Let v= i+2 j-3 k, w=2 i+ j- k, and u be a vector with | u|=0.

Question

Let $\vec v=\alpha\hat i+2\hat j-3\hat k$, $\vec w=2\alpha\hat i+\hat j-\hat k$, and $\vec u$ be a vector with $|\vec u|=\alpha>0$. If the minimum value of the scalar triple product $[\vec u\ \vec v\ \vec w]$ is $-\alpha\sqrt{3401}$, and $(\vec u\cdot\hat i)^2=\dfrac mn$ with $m,n$ coprime, then $m+n$ is \_\_\_\_\_\_\_.

Answer: 3501

Accepted Answer

$[\vec u\ \vec v\ \vec w] = \vec u\cdot(\vec v\times\vec w)$; its minimum (over direction of $\vec u$) is $-|\vec u||\vec v\times\vec w|$, attained when $\vec u$ is antiparallel to $\vec v\times\vec w$. So $|\vec v\times\vec w|=\sqrt{3401}$.

$$\vec v\times\vec w = \begin{vmatrix}\hat i&\hat j&\hat k\ \alpha&2&-3\2\alpha&1&-1\end{vmatrix} = \hat i-5\alpha\hat j-3\alpha\hat k$$
$$1+25\alpha^2+9\alpha^2 = 3401 \implies 34\alpha^2=3400 \implies \alpha^2=100 \implies \alpha=10$$

$\vec u = -\alpha\cdot\dfrac{\vec v\times\vec w}{|\vec v\times\vec w|}$ (antiparallel), so $\vec u\cdot\hat i = \mp\dfrac{\alpha}{\sqrt{3401}}$, giving
$$(\vec u\cdot\hat i)^2 = \frac{\alpha^2}{3401} = \frac{100}{3401} = \frac mn$$

$m=100,\ n=3401$ (coprime), so $m+n=3501$.

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