1. Question: Find the directional derivatives of f(x, y, z) = x2+2xyz−yz2 at (1, 1, 2) in the directions parallel to the line (x−1)/2 = y − 1 = (z−2)/-3. 2. Solution: We have ∇f = (2x + 2yz)i + (2xz - z2)j + (2xy - 2yz)k. Therefore, ∇f(1, 1, 2) = 6i - 2k. The given line is parallel to the vector v = (2, 1, -3). The corresponding unit vectors are u = ± 1/||v|| and v = (±1/√14)(2, 1, -3). For the directional derivatives we find f'u±(1, 1, 2) = ∇f(1, 1, 2)dot(u) = ±18/√14 3. My Questions: The only part of this that I have no clue about is how do they get the vector v from the information given. Could someone please explain how they find that direction vector? I get that since the lines are parallel the direction vector is the same but how does one find the directional vector? Any help is greatly appreciated. Thanks.