## Bezier Curve-

Bezier Curve may be defined as-

• Bezier Curve is parametric curve defined by a set of control points.
• Two points are ends of the curve.
• Other points determine the shape of the curve.

The concept of bezier curves was given by Pierre Bezier.

## Bezier Curve Example-

The following curve is an example of a bezier curve- Here,

• This bezier curve is defined by a set of control points b0, b1, b2 and b3.
• Points b0 and b3 are ends of the curve.
• Points b1 and b2 determine the shape of the curve.

## Bezier Curve Properties-

Few important properties of a bezier curve are-

### Property-01:

Bezier curve is always contained within a polygon called as convex hull of its control points. ### Property-02:

• Bezier curve generally follows the shape of its defining polygon.
• The first and last points of the curve are coincident with the first and last points of the defining polygon.

### Property-03:

The degree of the polynomial defining the curve segment is one less than the total number of control points.

Degree = Number of Control Points – 1

### Property-04:

The order of the polynomial defining the curve segment is equal to the total number of control points.

Order = Number of Control Points

### Property-05:

• Bezier curve exhibits the variation diminishing property.
• It means the curve do not oscillate about any straight line more often than the defining polygon.

## Bezier Curve Equation-

A bezier curve is parametrically represented by- Here,

• t is any parameter where 0 <= t <= 1
• P(t) = Any point lying on the bezier curve
• Bi = ith control point of the bezier curve
• n = degree of the curve
• Jn,i(t) = Blending function = C(n,i)ti(1-t)n-i where C(n,i) = n! / i!(n-i)!

## Cubic Bezier Curve-

• Cubic bezier curve is a bezier curve with degree 3.
• The total number of control points in a cubic bezier curve is 4.

### Example-

The following curve is an example of a cubic bezier curve- Here,

• This curve is defined by 4 control points b0, b1, b2 and b3.
• The degree of this curve is 3.
• So, it is a cubic bezier curve.

### Cubic Bezier Curve Equation-

The parametric equation of a bezier curve is- Substituting n = 3 for a cubic bezier curve, we get- Expanding the above equation, we get-

P (t) = B0J3,0(t) + B1J3,1(t) + B2J3,2(t) + B3J3,3(t) ………..(1)

Now, Using (2), (3), (4) and (5) in (1), we get-

P(t) = B0(1-t)3 + B13t(1-t)2 + B23t2(1-t) + B3t3

This is the required parametric equation for a cubic bezier curve.

## Applications of Bezier Curves-

Bezier curves have their applications in the following fields-

### 1. Computer Graphics-

• Bezier curves are widely used in computer graphics to model smooth curves.
• The curve is completely contained in the convex hull of its control points.
• So, the points can be graphically displayed & used to manipulate the curve intuitively.

### 2. Animation-

• Bezier curves are used to outline movement in animation applications such as Adobe Flash and synfig.
• Users outline the wanted path in bezier curves.
• The application creates the needed frames for the object to move along the path.
• For 3D animation, bezier curves are often used to define 3D paths as well as 2D curves.

### 3. Fonts-

• True type fonts use composite bezier curves composed of quadratic bezier curves.
• Modern imaging systems like postscript, asymptote etc use composite bezier curves composed of cubic bezier curves for drawing curved shapes.

## Problem-01:

Given a bezier curve with 4 control points-

B0[1 0] , B1[3 3] , B2[6 3] , B3[8 1]

Determine any 5 points lying on the curve. Also, draw a rough sketch of the curve.

## Solution-

We have-

• The given curve is defined by 4 control points.
• So, the given curve is a cubic bezier curve.

The parametric equation for a cubic bezier curve is-

P(t) = B0(1-t)3 + B13t(1-t)2 + B23t2(1-t) + B3t3

Substituting the control points B0, B1, B2 and B3, we get-

P(t) = [1 0](1-t)3 + [3 3]3t(1-t)2 + [6 3]3t2(1-t) + [8 1]t3 ……..(1)

Now,

To get 5 points lying on the curve, assume any 5 values of t lying in the range 0 <= t <= 1.

Let 5 values of t are 0, 0.2, 0.5, 0.7, 1

### For t = 0:

Substituting t=0 in (1), we get-

P(0) = [1 0](1-0)3 + [3 3]3(0)(1-t)2 + [6 3]3(0)2(1-0) + [8 1](0)3

P(0) = [1 0] + 0 + 0 + 0

P(0) = [1 0]

### For t = 0.2:

Substituting t=0.2 in (1), we get-

P(0.2) = [1 0](1-0.2)3 + [3 3]3(0.2)(1-0.2)2 + [6 3]3(0.2)2(1-0.2) + [8 1](0.2)3

P(0.2) = [1 0](0.8)3 + [3 3]3(0.2)(0.8)2 + [6 3]3(0.2)2(0.8) + [8 1](0.2)3

P(0.2) = [1 0] x 0.512 + [3 3] x 3 x 0.2 x 0.64 + [6 3] x 3 x 0.04 x 0.8 + [8 1] x 0.008

P(0.2) = [1 0] x 0.512 + [3 3] x 0.384 + [6 3] x 0.096 + [8 1] x 0.008

P(0.2) = [0.512 0] + [1.152 1.152] + [0.576 0.288] + [0.064 0.008]

P(0.2) = [2.304 1.448]

### For t = 0.5:

Substituting t=0.5 in (1), we get-

P(0.5) = [1 0](1-0.5)3 + [3 3]3(0.5)(1-0.5)2 + [6 3]3(0.5)2(1-0.5) + [8 1](0.5)3

P(0.5) = [1 0](0.5)3 + [3 3]3(0.5)(0.5)2 + [6 3]3(0.5)2(0.5) + [8 1](0.5)3

P(0.5) = [1 0] x 0.125 + [3 3] x 3 x 0.5 x 0.25 + [6 3] x 3 x 0.25 x 0.5 + [8 1] x 0.125

P(0.5) = [1 0] x 0.125 + [3 3] x 0.375 + [6 3] x 0.375 + [8 1] x 0.125

P(0.5) = [0.125 0] + [1.125 1.125] + [2.25 1.125] + [1 0.125]

P(0.5) = [4.5 2.375]

### For t = 0.7:

Substituting t=0.7 in (1), we get-

P(t) = [1 0](1-t)3 + [3 3]3t(1-t)2 + [6 3]3t2(1-t) + [8 1]t3

P(0.7) = [1 0](1-0.7)3 + [3 3]3(0.7)(1-0.7)2 + [6 3]3(0.7)2(1-0.7) + [8 1](0.7)3

P(0.7) = [1 0](0.3)3 + [3 3]3(0.7)(0.3)2 + [6 3]3(0.7)2(0.3) + [8 1](0.7)3

P(0.7) = [1 0] x 0.027 + [3 3] x 3 x 0.7 x 0.09 + [6 3] x 3 x 0.49 x 0.3 + [8 1] x 0.343

P(0.7) = [1 0] x 0.027 + [3 3] x 0.189 + [6 3] x 0.441 + [8 1] x 0.343

P(0.7) = [0.027 0] + [0.567 0.567] + [2.646 1.323] + [2.744 0.343]

P(0.7) = [5.984 2.233]

### For t = 1:

Substituting t=1 in (1), we get-

P(1) = [1 0](1-1)3 + [3 3]3(1)(1-1)2 + [6 3]3(1)2(1-1) + [8 1](1)3

P(1) = [1 0] x 0 + [3 3] x 3 x 1 x 0 + [6 3] x 3 x 1 x 0 + [8 1] x 1

P(1) = 0 + 0 + 0 + [8 1]

P(1) = [8 1]

Following is the required rough sketch of the curve- To gain better understanding about Bezier Curves in Computer Graphics,

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## 3D Transformations in Computer Graphics-

We have discussed-

• Transformation is a process of modifying and re-positioning the existing graphics.
• 3D Transformations take place in a three dimensional plane.

In computer graphics, various transformation techniques are- ## 3D Shearing in Computer Graphics-

 In Computer graphics, 3D Shearing is an ideal technique to change the shape of an existing object in a three dimensional plane.

In a three dimensional plane, the object size can be changed along X direction, Y direction as well as Z direction.

So, there are three versions of shearing- 1. Shearing in X direction
2. Shearing in Y direction
3. Shearing in Z direction

Consider a point object O has to be sheared in a 3D plane.

Let-

• Initial coordinates of the object O = (Xold, Yold, Zold)
• Shearing parameter towards X direction = Shx
• Shearing parameter towards Y direction = Shy
• Shearing parameter towards Z direction = Shz
• New coordinates of the object O after shearing = (Xnew, Ynew, Znew)

### Shearing in X Axis-

Shearing in X axis is achieved by using the following shearing equations-

• Xnew = Xold
• Ynew = Yold + Shy x Xold
• Znew = Zold + Shz x Xold

In Matrix form, the above shearing equations may be represented as- ### Shearing in Y Axis-

Shearing in Y axis is achieved by using the following shearing equations-

• Xnew = Xold + Shx x Yold
• Ynew = Yold
• Znew = Zold + Shz x Yold

In Matrix form, the above shearing equations may be represented as- ### Shearing in Z Axis-

Shearing in Z axis is achieved by using the following shearing equations-

• Xnew = Xold + Shx x Zold
• Ynew = Yold + Shy x Zold
• Znew = Zold

In Matrix form, the above shearing equations may be represented as- ## Problem-01:

Given a 3D triangle with points (0, 0, 0), (1, 1, 2) and (1, 1, 3). Apply shear parameter 2 on X axis, 2 on Y axis and 3 on Z axis and find out the new coordinates of the object.

## Solution-

Given-

• Old corner coordinates of the triangle = A (0, 0, 0), B(1, 1, 2), C(1, 1, 3)
• Shearing parameter towards X direction (Shx) = 2
• Shearing parameter towards Y direction (Shy) = 2
• Shearing parameter towards Y direction (Shz) = 3

## Shearing in X Axis-

### For Coordinates A(0, 0, 0)

Let the new coordinates of corner A after shearing = (Xnew, Ynew, Znew).

Applying the shearing equations, we have-

• Xnew = Xold = 0
• Ynew = Yold + Shy x Xold = 0 + 2 x 0 = 0
• Znew = Zold + Shz x Xold = 0 + 3 x 0 = 0

Thus, New coordinates of corner A after shearing = (0, 0, 0).

### For Coordinates B(1, 1, 2)

Let the new coordinates of corner B after shearing = (Xnew, Ynew, Znew).

Applying the shearing equations, we have-

• Xnew = Xold = 1
• Ynew = Yold + Shy x Xold = 1 + 2 x 1 = 3
• Znew = Zold + Shz x Xold = 2 + 3 x 1 = 5

Thus, New coordinates of corner B after shearing = (1, 3, 5).

### For Coordinates C(1, 1, 3)

Let the new coordinates of corner C after shearing = (Xnew, Ynew, Znew).

Applying the shearing equations, we have-

• Xnew = Xold = 1
• Ynew = Yold + Shy x Xold = 1 + 2 x 1 = 3
• Znew = Zold + Shz x Xold = 3 + 3 x 1 = 6

Thus, New coordinates of corner C after shearing = (1, 3, 6).

Thus, New coordinates of the triangle after shearing in X axis = A (0, 0, 0), B(1, 3, 5), C(1, 3, 6).

## Shearing in Y Axis-

### For Coordinates A(0, 0, 0)

Let the new coordinates of corner A after shearing = (Xnew, Ynew, Znew).

Applying the shearing equations, we have-

• Xnew = Xold + Shx x Yold = 0 + 2 x 0 = 0
• Ynew = Yold = 0
• Znew = Zold + Shz x Yold = 0 + 3 x 0 = 0

Thus, New coordinates of corner A after shearing = (0, 0, 0).

### For Coordinates B(1, 1, 2)

Let the new coordinates of corner B after shearing = (Xnew, Ynew, Znew).

Applying the shearing equations, we have-

• Xnew = Xold + Shx x Yold = 1 + 2 x 1 = 3
• Ynew = Yold = 1
• Znew = Zold + Shz x Yold = 2 + 3 x 1 = 5

Thus, New coordinates of corner B after shearing = (3, 1, 5).

### For Coordinates C(1, 1, 3)

Let the new coordinates of corner C after shearing = (Xnew, Ynew, Znew).

Applying the shearing equations, we have-

• Xnew = Xold + Shx x Yold = 1 + 2 x 1 = 3
• Ynew = Yold = 1
• Znew = Zold + Shz x Yold = 3 + 3 x 1 = 6

Thus, New coordinates of corner C after shearing = (3, 1, 6).

Thus, New coordinates of the triangle after shearing in Y axis = A (0, 0, 0), B(3, 1, 5), C(3, 1, 6).

## Shearing in Z Axis-

### For Coordinates A(0, 0, 0)

Let the new coordinates of corner A after shearing = (Xnew, Ynew, Znew).

Applying the shearing equations, we have-

• Xnew = Xold + Shx x Zold = 0 + 2 x 0 = 0
• Ynew = Yold + Shy x Zold = 0 + 2 x 0 = 0
• Znew = Zold = 0

Thus, New coordinates of corner A after shearing = (0, 0, 0).

### For Coordinates B(1, 1, 2)

Let the new coordinates of corner B after shearing = (Xnew, Ynew, Znew).

Applying the shearing equations, we have-

• Xnew = Xold + Shx x Zold = 1 + 2 x 2 = 5
• Ynew = Yold + Shy x Zold = 1 + 2 x 2 = 5
• Znew = Zold = 2

Thus, New coordinates of corner B after shearing = (5, 5, 2).

### For Coordinates C(1, 1, 3)

Let the new coordinates of corner C after shearing = (Xnew, Ynew, Znew).

Applying the shearing equations, we have-

• Xnew = Xold + Shx x Zold = 1 + 2 x 3 = 7
• Ynew = Yold + Shy x Zold = 1 + 2 x 3 = 7
• Znew = Zold = 3

Thus, New coordinates of corner C after shearing = (7, 7, 3).

Thus, New coordinates of the triangle after shearing in Z axis = A (0, 0, 0), B(5, 5, 2), C(7, 7, 3).

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## 3D Transformations in Computer Graphics-

We have discussed-

• Transformation is a process of modifying and re-positioning the existing graphics.
• 3D Transformations take place in a three dimensional plane.

In computer graphics, various transformation techniques are- ## 3D Reflection in Computer Graphics-

• Reflection is a kind of rotation where the angle of rotation is 180 degree.
• The reflected object is always formed on the other side of mirror.
• The size of reflected object is same as the size of original object.

Consider a point object O has to be reflected in a 3D plane.

Let-

• Initial coordinates of the object O = (Xold, Yold, Zold)
• New coordinates of the reflected object O after reflection = (Xnew, Ynew,Znew)

In 3 dimensions, there are 3 possible types of reflection- • Reflection relative to XY plane
• Reflection relative to YZ plane
• Reflection relative to XZ plane

### Reflection Relative to XY Plane:

This reflection is achieved by using the following reflection equations-

• Xnew = Xold
• Ynew = Yold
• Znew = -Zold

In Matrix form, the above reflection equations may be represented as- ### Reflection Relative to YZ Plane:

This reflection is achieved by using the following reflection equations-

• Xnew = -Xold
• Ynew = Yold
• Znew = Zold

In Matrix form, the above reflection equations may be represented as- ### Reflection Relative to XZ Plane:

This reflection is achieved by using the following reflection equations-

• Xnew = Xold
• Ynew = -Yold
• Znew = Zold

In Matrix form, the above reflection equations may be represented as- ## Problem-01:

Given a 3D triangle with coordinate points A(3, 4, 1), B(6, 4, 2), C(5, 6, 3). Apply the reflection on the XY plane and find out the new coordinates of the object.

## Solution-

Given-

• Old corner coordinates of the triangle = A (3, 4, 1), B(6, 4, 2), C(5, 6, 3)
• Reflection has to be taken on the XY plane

### For Coordinates A(3, 4, 1)

Let the new coordinates of corner A after reflection = (Xnew, Ynew, Znew).

Applying the reflection equations, we have-

• Xnew = Xold = 3
• Ynew = Yold = 4
• Znew = -Zold = -1

Thus, New coordinates of corner A after reflection = (3, 4, -1).

### For Coordinates B(6, 4, 2)

Let the new coordinates of corner B after reflection = (Xnew, Ynew, Znew).

Applying the reflection equations, we have-

• Xnew = Xold = 6
• Ynew = Yold = 4
• Znew = -Zold = -2

Thus, New coordinates of corner B after reflection = (6, 4, -2).

### For Coordinates C(5, 6, 3)

Let the new coordinates of corner C after reflection = (Xnew, Ynew, Znew).

Applying the reflection equations, we have-

• Xnew = Xold = 5
• Ynew = Yold = 6
• Znew = -Zold = -3

Thus, New coordinates of corner C after reflection = (5, 6, -3).

Thus, New coordinates of the triangle after reflection = A (3, 4, -1), B(6, 4, -2), C(5, 6, -3).

## Problem-02:

Given a 3D triangle with coordinate points A(3, 4, 1), B(6, 4, 2), C(5, 6, 3). Apply the reflection on the XZ plane and find out the new coordinates of the object.

## Solution-

Given-

• Old corner coordinates of the triangle = A (3, 4, 1), B(6, 4, 2), C(5, 6, 3)
• Reflection has to be taken on the XZ plane

### For Coordinates A(3, 4, 1)

Let the new coordinates of corner A after reflection = (Xnew, Ynew, Znew).

Applying the reflection equations, we have-

• Xnew = Xold = 3
• Ynew = -Yold = -4
• Znew = Zold = 1

Thus, New coordinates of corner A after reflection = (3, -4, 1).

### For Coordinates B(6, 4, 2)

Let the new coordinates of corner B after reflection = (Xnew, Ynew, Znew).

Applying the reflection equations, we have-

• Xnew = Xold = 6
• Ynew = -Yold = -4
• Znew = Zold = 2

Thus, New coordinates of corner B after reflection = (6, -4, 2).

### For Coordinates C(5, 6, 3)

Let the new coordinates of corner C after reflection = (Xnew, Ynew, Znew).

Applying the reflection equations, we have-

• Xnew = Xold = 5
• Ynew = -Yold = -6
• Znew = Zold = 3

Thus, New coordinates of corner C after reflection = (5, -6, 3).

Thus, New coordinates of the triangle after reflection = A (3, -4, 1), B(6, -4, 2), C(5, -6, 3).

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## 3D Transformations in Computer Graphics-

We have discussed-

• Transformation is a process of modifying and re-positioning the existing graphics.
• 3D Transformations take place in a three dimensional plane.

In computer graphics, various transformation techniques are- ## 3D Scaling in Computer Graphics-

 In computer graphics, scaling is a process of modifying or altering the size of objects.

• Scaling may be used to increase or reduce the size of object.
• Scaling subjects the coordinate points of the original object to change.
• Scaling factor determines whether the object size is to be increased or reduced.
• If scaling factor > 1, then the object size is increased.
• If scaling factor < 1, then the object size is reduced.

Consider a point object O has to be scaled in a 3D plane.

Let-

• Initial coordinates of the object O = (Xold, Yold,Zold)
• Scaling factor for X-axis = Sx
• Scaling factor for Y-axis = Sy
• Scaling factor for Z-axis = Sz
• New coordinates of the object O after scaling = (Xnew, Ynew, Znew)

This scaling is achieved by using the following scaling equations-

• Xnew = Xold x Sx
• Ynew = Yold x Sy
• Znew = Zold x Sz

In Matrix form, the above scaling equations may be represented as- ## Problem-01:

Given a 3D object with coordinate points A(0, 3, 3), B(3, 3, 6), C(3, 0, 1), D(0, 0, 0). Apply the scaling parameter 2 towards X axis, 3 towards Y axis and 3 towards Z axis and obtain the new coordinates of the object.

## Solution-

Given-

• Old coordinates of the object = A (0, 3, 3), B(3, 3, 6), C(3, 0, 1), D(0, 0, 0)
• Scaling factor along X axis = 2
• Scaling factor along Y axis = 3
• Scaling factor along Z axis = 3

### For Coordinates A(0, 3, 3)

Let the new coordinates of A after scaling = (Xnew, Ynew, Znew).

Applying the scaling equations, we have-

• Xnew = Xold x Sx = 0 x 2 = 0
• Ynew = Yold x Sy = 3 x 3 = 9
• Znew = Zold x Sz = 3 x 3 = 9

Thus, New coordinates of corner A after scaling = (0, 9, 9).

### For Coordinates B(3, 3, 6)

Let the new coordinates of B after scaling = (Xnew, Ynew, Znew).

Applying the scaling equations, we have-

• Xnew = Xold x Sx = 3 x 2 = 6
• Ynew = Yold x Sy = 3 x 3 = 9
• Znew = Zold x Sz = 6 x 3 = 18

Thus, New coordinates of corner B after scaling = (6, 9, 18).

### For Coordinates C(3, 0, 1)

Let the new coordinates of C after scaling = (Xnew, Ynew, Znew).

Applying the scaling equations, we have-

• Xnew = Xold x Sx = 3 x 2 = 6
• Ynew = Yold x Sy = 0 x 3 = 0
• Znew = Zold x Sz = 1 x 3 = 3

Thus, New coordinates of corner C after scaling = (6, 0, 3).

### For Coordinates D(0, 0, 0)

Let the new coordinates of D after scaling = (Xnew, Ynew, Znew).

Applying the scaling equations, we have-

• Xnew = Xold x Sx = 0 x 2 = 0
• Ynew = Yold x Sy = 0 x 3 = 0
• Znew = Zold x Sz = 0 x 3 = 0

Thus, New coordinates of corner D after scaling = (0, 0, 0).

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## 3D Transformations in Computer Graphics-

We have discussed-

• Transformation is a process of modifying and re-positioning the existing graphics.
• 3D Transformations take place in a three dimensional plane.

In computer graphics, various transformation techniques are- ## 3D Rotation in Computer Graphics-

 In Computer graphics, 3D Rotation is a process of rotating an object with respect to an angle in a three dimensional plane.

Consider a point object O has to be rotated from one angle to another in a 3D plane.

Let-

• Initial coordinates of the object O = (Xold, Yold, Zold)
• Initial angle of the object O with respect to origin = Φ
• Rotation angle = θ
• New coordinates of the object O after rotation = (Xnew, Ynew, Znew)

In 3 dimensions, there are 3 possible types of rotation-

• X-axis Rotation
• Y-axis Rotation
• Z-axis Rotation

### For X-Axis Rotation-

This rotation is achieved by using the following rotation equations-

• Xnew = Xold
• Ynew = Yold x cosθ – Zold x sinθ
• Znew = Yold x sinθ + Zold x cosθ

In Matrix form, the above rotation equations may be represented as- ### For Y-Axis Rotation-

This rotation is achieved by using the following rotation equations-

• Xnew = Zold x sinθ + Xold x cosθ
• Ynew = Yold
• Znew = Yold x cosθ – Xold x sinθ

In Matrix form, the above rotation equations may be represented as- ### For Z-Axis Rotation-

This rotation is achieved by using the following rotation equations-

• Xnew = Xold x cosθ – Yold x sinθ
• Ynew = Xold x sinθ + Yold x cosθ
• Znew = Zold

In Matrix form, the above rotation equations may be represented as- ## Problem-01:

Given a homogeneous point (1, 2, 3). Apply rotation 90 degree towards X, Y and Z axis and find out the new coordinate points.

## Solution-

Given-

• Old coordinates = (Xold, Yold, Zold) = (1, 2, 3)
• Rotation angle = θ = 90º

### For X-Axis Rotation-

Let the new coordinates after rotation = (Xnew, Ynew, Znew).

Applying the rotation equations, we have-

• Xnew = Xold = 1
• Ynew = Yold x cosθ – Zold x sinθ = 2 x cos90° – 3 x sin90° = 2 x 0 – 3 x 1 = -3
• Znew = Yold x sinθ + Zold x cosθ = 2 x sin90° + 3 x cos90° = 2 x 1 + 3 x 0 = 2

Thus, New coordinates after rotation = (1, -3, 2).

### For Y-Axis Rotation-

Let the new coordinates after rotation = (Xnew, Ynew, Znew).

Applying the rotation equations, we have-

• Xnew = Zold x sinθ + Xold x cosθ = 3 x sin90° + 1 x cos90° = 3 x 1 + 1 x 0 = 3
• Ynew = Yold = 2
• Znew = Yold x cosθ – Xold x sinθ = 2 x cos90° – 1 x sin90° = 2 x 0 – 1 x 1 = -1

Thus, New coordinates after rotation = (3, 2, -1).

### For Z-Axis Rotation-

Let the new coordinates after rotation = (Xnew, Ynew, Znew).

Applying the rotation equations, we have-

• Xnew = Xold x cosθ – Yold x sinθ = 1 x cos90° – 2 x sin90° = 1 x 0 – 2 x 1 = -2
• Ynew = Xold x sinθ + Yold x cosθ = 1 x sin90° + 2 x cos90° = 1 x 1 + 2 x 0 = 1
• Znew = Zold = 3

Thus, New coordinates after rotation = (-2, 1, 3).

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