1. **Problem statement:** Given that $\mathbf{a}$ and $\mathbf{c}$ are unit vectors, the magnitude of $\mathbf{b}$ is 4, and the angle between $\mathbf{b}$ and $\mathbf{c}$ is 60 degrees, find the two possible values of $k$ using a suitable scalar product. 2. **Recall the scalar product formula:** For any vectors $\mathbf{u}$ and $\mathbf{v}$, the scalar (dot) product is given by: $$\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos \theta$$ where $\theta$ is the angle between $\mathbf{u}$ and $\mathbf{v}$. 3. **Apply the formula to $\mathbf{b}$ and $\mathbf{c}$:** Since $|\mathbf{b}| = 4$, $|\mathbf{c}| = 1$ (unit vector), and $\theta = 60^\circ$, we have: $$\mathbf{b} \cdot \mathbf{c} = 4 \times 1 \times \cos 60^\circ = 4 \times \frac{1}{2} = 2$$ 4. **Express $\mathbf{b}$ in terms of $k$:** Assuming $\mathbf{b} = k \mathbf{a} + \mathbf{c}$ (or a similar linear combination), then: $$\mathbf{b} \cdot \mathbf{c} = (k \mathbf{a} + \mathbf{c}) \cdot \mathbf{c} = k (\mathbf{a} \cdot \mathbf{c}) + \mathbf{c} \cdot \mathbf{c}$$ Since $\mathbf{c}$ is a unit vector, $\mathbf{c} \cdot \mathbf{c} = 1$. 5. **Let $\mathbf{a} \cdot \mathbf{c} = m$:** Since $\mathbf{a}$ and $\mathbf{c}$ are unit vectors, $m$ is the cosine of the angle between them, so $-1 \leq m \leq 1$. 6. **Set up the equation:** $$k m + 1 = 2 \implies k m = 1 \implies k = \frac{1}{m}$$ 7. **Use the magnitude of $\mathbf{b}$:** $$|\mathbf{b}|^2 = \mathbf{b} \cdot \mathbf{b} = (k \mathbf{a} + \mathbf{c}) \cdot (k \mathbf{a} + \mathbf{c}) = k^2 (\mathbf{a} \cdot \mathbf{a}) + 2k (\mathbf{a} \cdot \mathbf{c}) + (\mathbf{c} \cdot \mathbf{c})$$ Since $\mathbf{a}$ and $\mathbf{c}$ are unit vectors: $$= k^2 + 2k m + 1$$ 8. **Given $|\mathbf{b}| = 4$, so:** $$k^2 + 2k m + 1 = 16$$ 9. **Substitute $k = \frac{1}{m}$ into the equation:** $$\left(\frac{1}{m}\right)^2 + 2 \times \frac{1}{m} \times m + 1 = 16$$ Simplify: $$\frac{1}{m^2} + 2 + 1 = 16$$ $$\frac{1}{m^2} + 3 = 16$$ $$\frac{1}{m^2} = 13$$ 10. **Solve for $m$:** $$m^2 = \frac{1}{13}$$ $$m = \pm \frac{1}{\sqrt{13}}$$ 11. **Find the two possible values of $k$:** $$k = \frac{1}{m} = \pm \sqrt{13}$$ **Final answer:** The two possible values of $k$ are $$k = \sqrt{13} \quad \text{and} \quad k = -\sqrt{13}$$