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Lyryx Assignment 6 (Question 6)

Name: Shaunak Dogra Date: 2024-11-10

Official Time: 14:39:15     Question 6 [10 points]

Let L be the line passing through the point P=(5, −1, −1) with direction vector →d=[0, 1, −2]T. Find the shortest distance d from the point P0=(−4, 4, 3) to L, and the point Q on L that is closest to P0. Use the square root symbol '√' where needed to give an exact value for your answer.

d = 

√120

Q = (

0

0

0

)  

Solution

To find the shortest distance dd from the point P0=(4,4,3)P_0 = (-4, 4, 3) to the line LL, which passes through the point P=(5,1,1)P = (5, -1, -1) and has direction vector d=[0,1,2]T\vec{d} = [0, 1, -2]^T, we can use the following approach.

Step 1: Compute the vector PP0\vec{PP_0}

This vector goes from PP to P0P_0: PP0=P0P=(45,4+1,3+1)=(9,5,4)\vec{PP_0} = P_0 - P = (-4 - 5, 4 + 1, 3 + 1) = (-9, 5, 4)

Step 2: Find the projection of PP0\vec{PP_0} onto d\vec{d}

The projection formula is: projdPP0=PP0dddd\text{proj}_{\vec{d}} \, \vec{PP_0} = \frac{\vec{PP_0} \cdot \vec{d}}{\vec{d} \cdot \vec{d}} \, \vec{d}

  1. Calculate PP0d\vec{PP_0} \cdot \vec{d}: PP0d=(9)(0)+(5)(1)+(4)(2)=58=3\vec{PP_0} \cdot \vec{d} = (-9)(0) + (5)(1) + (4)(-2) = 5 - 8 = -3

  2. Calculate dd\vec{d} \cdot \vec{d}: dd=(0)2+(1)2+(2)2=0+1+4=5\vec{d} \cdot \vec{d} = (0)^2 + (1)^2 + (-2)^2 = 0 + 1 + 4 = 5

  3. So, [ \text{proj}_{\vec{d}} , \vec{PP_0} = \frac{-3}{5} \vec{d} = \frac{-3}{5} [0, 1, -2]^T = \left[ 0, -\frac{3}{5}, \frac{6}{5} \right] ]

Step 3: Find the perpendicular (shortest) distance vector

The perpendicular distance vector from P0P_0 to the line is: [ \vec{PP_0} - \text{proj}_{\vec{d}} , \vec{PP_0} = \left[ -9, 5, 4 \right] - \left[ 0, -\frac{3}{5}, \frac{6}{5} \right] = \left[ -9, \frac{28}{5}, \frac{14}{5} \right] ]

Step 4: Calculate the magnitude of the perpendicular distance vector

The distance dd is the magnitude of this vector: d=(9)2+(285)2+(145)2=81+78425+19625=120d = \sqrt{\left(-9\right)^2 + \left(\frac{28}{5}\right)^2 + \left(\frac{14}{5}\right)^2} = \sqrt{81 + \frac{784}{25} + \frac{196}{25}} = \sqrt{120}

So, the shortest distance is: d=120d = \sqrt{120}

Step 5: Find the point QQ on LL that is closest to P0P_0

The point QQ can be found by starting at PP and moving along d\vec{d} by the scalar multiple found in the projection: Q=P+35d=(5,1,1)+(0,35,65)=(5,85,15)Q = P + \frac{-3}{5} \vec{d} = \left( 5, -1, -1 \right) + \left( 0, -\frac{3}{5}, \frac{6}{5} \right) = \left( 5, -\frac{8}{5}, \frac{1}{5} \right)

Final Answers

d=120d = \sqrt{120} Q=(5,85,15)Q = \left( 5, -\frac{8}{5}, \frac{1}{5} \right)

Would you like further explanation on any part of this solution?


  1. How do you determine the shortest distance from a point to a line in 3D?
  2. What role does the direction vector play in finding the point closest to P0P_0?
  3. Why is the projection operation used here?
  4. How would you find the distance if LL were a plane instead of a line?
  5. What is the geometric interpretation of the perpendicular distance vector?

Tip: When finding distances to lines or planes, visualizing projections can greatly simplify understanding the shortest paths.

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Math Problem Analysis

Mathematical Concepts

Analytic Geometry
Vectors
Distance in 3D Space

Formulas

Projection of vector: proj_d a = (a · d / d · d) * d
Distance between a point and a line in 3D

Theorems

Vector Projection Theorem
Pythagorean Theorem for distances in space

Suitable Grade Level

Grades 11-12