The eight vertices of a rectangular prism are as follows, Find the coordinates of the vertices after the prism is rotated counterclockwise about the z- axis.
Question
The eight vertices of a rectangular prism are as follows:
`\V_1 = \left(0,0,0\right), V_2 = \left(1,0,0\right), V_3 = \left(1,2,0\right), V_4 = \left(0,2,0\right)`
`\V_5 = \left(0,0,3\right), V_6 = \left(1,0,3\right), V_7 = \left(1,2,3\right), V_8 = \left(0,2,3\right)`
Find the coordinates of the vertices after the prism is rotated counterclockwise about the z- axis through `\theta = 60^\circ`.
Solution
Here `\theta = 60`
and
The prism is rotated counterclockwise about the z -axis
then
now we first find `V_1^ - = (x_1^ - ,y_1^ - ,z_1^ - )`
`Here V_1 = \left(0,0,0\right)`
then
`V_1^ - = (0,0,0)`
Now `V_2^ - = = (x_2^ - ,y_2^ - ,z_2^ - )`
here `V_2 = \left(1,0,0\right)`
`x_2 = 1, y_2 = 0, and z_2 = 0`
`x_2^ - = x_2\cos 60 - y_2\sin 60 = (1)\left(\frac{1}{2}\right) -
(0)\left(\frac{\sqrt 3}{2}\right) = \frac{1}{2}`
`y_2^ - = x_2\sin 60 + y_2\cos 60 = (1)\left(\frac{\sqrt 3
}{2}\right) + (0)\left(\frac{1}{2}\right) = \frac{\sqrt 3 }{2}`
`z_2^ - = z_2 = 0`
So,
`V_2^ - = \left(\frac{1}{2},\frac{\sqrt 3 }{2},0\right)`
Now
`V_3^ - = (x_3^ - ,y_3^ - ,z_3^ - )`
here
`V_3 = \left(1,2,0\right)`
`x_3 = 1, y_3 = 2, z_3 = 0`
`x_3^ - = x_3\cos 60 - y_3\sin 60 = (1)\left(\frac{1}{2} \right) -
(2)\left(\frac{\sqrt 3}{2}\right) = \frac{1}{2}- \sqrt 3`
`y_3^ - = x_3\sin60 + y_3\cos 60 = (1)\left(\frac{\sqrt
3}{2}\right) + (2)\left(\frac{1}{2}\right) = \frac{\sqrt 3 }{2} - 1`
`z_3^ - = z_3 = 0`
So,
`V_3^ - = \left(\frac{1}{2} - \sqrt 3 ,\frac{\sqrt 3}{2} -
1,0)`
Now
`V_4^ - = (x_4^ - ,y_4^ - ,z_4^ - )`
`V_4 = \left(0,2,0\right)`
`x_4 = 0, y_4 = 2, z_4 = 0`
`x_4^ - = x_4\cos 60 - y_4\sin 60 = (0)\left(\frac{1}{2}\right) -
(2)\left(\frac{\sqrt 3}{2}\right) = - \sqrt 3`
`y_4^ - = x_4\sin 60 + y_4\cos 60 = (0)\left(\frac{\sqrt 3}{2}
\right) + (2)\left(\frac{1}{2}\right) = - 1`
`z_4^ - = z_4 = 0`
So,
`V_4^ - = \left(- \sqrt 3 , - 1,0\right)`
Now `V_5^ - = (x_5^ - ,y_5^ - ,z_5^ - )`
here
`V_5 = \left(0,0,3\right)`
`x_5 = 0, y_5 = 0, z_5 = 3`
`x_5^ - = x_5\cos 60 - y_5\sin 60 = (0)\left(\frac{1}{2} \right) -
(0)\left(\frac{\sqrt 3}{2}\right) = 0`
`y_5^ - = x_5\sin 60 + y_5\cos 60 = (0)\left(\frac{\sqrt 3}{2}
\right) + (0)\left(\frac{1}{2}\right) = 0`
`z_5^ - = z_5 = 3`
So,
`V_5^ - = \left(0,0,3\right)`
Now
`V_6^ - = (x_6^ - ,y_6^ - ,z_6^ - )`
here `V_6 = \left(1,0,3\right)`
`x_6 = 1, y_6 = 0, z_6 = 3`
`x_6^ - = x_6\cos 60 - y_6\sin 60 = (1)\left(\frac{1}{2} \right) -
(0)\left(\frac{\sqrt 3}{2}\right) = \frac{1}{2}`
`y_6^ - = x_6\sin 60 + y_6\cos 60 = (1)\left(\frac{\sqrt
3}{2}\right) + (0)\left(\frac{1}{2}\right) = \frac{\sqrt 3}{2}`
`z_6^ - = z_6 = 3`
So,
`V_6^ - = \left(\frac{1}{2},\frac{\sqrt 3}{2},3\right)`
Now `V_7^ - = (x_7^ - ,y_7^ - ,z_7^ - )`
here `V_7 = \left(1,2,3\right)`
`x_7 = 1, y_7 = 2, z_7 = 3`
`x_7^ - = x_7\cos 60 - x_7\sin 60 = (1)\left(\frac{1}{2}\right) -
(2)\left(\frac{\sqrt 3}{2} \right) = \frac{1}{2} - \sqrt 3`
`y_7^ - = x_7\sin 60 + x_7\cos 60 = (1)\left(\frac{\sqrt
3}{2}\right) + (2)\left(\frac{1}{2}\right) = \frac{\sqrt 3 }{2} - 1`
`z_7^ - = z_7 = 3`
So,
`V_7^ - = \left(\frac{1}{2} - \sqrt 3 ,\frac{\sqrt 3}{2} - ,3)`
Now `V_8^ - = (x_8^ - ,y_8^ - ,z_8^ - )`
`here V_8 = \left(0,2,3\right)`
`x_8 = 0, y_8 = 2, and z_8 = 3`
`x_8^ - = x_8\cos 60 - y_8\sin 60 = (0)\left(\frac{1}{2}\right) -
(2)\left(\frac{\sqrt 3}{2}\right) = - \sqrt 3`
`y_8^ - = x_8\sin 60 + y_8\cos 60 = (0)\left(\frac{\sqrt
3}{2}\right) + (2)\left( \frac{1}{2}\right) = - 1`
`z_8^ - = z_8 = 3`
So,
`V_8^ - = \left( - \sqrt 3 , - 1,3\right)`
\]
Hence,
The coordinates of the vertices after the prism are rotated
counterclockwise about the z-axis through `\theta = 60^\circ`
are
`V_1^ - = (0,0,0), V_2^ - = \left(\frac{1}{2},\frac{\sqrt
3}{2},0\right),`
`V_3^ - = \left(\frac{1}{2} - \sqrt 3 ,\frac{\sqrt 3 }{2} -
1,0\right), V_4^ - = \left(- \sqrt 3 , - 1,0\right),`
`V_5^ - = \left(0,0,3\right), V_6^ - =
\left(\frac{1}{2},\frac{\sqrt 3}{2},3\right), `
`V_7^ - = \left(\frac{1}{2} - \sqrt 3 ,\frac{\sqrt 3}{2} -
,3\right), V_8^ - = \left(- \sqrt 3 , - 1,3\right)`
FAQ
Q1: What is the rotation formula about the Z-axis?
x' = x cosθ - y sinθ, y' = x sinθ + y cosθ, z' = z
Q2: Are the z-coordinates affected by rotation?
No, z remains unchanged because rotation occurs in the xy-plane.
Q3: How to rotate a 3D point counterclockwise about Z-axis?
Use the formulas above with positive θ (counterclockwise angle in degrees).
Explore more about TENSOR ANALYSIS AND ITS APPLICATIONS in Mathematics Notes & MCQs.
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