What's the distance from this point to this line?

Question

point: (1,1,1)

line: x = y-1 = 2z/3

using cross product

Answer 1

Find the distance from point P(1, 1, 1) to line
t = x = y - 1 = 2z/3.

Write the equation of the line in parametric form to make it more user friendly.
L(t)
x = t
y = 1 + t
z = 3t/2

The directional vector v, of the line can be taken from the parametric equations as the coefficients of t.

v = <1, 1, 3/2>

Any non-zero multiple of v is also a directional vector of the line. Multiply by 2.

v = <2, 2, 3>

Calculate the magnitude of v. We will need it later.
| v | = √(2² + 2² + 3²) = √(4 + 4 + 9) = √17

We can also calculate a point on the line by letting t = 0. The point is Q(0, 1, 0).

Now calculate vector PQ.
PQ = = <0-1, 1-1, 0-1> = <-1, 0, -1>

Calculate the cross product w, of AB and the directional vector v, of the line.

AB X v = <-1, 0, -1> X <2, 2, 3> = <2, 1, -2>

Calculate the magnitude of |AB X v|. We will need it later.
|AB X v| = √[2² + 1² + (-2)²] = √(4 + 1 + 4) = √9 = 3
_________

Let
h = distance from point P to line L
θ = angle between vectors AB and v

The magnitude of the cross product can be represented as:

| AB X v | = | AB | | v | sinθ
| AB X v | = | AB | | v | (h / | AB |) = | v | h

h = | AB X v | / | v | = 3/√17 ≈ 0.7276068

Answer 2

The parametric form of the line equation should be more like:
r = (0,-1,0) + t (3,3,2)
so the direction unit vector would be sqrt(22).(3,3,2)

first we get the coordinates of the point in relation to the origin of the line (say point P=(1,1,1) and A=(0,-1,0) )
then to get from P to A we follow the vector P minus A. Call this vector D.

Now to find the perpendicular from the line to the point:
take the cross product of D with the line direction (3,3,2) to get the normal (N) to the plane containing the line and D.
then take the cross product of (3,3,2) with N to give the perpendicular, L.

Normalise L to get the unit vector, and take the dot product of L^ with D to get the perpendicular distance from the line to P.

I got 1/sqrt(2). There's probably a better way of doing this, but I'm pretty sure this works.

Answer 3

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