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**1. Question:**

Find the directional derivatives of f(x, y, z) = x

^{2}+2xyz−yz

^{2}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.