Math Problem Statement

A its at (8,0,8), B its at (0,4,0), P its at (-6,10,6) and F goes from A to B

Solution

Given the coordinates:

  • Point A=(8,0,8)A = (8, 0, 8)
  • Point B=(0,4,0)B = (0, 4, 0)
  • Point P=(6,10,6)P = (-6, 10, 6)

The vector F\mathbf{F} goes from point AA to point BB, so:

F=AB=BA\mathbf{F} = \overrightarrow{AB} = B - A

Step-by-Step Solution:

  1. Calculate Vector AB\overrightarrow{AB}:

AB=(xBxA,yByA,zBzA)=(08,40,08)\overrightarrow{AB} = (x_B - x_A, y_B - y_A, z_B - z_A) = (0 - 8, 4 - 0, 0 - 8)

AB=(8,4,8)\overrightarrow{AB} = (-8, 4, -8)

  1. Magnitude of Vector AB\overrightarrow{AB}:

AB=(8)2+42+(8)2|\overrightarrow{AB}| = \sqrt{(-8)^2 + 4^2 + (-8)^2}

AB=64+16+64=144=12|\overrightarrow{AB}| = \sqrt{64 + 16 + 64} = \sqrt{144} = 12

  1. Unit Vector in the Direction of AB\overrightarrow{AB}:

uAB=ABAB=(812,412,812)\mathbf{u}_{AB} = \frac{\overrightarrow{AB}}{|\overrightarrow{AB}|} = \left( \frac{-8}{12}, \frac{4}{12}, \frac{-8}{12} \right)

uAB=(23,13,23)\mathbf{u}_{AB} = \left( -\frac{2}{3}, \frac{1}{3}, -\frac{2}{3} \right)

  1. Vector from Point AA to Point PP:

AP=PA=(68,100,68)\overrightarrow{AP} = P - A = (-6 - 8, 10 - 0, 6 - 8)

AP=(14,10,2)\overrightarrow{AP} = (-14, 10, -2)

  1. Projection of Vector AP\overrightarrow{AP} onto AB\overrightarrow{AB}:

The projection formula is:

ProjABAP=(APuAB)uAB\text{Proj}_{\overrightarrow{AB}} \, \overrightarrow{AP} = (\overrightarrow{AP} \cdot \mathbf{u}_{AB}) \mathbf{u}_{AB}

First, compute the dot product:

APuAB=(14)(23)+10(13)+(2)(23)\overrightarrow{AP} \cdot \mathbf{u}_{AB} = (-14) \left( -\frac{2}{3} \right) + 10 \left( \frac{1}{3} \right) + (-2) \left( -\frac{2}{3} \right)

APuAB=283+103+43=423=14\overrightarrow{AP} \cdot \mathbf{u}_{AB} = \frac{28}{3} + \frac{10}{3} + \frac{4}{3} = \frac{42}{3} = 14

  1. Final Projection Value:

ProjABAP=14(23,13,23)\text{Proj}_{\overrightarrow{AB}} \, \overrightarrow{AP} = 14 \cdot \left( -\frac{2}{3}, \frac{1}{3}, -\frac{2}{3} \right)

ProjABAP=(283,143,283)\text{Proj}_{\overrightarrow{AB}} \, \overrightarrow{AP} = \left( -\frac{28}{3}, \frac{14}{3}, -\frac{28}{3} \right)

Conclusion:

The projection of vector AP\overrightarrow{AP} onto the direction of AB\overrightarrow{AB} is:

(283,143,283)\left( -\frac{28}{3}, \frac{14}{3}, -\frac{28}{3} \right)

Would you like further details on this solution, or have any questions?


Here are some related questions for you:

  1. How is the dot product used to find projections?
  2. What is the geometric interpretation of a vector projection?
  3. Can you calculate the angle between vectors AB\overrightarrow{AB} and AP\overrightarrow{AP}?
  4. How does the magnitude of a vector affect its unit vector?
  5. What are some practical applications of vector projections in physics?

Tip: When working with vectors, always visualize them in 3D space to understand their relative directions and magnitudes better!

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Math Problem Analysis

Mathematical Concepts

Vectors
Dot Product
Vector Projection

Formulas

Vector addition
Vector subtraction
Dot product formula
Vector projection formula

Theorems

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Suitable Grade Level

High School