Math Problem Statement

Find the distance from the point to the line: (6, 2, 4); x = 3 - t, y = 6 + 4t, z = 2 + 3t

Solution

To find the distance from the point P(6,2,4)P(6, 2, 4) to the line given by the parametric equations: x=3t,y=6+4t,z=2+3t,x = 3 - t, \quad y = 6 + 4t, \quad z = 2 + 3t, we can follow this procedure:

1. Vector Form of the Line:

The parametric equations describe a line. To write the line in vector form, we have: r(t)=3,6,2+t1,4,3.\mathbf{r}(t) = \langle 3, 6, 2 \rangle + t \langle -1, 4, 3 \rangle. Thus, the line has direction vector d=1,4,3\mathbf{d} = \langle -1, 4, 3 \rangle and passes through the point Q(3,6,2)Q(3, 6, 2).

2. Vector from Point on Line to Given Point:

Let PQ\mathbf{PQ} be the vector from the point P(6,2,4)P(6, 2, 4) to the point Q(3,6,2)Q(3, 6, 2) on the line. This vector is: PQ=PQ=63,26,42=3,4,2.\mathbf{PQ} = P - Q = \langle 6 - 3, 2 - 6, 4 - 2 \rangle = \langle 3, -4, 2 \rangle.

3. Project PQ\mathbf{PQ} onto the Line Direction:

To find the perpendicular distance, we need to project PQ\mathbf{PQ} onto the direction vector d=1,4,3\mathbf{d} = \langle -1, 4, 3 \rangle. The formula for the projection of a vector v\mathbf{v} onto another vector d\mathbf{d} is: projdPQ=PQdddd.\text{proj}_{\mathbf{d}} \mathbf{PQ} = \frac{\mathbf{PQ} \cdot \mathbf{d}}{\mathbf{d} \cdot \mathbf{d}} \mathbf{d}. First, compute the dot products: PQd=(3)(1)+(4)(4)+(2)(3)=316+6=13,\mathbf{PQ} \cdot \mathbf{d} = (3)(-1) + (-4)(4) + (2)(3) = -3 - 16 + 6 = -13, dd=(1)2+42+32=1+16+9=26.\mathbf{d} \cdot \mathbf{d} = (-1)^2 + 4^2 + 3^2 = 1 + 16 + 9 = 26. Thus, the projection is: projdPQ=1326d=121,4,3=12,2,32.\text{proj}_{\mathbf{d}} \mathbf{PQ} = \frac{-13}{26} \mathbf{d} = \frac{-1}{2} \langle -1, 4, 3 \rangle = \langle \frac{1}{2}, -2, -\frac{3}{2} \rangle.

4. Find the Perpendicular Vector:

The vector from PP to the line is the difference between PQ\mathbf{PQ} and the projection: v=PQprojdPQ=3,4,212,2,32=52,2,72.\mathbf{v}_{\perp} = \mathbf{PQ} - \text{proj}_{\mathbf{d}} \mathbf{PQ} = \langle 3, -4, 2 \rangle - \langle \frac{1}{2}, -2, -\frac{3}{2} \rangle = \langle \frac{5}{2}, -2, \frac{7}{2} \rangle.

5. Distance Calculation:

The distance from the point to the line is the magnitude of the perpendicular vector v\mathbf{v}_{\perp}: d=v=(52)2+(2)2+(72)2=254+4+494=744+4=904=902.d = \left| \mathbf{v}_{\perp} \right| = \sqrt{\left( \frac{5}{2} \right)^2 + (-2)^2 + \left( \frac{7}{2} \right)^2} = \sqrt{\frac{25}{4} + 4 + \frac{49}{4}} = \sqrt{\frac{74}{4} + 4} = \sqrt{\frac{90}{4}} = \frac{\sqrt{90}}{2}. Simplifying: d=3102.d = \frac{3\sqrt{10}}{2}.

So, the distance from the point to the line is 3102\frac{3\sqrt{10}}{2}.

Would you like more details or have any questions?

Here are some related questions you might explore:

  1. How do you find the parametric equations of a line given two points?
  2. How is the formula for projecting one vector onto another derived?
  3. What are the different methods to find the distance between a point and a plane?
  4. How do you calculate the shortest distance between two skew lines?
  5. How do you determine if a point lies on a given line in 3D space?

Tip: Always verify if the projection vector lies in the same direction as the line, as this ensures you're calculating the correct perpendicular distance.

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Math Problem Analysis

Mathematical Concepts

Vector Projection
Distance from Point to Line in 3D
Parametric Equations

Formulas

Distance from a point to a line: d = |PQ x d| / |d|
Vector projection: proj_d PQ = (PQ ⋅ d / d ⋅ d) * d
Magnitude of a vector: |v| = sqrt(v1^2 + v2^2 + v3^2)

Theorems

Pythagorean Theorem
Vector Cross Product

Suitable Grade Level

Grades 11-12 or early college level