Math Problem Statement

Given v⃗ v→ with an initial point of ⟨−4,−2,1⟩⟨−4,−2,1⟩ and a terminal point of ⟨3,6,−6⟩⟨3,6,−6⟩ , find a unit vector in the same direction as v⃗ v→.

Solution

To find a unit vector in the same direction as v\mathbf{v} with an initial point 4,2,1\langle -4, -2, 1 \rangle and a terminal point 3,6,6\langle 3, 6, -6 \rangle, we need to follow these steps:

  1. Find the vector v\mathbf{v} from the initial and terminal points.
  2. Calculate the magnitude of v\mathbf{v}.
  3. Divide the vector v\mathbf{v} by its magnitude to get the unit vector.

Step 1: Find the vector v\mathbf{v}

The vector v\mathbf{v} is given by the difference between the terminal point and the initial point: v=3,6,64,2,1\mathbf{v} = \langle 3, 6, -6 \rangle - \langle -4, -2, 1 \rangle v=3(4),6(2),61\mathbf{v} = \langle 3 - (-4), 6 - (-2), -6 - 1 \rangle v=3+4,6+2,61\mathbf{v} = \langle 3 + 4, 6 + 2, -6 - 1 \rangle v=7,8,7\mathbf{v} = \langle 7, 8, -7 \rangle

Step 2: Calculate the magnitude of v\mathbf{v}

The magnitude of v\mathbf{v}, denoted as v\|\mathbf{v}\|, is given by: v=72+82+(7)2\|\mathbf{v}\| = \sqrt{7^2 + 8^2 + (-7)^2} v=49+64+49\|\mathbf{v}\| = \sqrt{49 + 64 + 49} v=162\|\mathbf{v}\| = \sqrt{162} v=81×2\|\mathbf{v}\| = \sqrt{81 \times 2} v=92\|\mathbf{v}\| = 9\sqrt{2}

Step 3: Divide the vector v\mathbf{v} by its magnitude to get the unit vector

The unit vector u\mathbf{u} in the direction of v\mathbf{v} is given by: u=vv\mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|} u=7,8,792\mathbf{u} = \frac{\langle 7, 8, -7 \rangle}{9\sqrt{2}} u=792,892,792\mathbf{u} = \left\langle \frac{7}{9\sqrt{2}}, \frac{8}{9\sqrt{2}}, \frac{-7}{9\sqrt{2}} \right\rangle

To rationalize the denominator: u=7218,8218,7218\mathbf{u} = \left\langle \frac{7\sqrt{2}}{18}, \frac{8\sqrt{2}}{18}, \frac{-7\sqrt{2}}{18} \right\rangle u=7218,429,7218\mathbf{u} = \left\langle \frac{7\sqrt{2}}{18}, \frac{4\sqrt{2}}{9}, \frac{-7\sqrt{2}}{18} \right\rangle

Thus, the unit vector in the same direction as v\mathbf{v} is: u=7218,429,7218\mathbf{u} = \left\langle \frac{7\sqrt{2}}{18}, \frac{4\sqrt{2}}{9}, \frac{-7\sqrt{2}}{18} \right\rangle

Would you like any more details or have any questions about this solution?

Here are some related questions:

  1. How do you find the direction cosines of a vector?
  2. What is the difference between a unit vector and a normal vector?
  3. How can unit vectors be used in physics?
  4. What is the significance of the dot product of unit vectors?
  5. How do you normalize a vector in higher dimensions?
  6. What is the role of unit vectors in coordinate systems?
  7. Can you explain the concept of vector projection?
  8. How do you calculate the angle between two vectors using their unit vectors?

Tip: Unit vectors are especially useful in defining directions and are often used to simplify vector calculations in physics and engineering.

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Math Problem Analysis

Mathematical Concepts

Vectors
Vector Operations
Unit Vectors

Formulas

Vector subtraction
Magnitude of a vector
Unit vector formula

Theorems

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Suitable Grade Level

High School