1. **Problem statement:** Given a regular quadrilateral pyramid $S.ABCD$ with base side length $a$ and center $O$. Points $M$ and $N$ are midpoints of edges $SA$ and $BC$ respectively. The angle between line $MN$ and the base plane $(ABCD)$ is $60^\circ$. Find: (a) The length of $MN$. (b) The cosine of the angle between $MN$ and the plane $(SBD)$. --- 2. **Known data and notation:** - Base $ABCD$ is a square with side length $a$. - $M$ is midpoint of $SA$, so $SM = MA = \frac{SA}{2}$. - $N$ is midpoint of $BC$, so $BN = NC = \frac{BC}{2} = \frac{a}{2}$. - The angle between $MN$ and $(ABCD)$ is $60^\circ$. --- 3. **Step (a): Find length $MN$** - Let’s place the base $ABCD$ in the $xy$-plane for convenience. - Assume $A=(0,0,0)$, $B=(a,0,0)$, $C=(a,a,0)$, $D=(0,a,0)$. - The apex $S$ is above the center $O$ of the square base. Since $O$ is midpoint of $AC$ and $BD$, $O=(\frac{a}{2}, \frac{a}{2}, 0)$. - Let the height of the pyramid be $h$, so $S=(\frac{a}{2}, \frac{a}{2}, h)$. - Point $M$ is midpoint of $SA$: $$M = \left(\frac{0 + \frac{a}{2}}{2}, \frac{0 + \frac{a}{2}}{2}, \frac{0 + h}{2}\right) = \left(\frac{a}{4}, \frac{a}{4}, \frac{h}{2}\right)$$ - Point $N$ is midpoint of $BC$: $$N = \left(\frac{a + a}{2}, \frac{0 + a}{2}, 0\right) = (a, \frac{a}{2}, 0)$$ - Vector $\overrightarrow{MN} = N - M = \left(a - \frac{a}{4}, \frac{a}{2} - \frac{a}{4}, 0 - \frac{h}{2}\right) = \left(\frac{3a}{4}, \frac{a}{4}, -\frac{h}{2}\right)$ - Length $MN$: $$MN = \sqrt{\left(\frac{3a}{4}\right)^2 + \left(\frac{a}{4}\right)^2 + \left(-\frac{h}{2}\right)^2} = \sqrt{\frac{9a^2}{16} + \frac{a^2}{16} + \frac{h^2}{4}} = \sqrt{\frac{10a^2}{16} + \frac{h^2}{4}} = \sqrt{\frac{5a^2}{8} + \frac{h^2}{4}}$$ --- 4. **Use the angle between $MN$ and $(ABCD)$ to find $h$** - The angle between $MN$ and the base plane $(ABCD)$ is $60^\circ$. - The base plane $(ABCD)$ is the $xy$-plane, so its normal vector is $\vec{n} = (0,0,1)$. - The angle $\theta$ between $MN$ and the plane satisfies: $$\sin \theta = \frac{|\overrightarrow{MN} \cdot \vec{n}|}{|\overrightarrow{MN}|}$$ - Dot product: $$\overrightarrow{MN} \cdot \vec{n} = -\frac{h}{2}$$ - So: $$\sin 60^\circ = \frac{\left| -\frac{h}{2} \right|}{MN} = \frac{\frac{h}{2}}{MN}$$ - Substitute $MN$: $$\sin 60^\circ = \frac{\frac{h}{2}}{\sqrt{\frac{5a^2}{8} + \frac{h^2}{4}}}$$ - Square both sides: $$\sin^2 60^\circ = \frac{h^2/4}{\frac{5a^2}{8} + \frac{h^2}{4}}$$ - Multiply both sides by denominator: $$\sin^2 60^\circ \left(\frac{5a^2}{8} + \frac{h^2}{4}\right) = \frac{h^2}{4}$$ - Expand: $$\frac{5a^2}{8} \sin^2 60^\circ + \frac{h^2}{4} \sin^2 60^\circ = \frac{h^2}{4}$$ - Rearrange terms: $$\frac{5a^2}{8} \sin^2 60^\circ = \frac{h^2}{4} - \frac{h^2}{4} \sin^2 60^\circ = \frac{h^2}{4} (1 - \sin^2 60^\circ)$$ - Recall $1 - \sin^2 \theta = \cos^2 \theta$: $$\frac{5a^2}{8} \sin^2 60^\circ = \frac{h^2}{4} \cos^2 60^\circ$$ - Solve for $h^2$: $$h^2 = \frac{5a^2}{8} \cdot \frac{\sin^2 60^\circ}{\cos^2 60^\circ} \cdot 4 = \frac{5a^2}{2} \cdot \frac{\sin^2 60^\circ}{\cos^2 60^\circ}$$ - Substitute $\sin 60^\circ = \frac{\sqrt{3}}{2}$ and $\cos 60^\circ = \frac{1}{2}$: $$h^2 = \frac{5a^2}{2} \cdot \frac{\left(\frac{\sqrt{3}}{2}\right)^2}{\left(\frac{1}{2}\right)^2} = \frac{5a^2}{2} \cdot \frac{\frac{3}{4}}{\frac{1}{4}} = \frac{5a^2}{2} \cdot 3 = \frac{15a^2}{2}$$ - Therefore: $$h = a \sqrt{\frac{15}{2}}$$ --- 5. **Calculate $MN$ length with $h$ found:** - Recall: $$MN = \sqrt{\frac{5a^2}{8} + \frac{h^2}{4}} = \sqrt{\frac{5a^2}{8} + \frac{1}{4} \cdot \frac{15a^2}{2}} = \sqrt{\frac{5a^2}{8} + \frac{15a^2}{8}} = \sqrt{\frac{20a^2}{8}} = \sqrt{\frac{5a^2}{2}} = a \sqrt{\frac{5}{2}}$$ --- 6. **Step (b): Find $\cos$ of angle between $MN$ and plane $(SBD)$** - The plane $(SBD)$ is defined by points $S$, $B$, and $D$. - Coordinates: - $S = \left(\frac{a}{2}, \frac{a}{2}, h\right)$ - $B = (a, 0, 0)$ - $D = (0, a, 0)$ - Vectors in plane $(SBD)$: $$\overrightarrow{SB} = B - S = \left(a - \frac{a}{2}, 0 - \frac{a}{2}, 0 - h\right) = \left(\frac{a}{2}, -\frac{a}{2}, -h\right)$$ $$\overrightarrow{SD} = D - S = \left(0 - \frac{a}{2}, a - \frac{a}{2}, 0 - h\right) = \left(-\frac{a}{2}, \frac{a}{2}, -h\right)$$ - Normal vector to plane $(SBD)$: $$\vec{n}_{SBD} = \overrightarrow{SB} \times \overrightarrow{SD}$$ - Compute cross product: $$\vec{n}_{SBD} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{a}{2} & -\frac{a}{2} & -h \\ -\frac{a}{2} & \frac{a}{2} & -h \end{vmatrix} = \mathbf{i} \left(-\frac{a}{2} \cdot (-h) - (-h) \cdot \frac{a}{2}\right) - \mathbf{j} \left(\frac{a}{2} \cdot (-h) - (-h) \cdot (-\frac{a}{2})\right) + \mathbf{k} \left(\frac{a}{2} \cdot \frac{a}{2} - (-\frac{a}{2}) \cdot (-\frac{a}{2})\right)$$ - Simplify each component: - $\mathbf{i}$ component: $$-\frac{a}{2} \cdot (-h) - (-h) \cdot \frac{a}{2} = \frac{ah}{2} + \frac{ah}{2} = ah$$ - $\mathbf{j}$ component: $$\frac{a}{2} \cdot (-h) - (-h) \cdot (-\frac{a}{2}) = -\frac{ah}{2} - \frac{ah}{2} = -ah$$ - $\mathbf{k}$ component: $$\frac{a}{2} \cdot \frac{a}{2} - (-\frac{a}{2}) \cdot (-\frac{a}{2}) = \frac{a^2}{4} - \frac{a^2}{4} = 0$$ - So: $$\vec{n}_{SBD} = (ah, -ah, 0) = ah(1, -1, 0)$$ - Normalize $\vec{n}_{SBD}$: $$|\vec{n}_{SBD}| = ah \sqrt{1^2 + (-1)^2 + 0^2} = ah \sqrt{2}$$ - Unit normal vector: $$\hat{n}_{SBD} = \frac{1}{ah\sqrt{2}} (ah, -ah, 0) = \frac{1}{\sqrt{2}} (1, -1, 0)$$ --- 7. **Find cosine of angle between $MN$ and plane $(SBD)$** - The angle between line $MN$ and plane $(SBD)$ is complementary to the angle between $\overrightarrow{MN}$ and $\vec{n}_{SBD}$. - So: $$\cos(\text{angle between } MN \text{ and plane } SBD) = \sin(\text{angle between } \overrightarrow{MN} \text{ and } \vec{n}_{SBD})$$ - First find $\cos \phi$ where $\phi$ is angle between $\overrightarrow{MN}$ and $\vec{n}_{SBD}$: - Recall: $$\overrightarrow{MN} = \left(\frac{3a}{4}, \frac{a}{4}, -\frac{h}{2}\right)$$ - Dot product: $$\overrightarrow{MN} \cdot \hat{n}_{SBD} = \left(\frac{3a}{4}\right) \cdot \frac{1}{\sqrt{2}} + \left(\frac{a}{4}\right) \cdot \left(-\frac{1}{\sqrt{2}}\right) + \left(-\frac{h}{2}\right) \cdot 0 = \frac{3a}{4\sqrt{2}} - \frac{a}{4\sqrt{2}} = \frac{2a}{4\sqrt{2}} = \frac{a}{2\sqrt{2}}$$ - Length $|\overrightarrow{MN}| = a \sqrt{\frac{5}{2}}$ (from step 5) - Length $|\hat{n}_{SBD}| = 1$ - So: $$\cos \phi = \frac{\overrightarrow{MN} \cdot \hat{n}_{SBD}}{|\overrightarrow{MN}|} = \frac{\frac{a}{2\sqrt{2}}}{a \sqrt{\frac{5}{2}}} = \frac{1}{2\sqrt{2}} \cdot \frac{1}{\sqrt{\frac{5}{2}}} = \frac{1}{2\sqrt{2}} \cdot \sqrt{\frac{2}{5}} = \frac{1}{2\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{5}} = \frac{1}{2} \cdot \frac{1}{\sqrt{5}} = \frac{1}{2\sqrt{5}}$$ - Then: $$\cos(\text{angle between } MN \text{ and plane } SBD) = \sin \phi = \sqrt{1 - \cos^2 \phi} = \sqrt{1 - \left(\frac{1}{2\sqrt{5}}\right)^2} = \sqrt{1 - \frac{1}{20}} = \sqrt{\frac{19}{20}} = \frac{\sqrt{19}}{\sqrt{20}} = \frac{\sqrt{19}}{2\sqrt{5}}$$ --- **Final answers:** (a) $$MN = a \sqrt{\frac{5}{2}}$$ (b) $$\cos(\text{angle between } MN \text{ and plane } SBD) = \frac{\sqrt{19}}{2\sqrt{5}}$$