t)^2

This is minimised by a value of t such that (d/dt)D(t)=0,

Now:

(d/dt)D(t)=2(7-2t)(-2) + 2(-1-2t)(-2) +2(-3-t)(-1)=18t-18,

so if (d/dt)D(t)=0 t=1.

So the two points on the required line are the given point P=(3,1,-2), and
the point L(1)=(-2,4,3).

Then the line M(lambda)=lambda P + (1-lambda) L(1) is the required line,
or:

x=3 lambda - 2 (1-lambda)=-2 + 5 lambda,
y=1 lambda + 4 (1-lambda)= 4 - 3 lambda
z=-2 lambda +3 (1-lambda)= 3 - 5 lambda.

RonL

(you will need to check the algebra here, there is probably a mistake
somewhere but it should give you a idea of how to solve it).