Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am stuck on the following

Show that the vector equation $r\wedge a=b$ has a solution $$r=\lambda a + \frac {a \wedge b}{|a|^{2}}$$

Show that the vector $r\wedge a=b$ and $r\wedge c=d$, with $a\wedge c \neq 0$, $a\cdot b=0$ and $c\cdot d=0$, have a simultaneous solution for r provided that $a\cdot d+b\cdot c=0$.

I have got the first part with taking the cross product the $r\wedge a=b$ w.r.t a to both sides, but struggling on the second part, many thanks in advance.

share|cite|improve this question
up vote 2 down vote accepted

For the second part, if there exists a simultaneous solution there must exists constants $\lambda,\mu$ such that

$$ r = \lambda a + \frac{a\wedge b}{|a|^2} = \mu c + \frac{c\wedge d}{|c|^2} $$

by the first part of your question. Then

$$ r\wedge c = d \implies \lambda a \wedge c + \frac{1}{|a|^2}(a\wedge b) \wedge c = d \tag{1}$$


$$ r\wedge a = b \implies -\mu a \wedge c + \frac{1}{|c|^2}(c\wedge d)\wedge a = b $$

We consider (1). The second equation can be solved analogously.

Rewrite (1) as

$$ \lambda a\wedge c = d - \frac{1}{|a|^2}(a\wedge b)\wedge c $$

we see that to have a solution $\lambda$ we need that the vector $a\wedge c$ and the vector $d - \frac{1}{|a|^2}(a\wedge b)\wedge c$ are collinear, in which case we can "divide by $a\wedge c$" (since we assumed that $a\wedge c \neq 0$) and get the correct factor $\lambda$. Now, we know that $a\wedge c$ is orthogonal to both $a$ and $c$. Since we are working in three dimensions, to show that the right hand side is collinear with $a\wedge c$ it suffices to show that it is also orthogonal to both $a$ and $c$.

Now recall the vector triple product formula which gives $(a\wedge b)\wedge c = -(c\cdot b) a + (c\cdot a)b$.

Taking the dot products we require

$$ 0 = d\cdot c - \frac{1}{|a|^2} \left[ (c\cdot a)(b\cdot c) - (c\cdot b)(a\cdot c)\right] \tag{2}$$


$$ 0 = d\cdot a - \frac{1}{|a|^2} \left[ (c\cdot a)(b\cdot a) - (c\cdot b)(a\cdot a)\right] \tag{3}$$

Now, under the assumption that $c\cdot d = 0$, equation (2) holds automatically. Under the assumption that $b\cdot a = 0$, equation (3) holds if and only if $$ 0 = a\cdot d + b\cdot c$$ which is the condition you desire.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.