1. The size of the dihedral angle at the edge AB.
This angle is angle between planes ABC and ABD.
Equation of plane ABD:
∣ x − 94 52 − 94 128 − 94 y 56 64 z − 14 94 − 14 10 − 14 ∣ = ∣ x − 94 − 42 34 y 56 64 z − 14 80 − 4 ∣ = − 5344 ( x − 94 ) + 2552 y − − 4592 ( z − 14 ) = 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 ∣ ∣ x − 94 y z − 14 52 − 94 56 94 − 14 128 − 94 64 10 − 14 ∣ ∣ = ∣ ∣ x − 94 y z − 14 − 42 56 80 34 64 − 4 ∣ ∣ = − 5344 ( x − 94 ) + 2552 y − − 4592 ( z − 14 ) = 0
5344 x − 2552 y + 4592 z − 566624 = 0 5344x-2552y+4592z-566624=0 5344 x − 2552 y + 4592 z − 566624 = 0
c o s φ = 1424 ⋅ 5344 + 5292 ⋅ 2552 + 4452 ⋅ 4592 142 4 2 − 529 2 2 + 445 2 2 ⋅ 534 4 2 + 255 2 2 + 459 2 2 = 41558624 7060.7 ⋅ 7493.83 = 0.7854 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 cos φ = 142 4 2 − 529 2 2 + 445 2 2 ⋅ 534 4 2 + 255 2 2 + 459 2 2 1424 ⋅ 5344 + 5292 ⋅ 2552 + 4452 ⋅ 4592 = 7060.7 ⋅ 7493.83 41558624 = 0.7854
φ = 38.24 ° \varphi=38.24\degree φ = 38.24°
2.The shortest distance between ribs DA and BC.
Equations of lines:
DA: x − 128 − 34 = y − 64 − 64 = z − 10 4 \frac{x-128}{-34}=\frac{y-64}{-64}=\frac{z-10}{4} − 34 x − 128 = − 64 y − 64 = 4 z − 10
BC: x − 52 − 42 = y − 56 − 50 = z − 94 − 46 \frac{x-52}{-42}=\frac{y-56}{-50}=\frac{z-94}{-46} − 42 x − 52 = − 50 y − 56 = − 46 z − 94
The shortest distance:
d = ∣ p 1 ⋅ p 2 ⋅ M 1 M 2 ∣ ∣ p 1 × p 2 ∣ d=\frac{|p_1\cdot p_2\cdot M_1M_2|}{|p_1\times p_2|} d = ∣ p 1 × p 2 ∣ ∣ p 1 ⋅ p 2 ⋅ M 1 M 2 ∣
M 1 M 2 = ( 52 − 128 , 56 − 64 , 94 − 10 ) = ( − 76 , − 8 , 84 ) M_1M_2=(52-128,56-64,94-10)=(-76,-8,84) M 1 M 2 = ( 52 − 128 , 56 − 64 , 94 − 10 ) = ( − 76 , − 8 , 84 )
p 1 × p 2 = ( i j k − 34 − 64 4 − 42 − 50 − 46 ) = i ∣ − 64 4 − 50 − 46 ∣ − j ∣ − 34 4 − 42 − 46 ∣ + + k ∣ − 34 − 64 − 42 − 50 ∣ = 3144 i − 1732 j − 988 k 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 × p 2 = ⎝ ⎛ i − 34 − 42 j − 64 − 50 k 4 − 46 ⎠ ⎞ = i ∣ ∣ − 64 − 50 4 − 46 ∣ ∣ − j ∣ ∣ − 34 − 42 4 − 46 ∣ ∣ + + k ∣ ∣ − 34 − 42 − 64 − 50 ∣ ∣ = 3144 i − 1732 j − 988 k
∣ p 1 × p 2 ∣ = 314 4 2 + 173 2 2 + 98 8 2 = 3723 |p_1\times p_2|=\sqrt{3144^2+1732^2+988^2}=3723 ∣ p 1 × p 2 ∣ = 314 4 2 + 173 2 2 + 98 8 2 = 3723
∣ p 1 ⋅ p 2 ⋅ M 1 M 2 ∣ = ∣ − 34 − 42 − 76 − 64 − 50 − 8 4 − 46 84 ∣ = 308080 |p_1\cdot p_2\cdot M_1M_2|=\begin{vmatrix}
-34 & -42&-76 \\
-64 & -50&-8\\
4&-46&84
\end{vmatrix}=308080 ∣ p 1 ⋅ p 2 ⋅ M 1 M 2 ∣ = ∣ ∣ − 34 − 64 4 − 42 − 50 − 46 − 76 − 8 84 ∣ ∣ = 308080
d = 308080 / 3723 = 82.75 d=308080/3723=82.75 d = 308080/3723 = 82.75
3.Distance from vertex D to plane ABC.
Equation of plane ABC:
∣ x − 94 52 − 94 10 − 94 y 56 6 z − 14 94 − 14 48 − 14 ∣ = ∣ x − 94 − 42 − 84 y 56 6 z − 14 80 34 ∣ = 1424 ( x − 94 ) − 5292 y + + 4452 ( z − 14 ) = 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 ∣ ∣ x − 94 y z − 14 52 − 94 56 94 − 14 10 − 94 6 48 − 14 ∣ ∣ = ∣ ∣ x − 94 y z − 14 − 42 56 80 − 84 6 34 ∣ ∣ = 1424 ( x − 94 ) − 5292 y + + 4452 ( z − 14 ) = 0
1424 x − 5292 y + 4452 z − 196184 = 0 1424x-5292y+4452z-196184=0 1424 x − 5292 y + 4452 z − 196184 = 0
The distance:
d = ∣ 1424 ⋅ 128 − 5292 ⋅ 64 + 4452 ⋅ 10 − 196184 ∣ 142 4 2 − 529 2 2 + 445 2 2 = 308080 7060.7 = 43.63 d=\frac{|1424\cdot128-5292\cdot64+4452\cdot10-196184|}{\sqrt{1424^2-5292^2+4452^2}}=\frac{308080}{7060.7}=43.63 d = 142 4 2 − 529 2 2 + 445 2 2 ∣1424 ⋅ 128 − 5292 ⋅ 64 + 4452 ⋅ 10 − 196184∣ = 7060.7 308080 = 43.63
4. Natural size of triangle ABC.
∣ A B ∣ = ( 94 − 52 ) 2 + 5 6 2 + ( 94 − 14 ) 2 = 106.3 |AB|=\sqrt{(94-52)^2+56^2+(94-14)^2}=106.3 ∣ A B ∣ = ( 94 − 52 ) 2 + 5 6 2 + ( 94 − 14 ) 2 = 106.3
∣ B C ∣ = ( 52 − 10 ) 2 + ( 56 − 6 ) 2 + ( 94 − 48 ) 2 = 79.87 |BC|=\sqrt{(52-10)^2+(56-6)^2+(94-48)^2}=79.87 ∣ BC ∣ = ( 52 − 10 ) 2 + ( 56 − 6 ) 2 + ( 94 − 48 ) 2 = 79.87
∣ A C ∣ = ( 94 − 10 ) 2 + 6 2 + ( 48 − 14 ) 2 = 90.82 |AC|=\sqrt{(94-10)^2+6^2+(48-14)^2}=90.82 ∣ A C ∣ = ( 94 − 10 ) 2 + 6 2 + ( 48 − 14 ) 2 = 90.82
Comments
Dear Dee_june, please use the panel for submitting new questions. It is necessary to describe all requirements for this question to be done.
Please draw it’s Like geometry not maths