Solution of vector equation to find x

Question

Solve for $\mathbf{x}$ in the vector equation $\;\mathbf{a}\wedge\mathbf{x}+\left(\mathbf{a}\cdot\mathbf{x}\right)\mathbf{a}+\mathbf{b}=0$.

I attempted dot product with $\mathbf{x}$:

$$ \mathbf{a\wedge x\cdot x + \left(a\cdot x\right)\left(a\cdot x\right)+b\cdot x=0\cdot x} \\ \mathbf{0+\left\lvert a\cdot x\right\rvert ^{2}+b\cdot x=0\cdot x} \\ \mathbf{\left\lvert a\cdot x\right\rvert ^{2} + b\cdot x = 0} $$ This does not take me to the answer which is $$ \mathbf{x}=\mathbf{\frac{a\wedge b}{a^2}}-\mathbf{\dfrac{\left(a\cdot b\right)a}{a^4}} $$

Answer 1

We consider the following

Let $\mathbf{x}$=$\lambda\mathbf{a}+\mu\mathbf{b}+\vartheta\mathbf{{aΛb}}$[Note that this is the general solution for any unknown vector]

We substitute the $\mathbf{x}$ in our equation.

$\mathbf{a}\wedge\mathbf({\lambda\mathbf{a}+\mu\mathbf{b}+\vartheta\mathbf{{aΛb}}) }+\left(\mathbf{a}\cdot(\mathbf{\lambda\mathbf{a}+\mu\mathbf{b}+\vartheta\mathbf{{aΛb}})$ }\right)\mathbf{a}+\mathbf{b}=\mathbf{0}$.

$\mu (\mathbf{a}\wedge\mathbf{b})$+$\vartheta\mathbf{a\wedge aΛb}$ +$\left(\lambda\mathbf{a}\cdot\mathbf{\mathbf{a}+\mu\mathbf{a.b}+\vartheta\mathbf{{a.aΛb}} }\right)\mathbf{a}+\mathbf{b}=\mathbf{0}$.

$\mu (\mathbf{a}\wedge\mathbf{b})$+$\vartheta(\mathbf{(a.b)a-(a.a)b)}$ +$\left(\lambda\mathbf{a}\cdot\mathbf{\mathbf{a}+\mu\mathbf{a.b}+\vartheta\mathbf{{a.aΛb}} }\right)\mathbf{a}+\mathbf{b}=\mathbf{0}$.

We equate coefficient

$\mathbf{a}$: $\vartheta(\mathbf{(a.b)}+\lambda\mathbf{a}\cdot\mathbf{a}+\mu\mathbf{a.b}$=$(\mathbf{(a.b)}(\vartheta + \mu)+\lambda a^2=0$

$\mathbf{b}$: $-\vartheta a^2+1=0$

$\mathbf{a}\wedge\mathbf{b}$: $\mu=0$

Now, we just replace.

$\mu=0$; $\vartheta=\frac{1}{a^2}$; $\lambda= -\frac{\mathbf{(a.b)}}{a^4}$

By replacing that we get our $$ \mathbf{x}=\mathbf{\frac{a\wedge b}{a^2}}-\mathbf{\dfrac{\left(a\cdot b\right)a}{a^4}} $$

Answer 2

$$ \pmb a\wedge \pmb x+(\pmb a\cdot \pmb x)\pmb a+\pmb b=0\tag 1 $$ Multiplying by $\pmb a \cdot(1)$ we have $$ \underbrace{\pmb a \cdot(\pmb a\wedge \pmb x)}_0+\pmb a \cdot(\pmb a\cdot \pmb x)\pmb a+\pmb a \cdot\pmb b=0\Longrightarrow (\pmb a\cdot \pmb x) a^2+\pmb a \cdot\pmb b=0\tag 2 $$ Multiplying by $\pmb a \wedge (1)$ we have $$ \underbrace{\pmb a \wedge (\pmb a\wedge \pmb x)}_{(\pmb a\cdot \pmb x)\pmb a-a^2\pmb x}+\underbrace{\pmb a \wedge (\pmb a\cdot \pmb x)\pmb a}_0+\pmb a \wedge \pmb b=0\Longrightarrow (\pmb a\cdot \pmb x) \pmb a -a^2\pmb x+\pmb a \wedge \pmb b=0\tag 3 $$ using the identity $ \pmb{A\wedge }\left(\pmb{B}\wedge \pmb{C}\right)=\left(\pmb{A}\cdot\pmb{C}\right)\pmb{B}-\left(\pmb{A}\cdot\pmb{B}\right)\pmb{C} $.

From (2) we find $(\pmb a\cdot \pmb x)= -\frac{\pmb a \cdot\pmb b}{a^2}$ and subtituting in $(3)$ we have $$ -\frac{\pmb a \cdot\pmb b}{a^2}\pmb a -a^2\pmb x+\pmb a \wedge \pmb b=0 $$ and finally

$$ \pmb x=\frac{\pmb a \wedge \pmb b}{a^2}-\frac{\pmb a \cdot\pmb b}{a^4}\pmb a $$