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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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If the directional derivative of \[\phi(x,y,z)=ax^2y+by^2z+cz^2x\] at the point (1,1,1) has maximum magnitude 15 in the direction parallel to the line \[\frac{x-1}{2}=\frac{y-3}{-2}=\frac{z}{1}\] find the values of a,b, and c I've calculated \[\vec \nabla \phi(1,1,1)=(2a+c)\hat i+(a+2b)\hat j+(b+2c)\hat k\] The given line is parallel to the vector \[\vec b=2\hat i-2\hat j+\hat k\] Since \[\vec \nabla \phi(1,1,1) \parallel \vec b\] \[\implies (2a+c)\hat i+(a+2b)\hat j+(b+2c)\hat k=\lambda2\hat i+-\lambda2\hat j+\lambda \hat k\]\[\frac{2a+c}{2}=\frac{2b+a}{-2}=\frac{2c+b}{1}=\lambda\] Now using the equation for magnitude, \[|\vec \nabla \phi(1,1,1,)|^2=(2a+c)^2+(a+2b)^2+(b+2c)^2=15^2\] I've calculated \[\lambda=\pm5\] Thus 3 equations i've found \[2a+c=\pm10\]\[a+2b=\pm10\]\[b+2c=\pm5\] Now how can I solve this without the annoying determinant method or without having to remember that nasty cross multiply formula
Sorry 2nd equation is \[a+2b=\mp10\]

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Meh I ended up using the determinant method anyway, \[\left[\begin{matrix}2 & 0&1 \\ 1 & 2&0\\0&1&2\end{matrix}\right]\left[\begin{matrix}a \\ b\\c\end{matrix}\right]=\left[\begin{matrix}\pm10 \\ \mp10\\ \pm5\end{matrix}\right]\] AX=B \[\det(A)=9\] \[adj(A)=\left[\begin{matrix}4 & 1&-2 \\ -2 & 4&1\\1&-2&4\end{matrix}\right]\] \[X=\frac{adj(A)}{\det(A)}B\] \[X=\left[\begin{matrix}4/9 & 1/9&-2/9 \\ -2/9 & 4/9&1/9\\1/9&-2/9&4/9\end{matrix}\right]\left[\begin{matrix}\pm10 \\ \mp10\\ \pm5\end{matrix}\right]\] \[X=\left[\begin{matrix}\pm\frac{20}{9} \\ \mp\frac{55}{9}\\ \pm\frac{50}{9}\end{matrix}\right]\] \[a=\pm\frac{20}{9}\]\[b=\mp\frac{55}{9}\]\[c=\pm\frac{50}{9}\]
looks good to me!
Didn't take me as long as I thought
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@Nishant_Garg i posted that because i took a slightly different tack, but i am not really sure it is any easier admin-wise in terms of cranking out an answer
Great stuff, I think it would've been just as good!

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