Find a single matrix that defines a rotation of the plane through an angle of π/6 about the origin, while at the same time moves points to five times their original distance from the origin.
To find a single matrix that defines a rotation of the plane through an angle of π/6 about the origin while also scaling points to five times their original distance from the origin, we can use a composition of two transformation matrices: a rotation matrix and a scaling matrix.
Let's break down the process step by step:
1. Start by creating a rotation matrix. A rotation of angle θ about the origin can be represented by the following matrix:
R(θ) = [cos(θ) -sin(θ)]
[sin(θ) cos(θ)]
For our case, θ = π/6, so the rotation matrix becomes:
R(π/6) = [cos(π/6) -sin(π/6)]
[sin(π/6) cos(π/6)]
Simplifying the values:
R(π/6) = [√3/2 -1/2]
[ 1/2 √3/2]
2. Next, create a scaling matrix. Scaling points to five times their original distance from the origin can be achieved by multiplying the original point coordinates by a scaling factor of 5. Since we want the scaling to happen uniformly in all directions, the scaling matrix will be:
S = [5 0]
[0 5]
3. Now, we can find the desired single matrix by multiplying the rotation matrix (R) and the scaling matrix (S) in the appropriate order:
RS = R(π/6) * S
RS = [√3/2 -1/2] * [5 0]
[0 5]
Applying matrix multiplication:
RS = [√3/2 * 5 + (-1/2) * 0 √3/2 * 0 + (-1/2) * 5]
[1/2 * 5 + √3/2 * 0 1/2 * 0 + √3/2 * 5]
Simplifying further:
RS = [5√3/2 -5/2]
[5/2 5√3/2]
Hence, the single matrix that defines a rotation of the plane through an angle of π/6 about the origin, while at the same time moves points to five times their original distance from the origin, is:
RS = [5√3/2 -5/2]
[5/2 5√3/2]