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We have f(x,y,z) at point $m$ the directional derivative is maximum at the direction of vector $u=(1,1,1)$ And is equal to $\sqrt{27}$

find for which a the directional derivative for f in point m in direction of vector (1,2,a) is 5?

Since the D'D at point $m$ is maximum at the direction of vector $u=(1,1,1)$ that means $\triangledown f(m)$ is equal to the direction of vector $u$ and the direction of $u$ is $\frac{u}{\left \| u \right \|}=\frac{1}{ \sqrt{3}}$ that is supposed to mean that gardient vector for $\triangledown f(m)=\alpha \vec{u}$ for some $\alpha\epsilon \mathbb{Z}$

Am I correct so far? and if that's so i dont know how to continue from here

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If the directional derivative of $f$ is at a maximum in some direction then $\nabla f$ points in that direction.

What is $\nabla f$

$\nabla f = \mu (1,1,1)\\ \nabla f \cdot (\frac{1}{\sqrt 3},\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}) = \sqrt {27}\\ \mu\sqrt 3 = \sqrt {27}\\ \mu = 3\\ \nabla f = (3,3,3)$

What is the directional derivative in the direction $(1,2,a)$?

$\nabla f\cdot \frac {(1,2,a)}{\|(1,2,a)\|} = \frac {9+3a}{\sqrt {5+a^2}}$

We have been told that this equals $5$

$\frac {9+3a}{\sqrt {5+a^2}} = 5\\ 9+3a = 5\sqrt {5+a^2}$

Square both sides and solve the quadratic for $a.$

$81 + 54a + 9a^2 = 125 + 25a^2\\ 16a^2 - 54a + 44 = 0\\ 2(8a - 11)(a-2)=0\\ a = \frac {11}{8}, 2$

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