### Instruction: If $\|\mathbf{a}\| = 3$ and $\|\mathbf{b}\| = 4,$ then find all $k$ for which the vectors $\mathbf{a} + k \mathbf{b}$ and $\mathbf{a} - k \mathbf{b}$ are orthogonal.### Response: I know that two vectors are orthogonal if and only if their dot product is zero.
So, I want to find all $k$ that make $(\mathbf{a} + k \mathbf{b}) \cdot (\mathbf{a} - k \mathbf{b}) = 0.$
I can expand this dot product using the distributive property and the fact that $\mathbf{a} \cdot \mathbf{a} = \|\mathbf{a}\|^2$ and $\mathbf{b} \cdot \mathbf{b} = \|\mathbf{b}\|^2.$
I get $(\mathbf{a} + k \mathbf{b}) \cdot (\mathbf{a} - k \mathbf{b}) = \|\mathbf{a}\|^2 - k^2 \|\mathbf{b}\|^2 + k (\mathbf{a} \cdot \mathbf{b} - \mathbf{a} \cdot \mathbf{b}).$
Simplifying, I get $(\mathbf{a} + k \mathbf{b}) \cdot (\mathbf{a} - k \mathbf{b}) = \|\mathbf{a}\|^2 - k^2 \|\mathbf{b}\|^2.$
Now, I can plug in the values of $\|\mathbf{a}\|$ and $\|\mathbf{b}\|$ that are given in the problem.
I get $(\mathbf{a} + k \mathbf{b}) \cdot (\mathbf{a} - k \mathbf{b}) = 3^2 - k^2 4^2 = 9 - 16k^2.$
This is equal to zero when $9 = 16k^2,$ or when $k = \pm \frac{3}{4}.$
Therefore, the vectors $\mathbf{a} + k \mathbf{b}$ and $\mathbf{a} - k \mathbf{b}$ are orthogonal when $k = \pm \frac{3}{4}.$