Question #151407
A(94,0,14) B(52,56,94) C(10,6,48) D(128,64,10) Content and instructions for completing task number 3.
The goal of the assignment is to apply methods of transforming orthogonal projections to solving metric problems.
The coordinates of four points ABCD are given (see next table). Determine the following values by transforming projection planes:
• 1. The size of the dihedral angle at the edge AB.
• 2. The shortest distance between ribs DA and BC.
• 3. Distance from vertex D to plane ABC.
• 4. Natural size of triangle ABC.
An example how to complete the task is shown after the table
1
Expert's answer
2020-12-17T17:15:59-0500

1. The size of the dihedral angle at the edge AB.

This angle is angle between planes ABC and ABD.

Equation of plane ABD:

x94529412894y5664z1494141014=x944234y5664z14804=5344(x94)+2552y4592(z14)=0\begin{vmatrix} x-94 & 52-94&128-94 \\ y & 56&64\\ z-14&94-14&10-14 \end{vmatrix}=\begin{vmatrix} x-94 & -42&34 \\ y & 56&64\\ z-14&80&-4 \end{vmatrix}=-5344(x-94)+2552y--4592(z-14)=0

5344x2552y+4592z566624=05344x-2552y+4592z-566624=0


cosφ=14245344+52922552+445245921424252922+4452253442+25522+45922=415586247060.77493.83=0.7854cos\varphi=\frac{1424\cdot5344+5292\cdot2552+4452\cdot4592}{\sqrt{1424^2-5292^2+4452^2}\cdot\sqrt{5344^2+2552^2+4592^2}}=\frac{41558624}{7060.7\cdot7493.83}=0.7854


φ=38.24°\varphi=38.24\degree



2.The shortest distance between ribs DA and BC.

Equations of lines:

DA: x12834=y6464=z104\frac{x-128}{-34}=\frac{y-64}{-64}=\frac{z-10}{4}

BC: x5242=y5650=z9446\frac{x-52}{-42}=\frac{y-56}{-50}=\frac{z-94}{-46}

The shortest distance:

d=p1p2M1M2p1×p2d=\frac{|p_1\cdot p_2\cdot M_1M_2|}{|p_1\times p_2|}

M1M2=(52128,5664,9410)=(76,8,84)M_1M_2=(52-128,56-64,94-10)=(-76,-8,84)

p1×p2=(ijk34644425046)=i6445046j3444246++k34644250=3144i1732j988kp_1\times p_2=\begin{pmatrix} i & j &k\\ -34 & -64&4\\ -42&-50&-46 \end{pmatrix}=i\begin{vmatrix} -64 & 4 \\ -50 & -46 \end{vmatrix}-j\begin{vmatrix} -34 & 4 \\ -42 & -46 \end{vmatrix}++k\begin{vmatrix} -34 & -64 \\ -42 & -50 \end{vmatrix}=3144i-1732j-988k

p1×p2=31442+17322+9882=3723|p_1\times p_2|=\sqrt{3144^2+1732^2+988^2}=3723

p1p2M1M2=3442766450844684=308080|p_1\cdot p_2\cdot M_1M_2|=\begin{vmatrix} -34 & -42&-76 \\ -64 & -50&-8\\ 4&-46&84 \end{vmatrix}=308080

d=308080/3723=82.75d=308080/3723=82.75



3.Distance from vertex D to plane ABC.

Equation of plane ABC:

x9452941094y566z1494144814=x944284y566z148034=1424(x94)5292y++4452(z14)=0\begin{vmatrix} x-94 & 52-94&10-94 \\ y & 56&6\\ z-14&94-14&48-14 \end{vmatrix}=\begin{vmatrix} x-94 & -42&-84 \\ y & 56&6\\ z-14&80&34 \end{vmatrix}=1424(x-94)-5292y++4452(z-14)=0


1424x5292y+4452z196184=01424x-5292y+4452z-196184=0

The distance:

d=1424128529264+4452101961841424252922+44522=3080807060.7=43.63d=\frac{|1424\cdot128-5292\cdot64+4452\cdot10-196184|}{\sqrt{1424^2-5292^2+4452^2}}=\frac{308080}{7060.7}=43.63



4. Natural size of triangle ABC.

AB=(9452)2+562+(9414)2=106.3|AB|=\sqrt{(94-52)^2+56^2+(94-14)^2}=106.3

BC=(5210)2+(566)2+(9448)2=79.87|BC|=\sqrt{(52-10)^2+(56-6)^2+(94-48)^2}=79.87

AC=(9410)2+62+(4814)2=90.82|AC|=\sqrt{(94-10)^2+6^2+(48-14)^2}=90.82



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Comments

Assignment Expert
18.12.20, 18:33

Dear Dee_june, please use the panel for submitting new questions. It is necessary to describe all requirements for this question to be done.

Dee_june
18.12.20, 00:28

Please draw it’s Like geometry not maths

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