$ABCDEFGH$ shown below is a cube. Find $\sin \angle HAC$.

[asy]

import three;

triple A,B,C,D,EE,F,G,H;

A = (0,0,0);

B = (1,0,0);

C = (1,1,0);

D= (0,1,0);

EE = (0,0,1);

F = B+EE;

G = C + EE;

H = D + EE;

draw(B--C--D);

draw(B--A--D,dashed);

draw(EE--F--G--H--EE);

draw(A--EE,dashed);

draw(B--F);

draw(C--G);

draw(D--H);

label("$A$",A,S);

label("$B$",B,W);

label("$C$",C,S);

label("$D$",D,E);

label("$E$",EE,N);

label("$F$",F,W);

label("$G$",G,SW);

label("$H$",H,E);

[/asy]
Each side of $\triangle HAC$ is a face diagonal of the cube:

[asy]

import three;

triple A,B,C,D,EE,F,G,H;

A = (0,0,0);

B = (1,0,0);

C = (1,1,0);

D= (0,1,0);

EE = (0,0,1);

F = B+EE;

G = C + EE;

H = D + EE;

draw(B--C--D);

draw(B--A--D,dashed);

draw(EE--F--G--H--EE);

draw(A--EE,dashed);

draw(H--A--C,dashed);

draw(B--F);

draw(C--G);

draw(D--H--C);

label("$A$",A,NW);

label("$B$",B,W);

label("$C$",C,S);

label("$D$",D,E);

label("$E$",EE,N);

label("$F$",F,W);

label("$G$",G,SW);

label("$H$",H,E);

[/asy]

Therefore, $\triangle HAC$ is equilateral, so $\sin \angle HAC = \sin 60^\circ = \boxed{\frac{\sqrt{3}}{2}}$.
\frac{\sqrt{3}}{2}
Level 5
Geometry
Problem: $ABCDEFGH$ shown below is a cube. Find $\sin \angle HAC$.

[asy]

import three;

triple A,B,C,D,EE,F,G,H;

A = (0,0,0);

B = (1,0,0);

C = (1,1,0);

D= (0,1,0);

EE = (0,0,1);

F = B+EE;

G = C + EE;

H = D + EE;

draw(B--C--D);

draw(B--A--D,dashed);

draw(EE--F--G--H--EE);

draw(A--EE,dashed);

draw(B--F);

draw(C--G);

draw(D--H);

label("$A$",A,S);

label("$B$",B,W);

label("$C$",C,S);

label("$D$",D,E);

label("$E$",EE,N);

label("$F$",F,W);

label("$G$",G,SW);

label("$H$",H,E);

[/asy]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find $\sin \angle HAC$, we first identify the coordinates of points $A$, $H$, and $C$ in the cube. The coordinates are as follows: $A(0,0,0)$, $H(0,1,1)$, and $C(1,1,0)$. 

Next, we calculate the vectors $\overrightarrow{HA}$ and $\overrightarrow{CA}$. The vector $\overrightarrow{HA}$ is given by the difference of the coordinates of $A$ and $H$: 
$$\overrightarrow{HA} = A - H = (0,0,0) - (0,1,1) = (0,-1,-1).$$ 

The vector $\overrightarrow{CA}$ is calculated similarly:
$$\overrightarrow{CA} = A - C = (0,0,0) - (1,1,0) = (-1,-1,0).$$ 

Next, we find the magnitudes of these vectors:
$$|\overrightarrow{HA}| = \sqrt{0^2 + (-1)^2 + (-1)^2} = \sqrt{2},$$
$$|\overrightarrow{CA}| = \sqrt{(-1)^2 + (-1)^2 + 0^2} = \sqrt{2}.$$

Now, we compute the dot product of the vectors $\overrightarrow{HA}$ and $\overrightarrow{CA}$:
$$\overrightarrow{HA} \cdot \overrightarrow{CA} = (0)(-1) + (-1)(-1) + (-1)(0) = 1.$$

Using the dot product to find the cosine of the angle $\angle HAC$, we apply the formula:
$$\cos \angle HAC = \frac{\overrightarrow{HA} \cdot \overrightarrow{CA}}{|\overrightarrow{HA}| |\overrightarrow{CA}|} = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}.$$

To find $\sin \angle HAC$, we use the identity $\sin^2 \theta + \cos^2 \theta = 1$:
$$\sin^2 \angle HAC = 1 - \cos^2 \angle HAC = 1 - \left(\frac{1}{2}\right)^2 = 1 - \frac{1}{4} = \frac{3}{4}.$$

Taking the square root gives:
$$\sin \angle HAC = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}.$$

Thus, the final answer is $\frac{\sqrt{3}}{2}$.<|END_OF_LATENT|> The final answer is $\frac{\sqrt{3}}{2}$.