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MTH623 ASSIGNMENT NO. 1 FALL 2022 (PART-2) |
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MTH623 ASSIGNMENT NO. 1 FALL 2022 (PART-2) || 100% RIGHT SOLUTION || TENSOR ANALYSIS AND ITS APLICATIONS || BY VuTech
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Assignment # 01 MTH623 (Fall 2022)
Maximum Marks: 20
Due Date: 24 Nov, 2022
INSTRUCTIONS
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Question # 2:
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)`
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