Math Problem Statement

u = -4j + 6k v = -3i + j + 5k find the angle between them

Solution

To find the angle θ\theta between two vectors u\mathbf{u} and v\mathbf{v}, we use the following formula:

cosθ=uvuv\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|}

Where:

  • uv\mathbf{u} \cdot \mathbf{v} is the dot product of the two vectors,
  • u\|\mathbf{u}\| is the magnitude of u\mathbf{u},
  • v\|\mathbf{v}\| is the magnitude of v\mathbf{v}.

Given: u=4j+6k\mathbf{u} = -4j + 6k v=3i+j+5k\mathbf{v} = -3i + j + 5k

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

The dot product of two vectors is given by: uv=(uivi)+(ujvj)+(ukvk)\mathbf{u} \cdot \mathbf{v} = (u_i \cdot v_i) + (u_j \cdot v_j) + (u_k \cdot v_k)

Since u=0i4j+6k\mathbf{u} = 0i - 4j + 6k and v=3i+j+5k\mathbf{v} = -3i + j + 5k, their dot product is: uv=(03)+(41)+(65)\mathbf{u} \cdot \mathbf{v} = (0 \cdot -3) + (-4 \cdot 1) + (6 \cdot 5) uv=04+30=26\mathbf{u} \cdot \mathbf{v} = 0 - 4 + 30 = 26

Step 2: Compute the magnitudes of u\mathbf{u} and v\mathbf{v}

The magnitude of a vector a=aii+ajj+akk\mathbf{a} = a_i i + a_j j + a_k k is: a=ai2+aj2+ak2\|\mathbf{a}\| = \sqrt{a_i^2 + a_j^2 + a_k^2}

For u=0i4j+6k\mathbf{u} = 0i - 4j + 6k: u=02+(4)2+62=16+36=52\|\mathbf{u}\| = \sqrt{0^2 + (-4)^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52}

For v=3i+j+5k\mathbf{v} = -3i + j + 5k: v=(3)2+12+52=9+1+25=35\|\mathbf{v}\| = \sqrt{(-3)^2 + 1^2 + 5^2} = \sqrt{9 + 1 + 25} = \sqrt{35}

Step 3: Compute the cosine of the angle

Now, substitute the values into the formula: cosθ=265235=261820\cos \theta = \frac{26}{\sqrt{52} \cdot \sqrt{35}} = \frac{26}{\sqrt{1820}}

cosθ=2642.680.609\cos \theta = \frac{26}{42.68} \approx 0.609

Step 4: Find the angle θ\theta

To find θ\theta, take the inverse cosine (arccos): θ=cos1(0.609)52.7\theta = \cos^{-1}(0.609) \approx 52.7^\circ

Thus, the angle between the two vectors is approximately 52.752.7^\circ.

Would you like more details or have any questions?


Here are 5 related questions:

  1. How is the dot product of two vectors used in physics?
  2. What is the significance of the magnitude of a vector?
  3. How do you calculate the angle between vectors in higher dimensions?
  4. What happens if the dot product of two vectors is zero?
  5. How does the cross product differ from the dot product?

Tip: The dot product gives the projection of one vector onto another, helping you find angles between vectors.

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Math Problem Analysis

Mathematical Concepts

Vector Algebra
Dot Product
Angle Between Vectors

Formulas

cos(θ) = (u · v) / (||u|| ||v||)
Dot product: u · v = (u_i * v_i) + (u_j * v_j) + (u_k * v_k)
Magnitude: ||a|| = √(a_i^2 + a_j^2 + a_k^2)

Theorems

Cosine of the angle between two vectors

Suitable Grade Level

Grades 11-12