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```And, solving for v from Equation 3.6c leads to

v = − (u + t)

which, when substituted into the above expression for u’ yields

u' = v + 2u = − u − t + 2u = u − t

In solving for an expression for v’, we begin with the one of the above expressions for this parameter—i.e.,

v' = 2u' − 3u

Now, substitution of the above expression for u’ into this equation leads to

vÕ= 2uÕ− 3u = (2)(u − t) − 3u = − u −2t

And solving for u from Equation 3.6c gives

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3-43
u = − v − t

which, when substituted in the previous equation results in the following expression for v’

vÕ= − u −2t = − (− v − t) − 2t = v − t

And, of course from Equation 3.6d

w’ = w
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3-44
Crystallographic Planes

3.37 (a) We are asked to draw a (021 ) plane within an orthorhombic unit cell. First remove the three
indices from the parentheses, and take their reciprocals--i.e., ∞, 1/2, and -1. This means that the plane parallels the
x-axis, intersects the y-axis at b/2, and intersects the z-axis at -c. The plane that satisfies these requirements has
been drawn within the orthorhombic unit cell below. (For orthorhombic, a ≠ b ≠ c, and α = β = γ = 90°.)

(b) A (200) plane is drawn within the monoclinic cell shown below. We first remove the parentheses and
take the reciprocals of the indices; this gives 1/2, ∞, and ∞,. Thus, the (200) plane parallels both y- and z-axes, and
intercepts the x-axis at a/2, as indicated in the drawing. (For monoclinic, a ≠ b ≠ c, and α = γ = 90° ≠ β.)

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3-45
3.38 This problem calls for specification of the indices for the two planes that are drawn in the sketch.
Plane 1 is a (211) plane. The determination of its indices is summarized below.

x y z
Intercepts a/2 b c
Intercepts in terms of a, b, and c 1/2 1 1
Reciprocals of intercepts 2 1 1
Enclosure (211)

Plane 2 is a (02 0) plane, as summarized below.

x y z

Intercepts ∞a -b/2 ∞c
Intercepts in terms of a, b, and c ∞ -1/2 ∞
Reciprocals of intercepts 0 -2 0
Enclosure (02 0)
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3-46
3.39 The planes called for are plotted in the cubic unit cells shown below.

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3-47
3.40 For plane A we will leave the origin at the unit cell as shown. If we extend this plane back into the
plane of the page, then it is a (111 ) plane, as summarized below.

x y z
Intercepts a b – c
Intercepts in terms of a, b, and c 1 1 – 1
Reciprocals of intercepts 1 1 – 1
Reduction not necessary
Enclosure (111 )

[Note: If we move the origin one unit cell distance parallel to the x axis and then one unit cell distance parallel to
the y axis, the direction becomes (1 1 1) ].

For plane B we will leave the origin of the unit cell as shown; this is a (230) plane, as summarized below.

x y z
Intercepts

a
2

b
3
∞c
Intercepts in terms of a, b, and c

1
2

1
3
∞
Reciprocals of intercepts 2 3 0
Enclosure (230)
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
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3-48
3.41 For plane A we will move the origin of the coordinate system one unit cell distance to the right along
the y axis; thus, this is a (11 0) plane, as summarized below.

x y z
Intercepts

a
2
–

b
2
∞ c
Intercepts in terms of a, b, and c

1
2
–

1
2
∞
Reciprocals of intercepts 2 – 2 0
Reduction 1 – 1 0
Enclosure (11 0)

For plane B we will leave the origin of the unit cell as shown; thus, this is a (122) plane, as summarized
below.

x y z
Intercepts a

b
2

c
2

Intercepts in terms of a, b, and c 1

1
2

1
2

Reciprocals of intercepts 1 2 2
Reduction not necessary
Enclosure (122)
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
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3-41

t = − (u+ v) = − 4
3
−
2
3
⎛
⎝
⎜
⎞
⎠
⎟ = −
2
3

w = w' = 1

Now, in order to get the lowest set of integers, it is necessary to multiply all indices by the factor 3, with the result
that the direction D is a [42 2 3] direction.
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
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3-49
3.42 For plane A since the plane passes through the origin of the coordinate system as shown, we will
move the origin of the coordinate system one unit cell distance vertically along the z axis; thus, this is a (211 )
plane, as summarized below.

x y z
Intercepts

a
2
b – c
Intercepts in terms of a, b, and c

1
2
1 – 1
Reciprocals of intercepts 2 1 – 1
Reduction not necessary
Enclosure (211 )

For plane B, since the plane passes through the origin of the coordinate system as shown, we will move the
origin one unit cell distance vertically along the z axis; this is a (021 ) plane, as summarized below.

x y z
Intercepts ∞ a```