MSE 331 Midterm-I Excercise -KEYS(2018).pdf

# It is 70 degree so it is consistent to what you

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those two poles can be measured directly. It is ~70 degree, so it is consistent to what you computed in (2), i.e., 70.53˚ (6). Find the crystallographic directions B to E. OE=(1/2,1/2,1), OB=(1,1,0) BE=OE-OB=(1/2, 1/2, 1)-(1,1,0)=(-1/2,-1/2,1) BE= ] 2 1 1 [ (7) Find the angle between AC and EO By symmetry, the angle=90 o

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Page. 3 (8). What is the zone axis for the planes ABO and ABC? (-111) (-100) (010) (00-1) (001) (100) (-11-1) (11-1) (111) (-110) (110) (01-1) (011) (-101) (-10-1) (10-1) (101) (ABO) (ACO) (ABC) Traces from ABO and ABC intercept at (01-1) pole. So the zone axis is [01-1] Verify: [01-1]•(-1 1 1)=0 [01-1]•(1 1 1)=0 [4]. For Hexagonal system shown; (1) Find the Miller index for the plane PQR (4 -1 -3 1) (2) Index the direction from P to R. Shift P°R to origin and the projection of this new vector is exactly the same as a 2 direction, therefore, [-1 2 -1 0]. Then considering the z-direction, therefore [-1 2 -1 3] a 1 a 2 a 3 C P Q R
Page. 4 Or you may do the following way Vector PR = PO + OR PO = 1/3[-1 2- 1 0] OR = [ 0 0 0 1] (+)____________________ [-1/3 2/3 -1/3 1] [ -1 2 -1 3] (3) Index the direction from P to Q. Shift P°Q to a parallel position O°R, therefore we can see the new direction OR, and OR is the vector addition of ½ OS and a2, while OS is the vector addition of a2 and –a3. OS = a2 – a3 = [-1 2 -1 0] – [1 1 -2 0] = [0 3 -3 0] ½ OS + a2 = OR = PQ = [0, 3/2, -3/2, 0] + [-1 2 -1 0] = [-1, 7/2, -5/2, 0] = [-2, 7, -5, 0]
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