Basic vectors operations question

Question

given: $\vec{u} = -5\vec{i} + 3\vec{j} -4\vec{k}$

find a unit vector $\vec{v}=\alpha\vec{i}+\beta\vec{j}+\gamma\vec{k}$ such that $|\vec{u}+\vec{v}|$ is maximal.

So I was thinking first of all we know $\sqrt{\alpha^2+\beta^2+\gamma^2}=1$

and we want $\sqrt{(-5+\alpha)^2+(3+\beta)^2+(-4+\gamma)^2}$ to be maximal

but I'm not sure how to continue

Answer 1

If $u$ and $v$ are vectors in $\mathbb{R}^n$, then the value of $\| u + v \|$ is maximized when $u$ and $v$ point in the same direction, or there exists $c \in \mathbb{R}_{\geq 0}$ such that $v=cu$. This can be seen geometrically, or by using the fact that equality in the triangle inequality $\|u + v\| \leq \|u \| + \| v \|$ holds when $u$ and $v$ are pointing in the same direction.

Now if we impose the restriction that $v$ is a unit vector, we can take $v= \frac{u}{\|u\|}$, which is a unit vector and points in the same direction as $u$.

Answer 2

You have two choices. One is Lagrange multipliers. The other is writing the unit vector in spherical coordinates. Both methods should give the same result.

Take the spherical coordinate method: $$\vec v = \sin\theta\cos\phi \vec i+\sin\theta\sin\phi\vec j+\cos\theta\vec k$$ Then $$\vec u+\vec v= (\sin\theta\cos\phi -5)\vec i+(\sin\theta\sin\phi+3)\vec j+(\cos\theta-4)\vec k$$ Finally, you want to maximize $|\vec u+\vec v|$. For simplicity, note that it's the same as maximizing $|\vec u+\vec v|^2$. $$f(\theta,\phi)=|\vec u+\vec v|^2=(\sin\theta\cos\phi -5)^2+(\sin\theta\sin\phi+3)^2+(\cos\theta-4)^2$$ Then set the derivatives with respect to $\theta$ and $\phi$ to zero: $$(\sin\theta\cos\phi-5)\cos\theta+(\sin\theta\sin\phi+3)\cos\theta-(\cos\theta-4)\sin\theta=0\\-(\sin\theta\cos\phi-5)\sin\phi+(\sin\theta\sin\phi+3)\cos\phi=0$$

Answer 3

Once you have that you want to maximize $\sqrt{(-5+ \alpha)^2+ (3+ \beta)^2+ (-4+ \gamma)^2}$ it should be obvious to do the arithmetic: $\sqrt{25- 10\alpha+ \alpha^2+ 9+ 6\beta+ \beta^2+ 16- 8\gamma+ \gamma^2}= \sqrt{51- 10\alpha+ 6\beta- 8\gamma}$ since $25+ 9+ 16+ \alpha^2+ \beta^2+ \gamma^2= 25+ 9+ 16+ 1= 51$. Now, to find the values of $\alpha$, $\beta$, and $\gamma$ that maximize that it is sufficient to find values that maximize its square, $51- 10\alpha+ 6\beta- 8\gamma$ subject to the condition that $\alpha^2+ \beta^2+ \gamma^2= 1$.

And the standard method for finding a max or min of a function of several variables, subject to a constraint is "Lagrange multipliers".

$\nabla 51- 10\alpha+ 6\beta- 8\gamma= <-10, 6, -8>$ while $\nabla \alpha^2+ \beta^2+ \gamma^2= <2\alpha, 2\beta, 2\gamma>$. At a max or min there will be a "Multiplier", M, such that $2\alpha= -10M$, $2\beta= 6M$, and $2\gamma= -8M$. Together with the condition that $\alpha^2+ \beta^2+ \gamma^2= 1$, that gives us 4 equations to solve for $\alpha$, $\beta$, $\gamma$, and M.

Since a value of M is not really necessary to solve this problem a common first step in problems like this is to eliminate M by dividing one equation by another. Dividing the first equation by the second, we get $\frac{\alpha}{\beta}= -\frac{5}{3}$ so $\alpha= -\frac{5}{3}\beta$. Dividing the third equation by the second we get $\frac{\gamma}{\beta}= -\frac{4}{3}$ so $\gamma= -\frac{4}{3}\beta$.

Then $\alpha^2+ \beta^2+ \gamma^2= \frac{25}{9}\beta^2+ \beta^2+ \frac{16}{9}\beta^2= \frac{50}{9}\beta^2= 1$. $\beta= \sqrt{\frac{9}{50}}= \frac{3}{5\sqrt{2}}= \frac{3\sqrt{2}}{10}$ since $\beta$ is positive.

Then $\alpha= -\frac{5}{3}\beta= -\frac{5\sqrt{2}}{10}= -\frac{\sqrt{2}}{2}$ and $\gamma=-\frac{4}{3}\beta= \frac{4\sqrt{2}}{10}=\frac{2\sqrt{2}}{5}$.