Dislocation Angle Calculation

To find the angle θ between the dislocation line direction u and the Burger's vector B, we utilize the vector dot product formula. The dot product relates the angle between two vectors to their components and magnitudes.

The formula is: $ \vec{B} \cdot \vec{u} = |\vec{B}| |\vec{u}| \cos(\theta) $

We are given:

• Burger's vector direction: Associated vector B' = $\langle 1, 1, 0 \rangle$. (The scalar $\frac{a}{2}$ does not affect the angle).
• Dislocation line direction: Vector u = $\langle 1, 1, 2 \rangle$.

Step 1: Calculate Dot Product

Compute the dot product of B' and u:

$ \vec{B'} \cdot \vec{u} = (1 \times 1) + (1 \times 1) + (0 \times 2) = 1 + 1 + 0 = 2 $

Step 2: Calculate Vector Magnitudes

Find the magnitude of each direction vector:

• Magnitude of B': $ |\vec{B'}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{1 + 1 + 0} = \sqrt{2} $
• Magnitude of u: $ |\vec{u}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6} $

Step 3: Solve for cos(θ)

Rearrange the dot product formula to solve for cos(θ):

$ \cos(\theta) = \frac{\vec{B'} \cdot \vec{u}}{|\vec{B'}| |\vec{u}|} = \frac{2}{\sqrt{2} \times \sqrt{6}} = \frac{2}{\sqrt{12}} $

Simplify the expression:

$ \cos(\theta) = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} $

Step 4: Find the Angle θ

Calculate the angle using the inverse cosine function:

$ \theta = \arccos\left(\frac{1}{\sqrt{3}}\right) $

Evaluating numerically:

$ \theta \approx \arccos(0.57735) \approx 54.74 \text{ degrees} $

The calculated angle of approximately 54.74° falls within the range of 54° to 55.5°.