MTH623 ASSIGNMENT NO. 1 FALL 2022 (PART-2) || 100% RIGHT SOLUTION || TENSOR ANALYSIS AND ITS APLICATIONS || BY VuTech

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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