Math Problem Statement

Mach 2a mit dem Beispiel

Mach 2a mit dem Beispiel

Solution

The task involves solving part 2a based on the provided example.

The first image:

The equation is: cos(α)=v1v2v1v2=6611\cos(\alpha) = \frac{\mathbf{v}_1 \cdot \mathbf{v}_2}{\|\mathbf{v}_1\| \|\mathbf{v}_2\|} = \frac{6}{\sqrt{6} \cdot \sqrt{11}} where: v1=(211),v2=(311)\mathbf{v}_1 = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}, \quad \mathbf{v}_2 = \begin{pmatrix} -3 \\ 1 \\ 1 \end{pmatrix} This calculates the cosine of the angle between two vectors.

The second image:

It gives a pair of parametric plane equations.

For 2a, let's extract the vectors and compute the angle between the normal vectors of two planes.

The parametric plane equations are given as:

  • E1:(x(120))(501)=0E_1: \left( \mathbf{x} - \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} \right) \cdot \begin{pmatrix} 5 \\ 0 \\ 1 \end{pmatrix} = 0
  • E2:(x(237))(610)=0E_2: \left( \mathbf{x} - \begin{pmatrix} 2 \\ 3 \\ 7 \end{pmatrix} \right) \cdot \begin{pmatrix} 6 \\ 1 \\ 0 \end{pmatrix} = 0

To solve this:

  1. Extract the normal vectors from the plane equations.

    • Normal vector to E1E_1: n1=(501)\mathbf{n}_1 = \begin{pmatrix} 5 \\ 0 \\ 1 \end{pmatrix}
    • Normal vector to E2E_2: n2=(610)\mathbf{n}_2 = \begin{pmatrix} 6 \\ 1 \\ 0 \end{pmatrix}
  2. Use the dot product formula to find the angle α\alpha between the two vectors: cos(α)=n1n2n1n2\cos(\alpha) = \frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{\|\mathbf{n}_1\| \|\mathbf{n}_2\|} Let's compute it:

    • Dot product n1n2=(56)+(01)+(10)=30\mathbf{n}_1 \cdot \mathbf{n}_2 = (5 \cdot 6) + (0 \cdot 1) + (1 \cdot 0) = 30
    • Magnitudes: n1=52+02+12=25+1=26\|\mathbf{n}_1\| = \sqrt{5^2 + 0^2 + 1^2} = \sqrt{25 + 1} = \sqrt{26} n2=62+12+02=36+1=37\|\mathbf{n}_2\| = \sqrt{6^2 + 1^2 + 0^2} = \sqrt{36 + 1} = \sqrt{37}
  3. Plugging these values into the formula: cos(α)=302637=309620.967\cos(\alpha) = \frac{30}{\sqrt{26} \cdot \sqrt{37}} = \frac{30}{\sqrt{962}} \approx 0.967

  4. Find the angle α\alpha using the inverse cosine: αcos1(0.967)15.05\alpha \approx \cos^{-1}(0.967) \approx 15.05^\circ

Thus, the angle between the two planes is approximately 15.0515.05^\circ.


Do you need more details on this process, or should I proceed with any further steps?

5 Questions for Further Exploration:

  1. How are dot products related to the angle between vectors?
  2. What is the geometric meaning of a normal vector to a plane?
  3. Can you solve the magnitude of vectors geometrically without using the formula?
  4. How does this concept extend to higher-dimensional spaces (beyond 3D)?
  5. What happens if the normal vectors are orthogonal? What is the angle between the planes?

Tip:

When computing angles between vectors, always ensure your vectors are in their simplest form and normalized if necessary.

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Math Problem Analysis

Mathematical Concepts

Vector Algebra
Dot Product
Angles between Vectors
Parametric Equations of Planes

Formulas

Dot product formula: \( \mathbf{v}_1 \cdot \mathbf{v}_2 = |\mathbf{v}_1||\mathbf{v}_2|\cos(\alpha) \)
Magnitude of a vector: \( \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2} \)

Theorems

Cosine theorem for vectors: \( \cos(\alpha) = \frac{\mathbf{v}_1 \cdot \mathbf{v}_2}{\|\mathbf{v}_1\| \|\mathbf{v}_2\|} \)

Suitable Grade Level

Undergraduate Math (Vectors and Linear Algebra)