Let $\mathbf{p}$ be the projection of $\mathbf{c}$ onto the plane containing $\mathbf{a}$ and $\mathbf{b}.$

[asy]
import three;

size(140);
currentprojection = perspective(6,3,2);

real t = 60, k = Cos(t);

triple A, B, C, O, P, Q;

A = (Cos(t/2),Sin(t/2),0);
B = (Cos(t/2),-Sin(t/2),0);
C = (k/Cos(t/2),0,sqrt(1 - k^2/Cos(t/2)^2));
O = (0,0,0);
P = (k/Cos(t/2),0,0);
Q = k/(k + 1)*A + k/(k + 1)*B;

draw(O--A,Arrow3(6));
draw(O--B,Arrow3(6));
draw(O--C,Arrow3(6));
draw(O--P,Arrow3(6));
draw(C--P,dashed);

label("$\mathbf{a}$", A, S, fontsize(10));
label("$\mathbf{b}$", B, W, fontsize(10));
label("$\mathbf{c}$", C, NW, fontsize(10));
label("$\mathbf{p}$", P, SW, fontsize(10));
[/asy]

Then
\[\mathbf{p} = s \mathbf{a} + t \mathbf{b}\]for some scalars $s$ and $t.$  Let $\mathbf{n}$ be the normal vector to the plane containing $\mathbf{a}$ and $\mathbf{b},$ so
\[\mathbf{c} = \mathbf{p} + u \mathbf{n} = s \mathbf{a} + t \mathbf{b} + u \mathbf{n}\]for some scalar $u.$

Taking the dot product with $\mathbf{a},$ we get
\[\mathbf{a} \cdot \mathbf{c} = s \mathbf{a} \cdot \mathbf{a} + t \mathbf{a} \cdot \mathbf{b} + u \mathbf{a} \cdot \mathbf{n}.\]Note that $\mathbf{a} \cdot \mathbf{a} = \|\mathbf{a}\|^2 = 4$ and $\mathbf{a} \cdot \mathbf{b} = \mathbf{a} \cdot \mathbf{c} = 2 \cdot 2 \cdot \frac{5}{8} = \frac{5}{2}.$  Also, $\mathbf{a} \cdot \mathbf{n} = 0,$ so
\[\frac{5}{2} = 4s + \frac{5t}{2}.\]Similarly, taking the dot product with $\mathbf{b},$ we get
\[\mathbf{b} \cdot \mathbf{c} = s \mathbf{a} \cdot \mathbf{b} + t \mathbf{b} \cdot \mathbf{b} + u \mathbf{b} \cdot \mathbf{n}.\]This reduces to $\frac{5}{2} = \frac{5s}{2} + 4t.$

Solving the equations $\frac{5}{2} = 4s + \frac{5t}{2}$ and $\frac{5}{2} = \frac{5s}{2} + 4t,$ we get $s = t = \frac{5}{13}.$  Hence,
\[\mathbf{p} = \frac{5}{13} (\mathbf{a} + \mathbf{b}).\]Then
\begin{align*}
\|\mathbf{p}\|^2 &= \frac{25}{169} (\mathbf{a} \cdot \mathbf{a} + 2 \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b}) \\
&= \frac{25}{169} \left( 4 + 2 \cdot \frac{5}{2} + 4 \right) = \frac{25}{13}.
\end{align*}By Pythagoras, the height of the parallelepiped is then given by
\[\sqrt{4 - \|\mathbf{p}\|^2} = \sqrt{4 - \frac{25}{13}} = \sqrt{\frac{27}{13}}.\]The base of the parallelepiped has area $2 \cdot 2 \cdot \sin \left( \arccos \frac{5}{8} \right) = 4 \sqrt{1 - \left( \frac{5}{8} \right)^2} = 4 \sqrt{\frac{39}{64}},$ so the volume of the parallelepiped is
\[\sqrt{\frac{27}{13}} \cdot 4 \sqrt{\frac{39}{64}} = \frac{9}{2}.\]
수학비프{p}$ 는 $\mathbf{a}$ 와 $\mathbf{b}.$ 를 포함하는 평면에 $\mathbf{c}$ 를 투영한 값입니다.

[asy]
3을 가져옵니다;

size(140);
currentprojection = perspective(6,3,2);

실수 t = 60, k = Cos(t);

트리플 A, B, C, O, P, Q;

A = (Cos(t/2),Sin(t/2),0);
B = (Cos(t/2),-Sin(t/2),0);
C = (k/Cos(t/2),0,sqrt(1 - k^2/Cos(t/2)^2));
O = (0,0,0);
P = (k/Cos(t/2),0,0);
Q = k/(k + 1)*A + k/(k + 1)*B;

draw(O--A,Arrow3(6));
draw(O--B,Arrow3(6));
draw(O--C,Arrow3(6));
draw(O--P,Arrow3(6));
draw(C--P,점선);

label("$\mathbf{a}$", A, S, fontsize(10));
label("$\mathbf{b}$", B, W, fontsize(10));
label("$\mathbf{c}$", C, NW, fontsize(10));
label("$\mathbf{p}$", P, SW, fontsize(10));
[/asy]

그러면
일부 스칼라 $s$ 및 $t$에 대해 \[\mathbf{p} = s \mathbf{a} + t \mathbf{b}\]는 $\mathbf{n}$를 $\mathbf{a}$ 및 $\mathbf{b}$를 포함하는 평면에 대한 법선 벡터로 합니다.
\[\mathbf{c} = \mathbf{p} + u \mathbf{n} = s \mathbf{a} + t \mathbf{b} + u \mathbf{n}\]는 일부 스칼라 $u.$의 경우입니다.

도트 곱을 $\mathbf{a},$로 취하면 다음과 같습니다.
\[\mathbf{a} \cdot \mathbf{c} = s \mathbf{a} \cdot \mathbf{a} + t \mathbf{a} \cdot \mathbf{b} + u \mathbf{a} \cdot \mathbf{n}.\]$\mathbf{a} \cdot \mathbf{a} = \|\mathbf{a}\|^2 = 4$ 및 $\mathbf{a} \cdot \mathbf{b} = \mathbf{a} \cdot \mathbf{c} = 2 \cdot 2 \cdot \frac{5}{8} = \frac{5}{2}에 유의합니다.또한, $\mathbf{a} \cdot \mathbf{n} = 0,$이므로 다음과 같습니다.
\[\frac{5}{2} = 4s + \frac{5t}{2}.\]마찬가지로, 도트 곱을 $\mathbf{b},$ 로 취하면 다음과 같이 됩니다.
\[\mathbf{b} \cdot \mathbf{c} = s \mathbf{a} \cdot \mathbf{b} + t \mathbf{b} \cdot \mathbf{b} + u \mathbf{b} \cdot \mathbf{n}.\]이것은 $\frac{5}{2} = \frac{5s}{2}로 감소합니다. + 4t.$

방정식 $\frac{5}{2} = 4s + \frac{5t}{2}$ 및 $\frac{5}{2} = \frac{5s}{2} + 4t,$ 는 $s = t = \frac{5}{13}$ 이므로 다음과 같습니다,
\[\mathbf{p} = \frac{5}{13} (\mathbf{a} + \mathbf{b}).\]그러면 다음과 같습니다.
\begin{align*}
\|\mathbf{p}\|^2 &= \frac{25}{169} (\mathbf{a} \cdot \mathbf{a} + 2 \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b}) \\
&= \frac{25}{169} \왼쪽(4 + 2 \cdot \frac{5}{2} + 4 \right) = \frac{25}{13}.
\end{align*}피타고라스에 따르면 평행 육면체의 높이는 다음과 같이 주어집니다.
\[\sqrt{4 - \|\mathbf{p}\|^2} = \sqrt{4 - \frac{25}{13}} = \sqrt{\frac{27}{13}}.\]평행 육면체의 밑변의 넓이는 $2 \cdot 2 \cdot \sin \left( \arccos \frac{5}{8} \right) = 4 \sqrt{1 - \left( \frac{5}{8} \right)^2} = 4 \sqrt{\frac{39}{64}},$이므로 평행 육면체의 체적은 다음과 같습니다.
\[\sqrt{\frac{27}{13}} \cdot 4 \sqrt{\frac{39}{64}} = \frac{9}{2}.\]