Find the image of (12,3√2)(12,32) under the rotation Rπ/6𝑅𝜋/6 through an anticlockwise angle of π6𝜋6 in R2

To find the image of a point under a rotation in R2, we can use the rotation matrix. The rotation matrix for a counterclockwise rotation of θ is given by:

[cos(θ) -sin(θ)]
[sin(θ)  cos(θ)]

Here, θ = π/6. So, the rotation matrix is:

[cos(π/6) -sin(π/6)]
[sin(π/6)  cos(π/6)]

This simplifies to:

[√3/2 -1/2]
[1/2   √3/2]

The given point is (12, 3√2). We can represent this as a column vector:

[12]
[3√2]

To find the image of the point, we multiply the rotation matrix by the column vector. This gives us:

[√3/2 -1/2] * [12]  =  [√3/2*12 - 1/2*3√2]
[1/2   √3/2]   [3√2]     [1/2*12 + √3/2*3√2]

Solving this gives us:

[6√3 - 3√2]
[6 + 3√6]

So, the image of the point (12, 3√2) under the rotation Rπ/6 through an anticlockwise angle of π/6 in R2 is (6√3 - 3√2, 6 + 3√6).