In the right rectangular prism , , , and . Find the angle between the space diagonal and the base plane .

If is the angle between and plane , then (so ).

Solid Geometry Solution – Problem 16: Find the angle between the space diagonal and the base plane

Question

In the right rectangular prism $ABCD-A_1B_1C_1D_1$ , $AB=2$ , $BC=1$ , and $AA_1=2$ . Find the angle between the space diagonal $AC_1$ and the base plane $ABCD$ .

Step-by-step solution

Set up coordinates: $A(0,0,0),\ B(2,0,0),\ C(2,1,0),\ C_1(2,1,2).$ A direction vector of $AC_1$ is $\overrightarrow{AC_1}=(2,1,2).$ Its length is $|\overrightarrow{AC_1}|=\sqrt{2^2+1^2+2^2}=3.$ Let $\theta$ be the angle between line $AC_1$ and the base plane $z=0$ . The perpendicular component to the base plane has magnitude $2$ . Hence $\sin\theta=\frac{2}{3}.$

Final answer

If $\theta$ is the angle between $AC_1$ and plane $ABCD$ , then $\sin\theta=\dfrac{2}{3}$ (so $\theta=\arcsin\dfrac{2}{3}$ ).

Marking scheme

1. Checkpoints (max 7 pts total)

Coordinate setup (2 pts): Assign correct box coordinates with edges $2,1,2$ .
Diagonal vector and norm (3 pts): Compute $\overrightarrow{AC_1}$ and $|\overrightarrow{AC_1}|=3$ .
Line-plane angle (2 pts): Use $\sin\theta=\frac{|v_\perp|}{|v|}=\frac{2}{3}$ .

2. Zero-credit items

Reporting $\sin\theta$ without computing the diagonal length.
Confusing the angle with the base diagonal $AC$ instead of $AC_1$ .

3. Deductions

Length error (-1): incorrect computation of $|\overrightarrow{AC_1}|$ .
Angle definition error (-1): using cosine with the plane normal but forgetting to convert to line-plane angle.

Solid Geometry – Problem 16: Find the angle between the space diagonal and the base plane