Answer to Question #151407 in Analytic Geometry for Dee_june

Question

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

Answer 1

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

This angle is angle between planes ABC and ABD.

Equation of plane ABD:

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

$5344x-2552y+4592z-566624=0$

$cos\varphi=\frac{1424\cdot5344+5292\cdot2552+4452\cdot4592}{\sqrt{1424^2-5292^2+4452^2}\cdot\sqrt{5344^2+2552^2+4592^2}}=\frac{41558624}{7060.7\cdot7493.83}=0.7854$

$\varphi=38.24\degree$

2.The shortest distance between ribs DA and BC.

Equations of lines:

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

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

The shortest distance:

$d=\frac{|p_1\cdot p_2\cdot M_1M_2|}{|p_1\times p_2|}$

$M_1M_2=(52-128,56-64,94-10)=(-76,-8,84)$

$p_1\times p_2=\begin{pmatrix} i & j &k\\ -34 & -64&4\\ -42&-50&-46 \end{pmatrix}=i\begin{vmatrix} -64 & 4 \\ -50 & -46 \end{vmatrix}-j\begin{vmatrix} -34 & 4 \\ -42 & -46 \end{vmatrix}++k\begin{vmatrix} -34 & -64 \\ -42 & -50 \end{vmatrix}=3144i-1732j-988k$

$|p_1\times p_2|=\sqrt{3144^2+1732^2+988^2}=3723$

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

$d=308080/3723=82.75$

3.Distance from vertex D to plane ABC.

Equation of plane ABC:

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

$1424x-5292y+4452z-196184=0$

The distance:

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

4. Natural size of triangle ABC.

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

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

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