In a cube with edge length , find , where is the angle between the line and the base plane .

Solid Geometry Advanced Solution – Problem 6: find , where is the angle between the line and the base plane

Question

In a cube $ABCD-A_1B_1C_1D_1$ with edge length $2$ , find $\sin\alpha$ , where $\alpha$ is the angle between the line $A_1C$ and the base plane $ABCD$ .

Step-by-step solution

(1) Place coordinates: $A(0,0,0),\ B(2,0,0),\ D(0,2,0),\ A_1(0,0,2),\ C(2,2,0).$ (2) A direction vector of $A_1C$ is $\mathbf u=\overrightarrow{A_1C}=(2,2,-2).$ A normal vector to plane $ABCD$ is $\mathbf n=(0,0,1)$ .

(3) For a line–plane angle $\alpha$ , we use $\sin\alpha=\frac{|\mathbf u\cdot\mathbf n|}{|\mathbf u||\mathbf n|}.$ Compute $|\mathbf u\cdot\mathbf n|=|-2|=2,\quad |\mathbf u|=\sqrt{12}=2\sqrt3.$ Thus $\sin\alpha=\frac{2}{2\sqrt3}=\frac{1}{\sqrt3}.$

Final answer

The sine of the line–plane angle is $\sin\alpha=\dfrac{1}{\sqrt3}$ .

Marking scheme

Step 1 — Setup

Checkpoint: assign coordinates and write a direction vector of $A_1C$ and a normal vector of plane $ABCD$ (2 pts)

Step 2 — Key Calculation

Checkpoint: apply $\sin\alpha=\frac{|\mathbf u\cdot\mathbf n|}{|\mathbf u||\mathbf n|}$ correctly (3 pts)

Step 3 — Final Answer

Checkpoint: simplify to $\sin\alpha=\frac{1}{\sqrt3}$ (2 pts)

Zero credit if: confuses $\sin$ and $\cos$ formulas for line–plane angles.

Deductions: -1 pt for a norm computation slip.

Solid Geometry Advanced – Problem 6: find , where is the angle between the line and the base plane