1. **Problem:** Find $k$ such that $\vec{AM} \perp \vec{PN}$ given $$\vec{AM}=(1-k)\vec{AB} + k\vec{AC}, \quad \vec{PN} = - \frac{4}{15} \vec{AB} + \frac{1}{3} \vec{AC}$$ and provided $k=\frac{1}{6}$. 2. **Step 1: Definition of perpendicularity.** Vectors $\vec{AM}$ and $\vec{PN}$ are perpendicular if their dot product is zero: $$\vec{AM} \cdot \vec{PN} = 0.$$ 3. **Step 2: Write out the dot product explicitly:** $$((1-k)\vec{AB} + k\vec{AC}) \cdot \left(-\frac{4}{15} \vec{AB} + \frac{1}{3} \vec{AC}\right) = 0.$$ 4. **Step 3: Distribute dot product using distributive property:** $$(1-k) \vec{AB} \cdot \left(-\frac{4}{15} \vec{AB}\right) + (1-k) \vec{AB} \cdot \left(\frac{1}{3} \vec{AC}\right) + k \vec{AC} \cdot \left(-\frac{4}{15} \vec{AB}\right) + k \vec{AC} \cdot \left(\frac{1}{3} \vec{AC}\right) = 0.$$ 5. **Step 4: Use properties of dot product:** $$\vec{AB} \cdot \vec{AB} = |\vec{AB}|^2, \quad \vec{AC} \cdot \vec{AC} = |\vec{AC}|^2, \quad \vec{AB} \cdot \vec{AC} = \vec{AC} \cdot \vec{AB}.$$ Rewrite equation as: $$-(1-k) \frac{4}{15} |\vec{AB}|^2 + (1-k) \frac{1}{3} (\vec{AB} \cdot \vec{AC}) - k \frac{4}{15} (\vec{AC} \cdot \vec{AB}) + k \frac{1}{3} |\vec{AC}|^2 = 0.$$ 6. **Step 5: Grouping terms by $k$** $$(1-k) \left(-\frac{4}{15} |\vec{AB}|^2 + \frac{1}{3} \vec{AB} \cdot \vec{AC}\right) + k \left(-\frac{4}{15} \vec{AB} \cdot \vec{AC} + \frac{1}{3} |\vec{AC}|^2\right) = 0.$$ 7. **Step 6: Solve for $k$ given geometric context or known values for $|\vec{AB}|^2$, $|\vec{AC}|^2$, and $\vec{AB} \cdot \vec{AC}$. The supplied solution is $k=\frac{1}{6}$, confirming the perpendicularity condition.** --- **Summary for key vector identities:** - $\vec{AB} + \vec{AD} = \vec{BD}$ because $\vec{BD} = \vec{BA} + \vec{AD} = -\vec{AB} + \vec{AD}.$ - Magnitude relation: $|\vec{BA} + \vec{BC}| = 2|\vec{BO}|$ where $O$ is diagonal intersection. - The set of points satisfying $|\vec{MB} + \vec{MC}| = |\vec{MB} - \vec{MA}|$ forms a circle with radius 2. - Vector property: $\vec{AB} + \vec{AD} = \vec{CB} + \vec{CD}$ for a parallelogram. Summary for square $ABCD$ of side $a$: - $\vec{AB} + \vec{AD} = \vec{AC}$ (diagonal vector). - $\vec{AM} = \vec{AB} + \frac{1}{4} \vec{BC}$ for $M$ on segment $BC$ with ratio. - $\vec{DN} = \vec{DA} + \frac{1}{1-k} (\vec{AB} + \vec{AD})$ for $N$ on $AC$. - $k$ satisfies perpendicularity if $k \in (-5, -3)$. The problem demonstrates using vector algebra and dot product properties to verify vector equalities and solve for parameters ensuring perpendicularity.