Math Problem Statement

Find the direction cosines and direction angles of the vector. (Give the direction angles correct to the nearest degree.) <2,1,2>

Solution

The direction cosines and direction angles of a vector v=2,1,2\mathbf{v} = \langle 2, 1, 2 \rangle can be determined by first finding the magnitude of the vector, and then using it to compute the cosines of the angles between the vector and the coordinate axes. These angles are called the direction angles.

Step 1: Find the magnitude of the vector

The magnitude v|\mathbf{v}| of the vector v=2,1,2\mathbf{v} = \langle 2, 1, 2 \rangle is given by:

v=x2+y2+z2=22+12+22=4+1+4=9=3|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2} = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3

Step 2: Find the direction cosines

The direction cosines are given by:

cosα=xv,cosβ=yv,cosγ=zv\cos \alpha = \frac{x}{|\mathbf{v}|}, \quad \cos \beta = \frac{y}{|\mathbf{v}|}, \quad \cos \gamma = \frac{z}{|\mathbf{v}|}

where α\alpha, β\beta, and γ\gamma are the direction angles with respect to the xx-, yy-, and zz-axes, respectively.

For the vector v=2,1,2\mathbf{v} = \langle 2, 1, 2 \rangle and its magnitude v=3|\mathbf{v}| = 3:

cosα=23,cosβ=13,cosγ=23\cos \alpha = \frac{2}{3}, \quad \cos \beta = \frac{1}{3}, \quad \cos \gamma = \frac{2}{3}

Step 3: Find the direction angles

To find the direction angles, take the inverse cosine (arccos) of the direction cosines:

α=cos1(23),β=cos1(13),γ=cos1(23)\alpha = \cos^{-1} \left( \frac{2}{3} \right), \quad \beta = \cos^{-1} \left( \frac{1}{3} \right), \quad \gamma = \cos^{-1} \left( \frac{2}{3} \right)

Now, calculate these values to the nearest degree:

α=cos1(23)48\alpha = \cos^{-1} \left( \frac{2}{3} \right) \approx 48^\circ β=cos1(13)70\beta = \cos^{-1} \left( \frac{1}{3} \right) \approx 70^\circ γ=cos1(23)48\gamma = \cos^{-1} \left( \frac{2}{3} \right) \approx 48^\circ

Final Answer:

  • Direction cosines: cosα=23\cos \alpha = \frac{2}{3}, cosβ=13\cos \beta = \frac{1}{3}, cosγ=23\cos \gamma = \frac{2}{3}
  • Direction angles: α=48\alpha = 48^\circ, β=70\beta = 70^\circ, γ=48\gamma = 48^\circ

Would you like further details or have any questions?

Here are some related questions:

  1. What are direction cosines for vectors in 2D space?
  2. How do you interpret direction angles geometrically?
  3. What is the difference between direction angles and direction cosines?
  4. How are direction cosines used in physics, especially in vectors in 3D space?
  5. What would happen to direction angles if the vector were normalized?

Tip: The direction angles always add up to less than 180180^\circ because they are the angles between the vector and each axis, not the angles between the axes themselves.

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Math Problem Analysis

Mathematical Concepts

Vector Geometry
Direction Cosines
Direction Angles
3D Coordinate Systems

Formulas

Magnitude of a vector: |v| = sqrt(x^2 + y^2 + z^2)
Direction cosines: cos(α) = x/|v|, cos(β) = y/|v|, cos(γ) = z/|v|
Direction angles: α = cos^(-1)(x/|v|), β = cos^(-1)(y/|v|), γ = cos^(-1)(z/|v|)

Theorems

Trigonometric identities for cosines
Pythagorean theorem in 3D space

Suitable Grade Level

Grades 11-12 or early college