Math Problem Statement

Laske vektorin u = (1, 1, 2) vektoriprojektio u_v suuntaan v = (1, 0, -1). Osoita laskemalla, että u - u_v on kohtisuorassa vektoria v vastaan (tämän tehtävän vektoreilla).

Solution

To solve the problem:

  1. Vector projection formula: The projection of u\mathbf{u} onto v\mathbf{v}, denoted as uv\mathbf{u}_{\mathbf{v}}, is calculated as:

    uv=uvv2v\mathbf{u}_{\mathbf{v}} = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2} \mathbf{v}

    Here:

    • u=(1,1,2)\mathbf{u} = (1, 1, 2)
    • v=(1,0,1)\mathbf{v} = (1, 0, -1)
    • uv\mathbf{u} \cdot \mathbf{v} is the dot product of u\mathbf{u} and v\mathbf{v},
    • v2\|\mathbf{v}\|^2 is the squared magnitude of v\mathbf{v}.
  2. Orthogonal component: After finding uv\mathbf{u}_{\mathbf{v}}, the orthogonal component uuv\mathbf{u} - \mathbf{u}_{\mathbf{v}} will be calculated. This vector should be perpendicular to v\mathbf{v}, meaning their dot product will be 0.


Step-by-step Solution:

Step 1: Compute uv\mathbf{u} \cdot \mathbf{v}

uv=(1)(1)+(1)(0)+(2)(1)=1+02=1\mathbf{u} \cdot \mathbf{v} = (1)(1) + (1)(0) + (2)(-1) = 1 + 0 - 2 = -1

Step 2: Compute v2\|\mathbf{v}\|^2

v2=(1)2+(0)2+(1)2=1+0+1=2\|\mathbf{v}\|^2 = (1)^2 + (0)^2 + (-1)^2 = 1 + 0 + 1 = 2

Step 3: Calculate uv\mathbf{u}_{\mathbf{v}}

uv=uvv2v=12(1,0,1)=(12,0,12)\mathbf{u}_{\mathbf{v}} = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2} \mathbf{v} = \frac{-1}{2}(1, 0, -1) = \left(-\frac{1}{2}, 0, \frac{1}{2}\right)

Step 4: Compute uuv\mathbf{u} - \mathbf{u}_{\mathbf{v}}

uuv=(1,1,2)(12,0,12)=(1+12,10,212)=(32,1,32)\mathbf{u} - \mathbf{u}_{\mathbf{v}} = (1, 1, 2) - \left(-\frac{1}{2}, 0, \frac{1}{2}\right) = \left(1 + \frac{1}{2}, 1 - 0, 2 - \frac{1}{2}\right) = \left(\frac{3}{2}, 1, \frac{3}{2}\right)

Step 5: Verify orthogonality

To confirm that uuv\mathbf{u} - \mathbf{u}_{\mathbf{v}} is perpendicular to v\mathbf{v}, compute their dot product: (32,1,32)(1,0,1)=(32)(1)+(1)(0)+(32)(1)=3232=0\left(\frac{3}{2}, 1, \frac{3}{2}\right) \cdot (1, 0, -1) = \left(\frac{3}{2}\right)(1) + (1)(0) + \left(\frac{3}{2}\right)(-1) = \frac{3}{2} - \frac{3}{2} = 0

Since the dot product is 0, uuv\mathbf{u} - \mathbf{u}_{\mathbf{v}} is orthogonal to v\mathbf{v}.


Final Results:

  1. uv=(12,0,12)\mathbf{u}_{\mathbf{v}} = \left(-\frac{1}{2}, 0, \frac{1}{2}\right)
  2. uuv=(32,1,32)\mathbf{u} - \mathbf{u}_{\mathbf{v}} = \left(\frac{3}{2}, 1, \frac{3}{2}\right), which is perpendicular to v\mathbf{v}.

Would you like further details on any step?


Relative Questions:

  1. What is the significance of vector projection in physics or engineering?
  2. How do you calculate the magnitude of a vector given its components?
  3. What are some real-world examples of orthogonal vectors?
  4. How is the concept of orthogonality used in computer graphics?
  5. Can projections be extended to spaces with more than three dimensions?

Tip: Always double-check dot products when working with projections; a small mistake can change the entire outcome.

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Math Problem Analysis

Mathematical Concepts

Vector Projection
Dot Product
Orthogonality

Formulas

Vector projection formula: u_v = (u · v / ||v||²) v
Dot product: u · v = Σ(u_i * v_i)
Magnitude squared: ||v||² = Σ(v_i²)

Theorems

Orthogonality via dot product (u · v = 0)

Suitable Grade Level

Undergraduate Mathematics or Advanced High School