Solution set to cross product

Question

If $\vec a,\vec b \in \mathbb{R}^3$ with $|\vec a|\ne0$ show that the equation $\vec a \times \vec u =\vec b$ has a solution if and only if $a \cdot b = 0$ and find all the solutions in this case.

The answer for the general solution to u is,

$-\dfrac{\vec a \times \vec b}{\lvert \vec a \rvert^2} + t \vec a $, where $t$ is a real parameter.

The first part is trivial, but I have no idea how to find the solution set. Could anybody shed any light on this matter? I would be very grateful.

Answer 1

Hint: $(\vec a\times \vec b)$ is $\perp$ to $\vec a$ and $\vec b$ both.So, you get $\vec a \cdot \vec b=0$ and $$\vec a\times \vec a=0$$ So, general solution for $\vec u=t\vec a+ \vec c$ . where $\vec a , \vec c $ are linearly independent (ie. $\vec a\cdot \vec c=0$) and $\vec a\times \vec c=\vec b$.

Now you can proceed. :) You have to find $\vec c$ satisfying it.

HINT: You need to use vector triple product. Pre-cross multiply with $\vec a$ on both sides of $\vec a\times \vec c=\vec b$.

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$\vec x\times (\vec y \times \vec z)=(\vec x \cdot \vec z) \vec y - (\vec x \cdot \vec y)\vec z$