Question: Let w = <1,1,1>. Find a unit vector v such that the angle between v and w is π/4.
My calculation: let v =
cos(pi/4) = sqrt(2)/2
sqrt(2)/2 = (a(1)+b(1)+c(1))/(sqrt(1^2 +1^2+1^2)*sqrt(a^2+b^2+c^2))
let a^2+b^2+c^2 = 1
=> sqrt(2)/2 = (a+b+c) / sqrt(3)
=> sqrt(2)*sqrt(3) / 2 = a + b + c
and from here I am stuck..
From my class right now (intro to lin arg), class covered projection last friday meaning we did not cover rotation.. etc.
Please help! Thank you.
The unit vectors making an angle of $\frac{\pi}{4}$ about the vector $\mathbf{w}=\langle 1,1,1\rangle$ will lie in a cone with center axis containing the origin and having direction $\mathbf{w}$. So one of them will lie directly "underneath" $\mathbf{w}$ in the sense that it will have coordinates of the form
$$ \mathbf{v}=\langle a,a,c\rangle $$
In other words, there will be a solution where $b=a$ in your approach.
Then you get two non-linear equations in two variables.
Graphed in the "$ac$-plane" these will be a straight line and an ellipse with possibly two solutions.
From the first equation set $c=\dfrac{\sqrt{6}}{2}-2a$ and substitute the value of $c^2$ into the second equation. This gives a somewhat messy quadratic equation in $a$, but it can be solved.
Select the appropriate solution for $a$ and find the corresponding value for $c$ and you will have a correct answer. Expect the answer to contain a square root or two.
ADDENDUM: Should you wish to continue your chosen approach to the problem and use my advice on how to continue down that path one of the solutions you obtain should be
$$ \mathbf{v}=\left\langle\frac{\sqrt{6}+\sqrt{3}}{6},\frac{\sqrt{6}+\sqrt{3}}{6},\frac{\sqrt{6}-2\sqrt{3}}{6} \right\rangle$$