Solution: To find $k$ such that $\mathbf{p} + k\mathbf{q}$ is perpendicular to $\mathbf{r}$, we use the dot product condition:

["Finding $k$ Such That $\mathbf{p} + k\mathbf{q}$ Is Perpendicular to $\mathbf{r}$: A Clear Solution Using the Dot Product", "In vector geometry and linear algebra, determining when one vector is perpendicular to another is a fundamental concept. One common scenario arises when finding the scalar $k$ such that the vector $\mathbf{p} + k\mathbf{q}$ is perpendicular to a given vector $\mathbf{r}$. The key insight lies in the dot product, a powerful tool that helps us leverage orthogonality conditions.", "The Perpendicularity Condition", "Two vectors are perpendicular if and only if their dot product equals zero. That is:\n$$\n(\mathbf{p} + k\mathbf{q}) \cdot \mathbf{r} = 0\n$$", "This equation formalizes the geometric requirement and allows us to solve for $k$ algebraically.", "Step-by-Step Derivation", "1. Expand the dot product using distributivity:\n$$\n\mathbf{p} \cdot \mathbf{r} + (k\mathbf{q}) \cdot \mathbf{r} = 0\n$$", "2. Apply the scalar multiplication property of dot products:\n$$\n\mathbf{p} \cdot \mathbf{r} + k(\mathbf{q} \cdot \mathbf{r}) = 0\n$$", "3. Isolate $k$:\n$$\nk(\mathbf{q} \cdot \mathbf{r}) = -(\mathbf{p} \cdot \mathbf{r})\n$$\n$$\nk = -\frac{\mathbf{p} \cdot \mathbf{r}}{\mathbf{q} \cdot \mathbf{r}}, \quad \ ext{provided } \mathbf{q} \cdot \mathbf{r} <br/>\ne 0\n$$", "This formula gives the value of $k$ that makes $\mathbf{p} + k\mathbf{q}$ perpendicular to $\mathbf{r}$, assuming the denominator is non-zero.", "When Does This Work?", "- The expression is valid only if $\mathbf{q} \cdot \mathbf{r} <br/>\ne 0$. If the dot product $\mathbf{q} \cdot \mathbf{r} = 0$, then $\mathbf{r}$ is already orthogonal to $\mathbf{q}$, and the expression breaks down. In that case, the equation $\mathbf{p} + k\mathbf{q} \perp \mathbf{r}$ may still hold for specific $k$ values depending on $\mathbf{p}$ and $\mathbf{r}$, but the above formula fails.\n- Hence, checking $\mathbf{q} \cdot \mathbf{r} = 0$ is a crucial preliminary step.", "Practical Example", "Let $\mathbf{p} = \langle 2, -1 \rangle$, $\mathbf{q} = \langle 3, 4 \rangle$, and $\mathbf{r} = \langle 5, 0 \rangle$.\n- Compute $\mathbf{p} \cdot \mathbf{r} = 2 \cdot 5 + (-1) \cdot 0 = 10$\n- Compute $\mathbf{q} \cdot \mathbf{r} = 3 \cdot 5 + 4 \cdot 0 = 15$\n- Then,\n$$\nk = -\frac{10}{15} = -\frac{2}{3}\n$$\nSo, $\mathbf{p} + k\mathbf{q} = \langle 2, -1 \rangle + \left(-\frac{2}{3}\right)\langle 3, 4 \rangle = \langle 0, -\frac{11}{3} \rangle$, which is perpendicular to $\mathbf{r} = \langle 5, 0 \rangle$, as their dot product is certainly zero.", "Conclusion", "Using the dot product condition provides a systematic and computationally efficient method to find $k$ ensuring orthogonality. The formula\n$$\nk = -\frac{\mathbf{p} \cdot \mathbf{r}}{\mathbf{q} \cdot \mathbf{r}}\n$$\nis valid whenever $\mathbf{q} \cdot \mathbf{r} <br/>\ne 0$, offering a direct algebraic path to solve geometric problems involving perpendicular vectors. This approach is widely applicable in physics, engineering, and computer graphics, where vector alignment and orthogonal projections are essential.", "---", "By understanding and applying this dot product condition, anyone can confidently find the required scalar $k$ and explore deeper applications in vector calculus and multivariable geometry."]