Solving vector equation 2

Question

Using vector method, show that the vector equation $$\bar{x}\times \bar{a}+(\bar{x}.\bar{b})\bar{c}=\bar{d}$$ is satisfied if $$\bar{x}=\lambda \bar{a}+\bar{a}\times \frac{\bar{a}\times (\bar{d}\times \bar{c})}{(\bar{a}.\bar{c}){a^2}}$$

Attempt

Taking dot product by $\bar{a}$

$0+(\bar{x}.\bar{b})(\bar{a}.\bar{c})=(\bar{a}.\bar{d})$

i.e $\bar{x}.\bar{b}=\frac{\bar{a}.\bar{d}}{\bar{a}.\bar{c}}$

Let $p=\frac{\bar{a}.\bar{d}}{\bar{a}.\bar{c}}$

Then $\bar{x}.\bar{b}=p =p \frac{\bar{b}.\bar{b}}{|\bar{b}|^2} \implies \left(x-p\frac{\bar{b}}{|\bar{b}|^2}\right).\bar{b}=0$

Its general solution is $\bar{x}=p\frac{\bar{b}}{|\bar{b}|^2}+\bar{t}\times \bar{b}=\frac{\bar{a}.\bar{d}}{\bar{a}.\bar{c}}\frac{\bar{b}}{|\bar{b}|^2}+\bar{t}\times \bar{b}$ where $\bar{t}$ is any vector.

Please help. My solution is not desired.

Answer 1

$\mathbf{a \times x} =\mathbf{u} \implies \displaystyle \mathbf{x}=\lambda \mathbf{a}-\frac{\mathbf{a\times u}}{a^2} $

Now,

\begin{align*} \mathbf{a\times x} &=(\mathbf{x\cdot b}) \mathbf{c-d} \\ &= \frac{\mathbf{a\cdot d}}{\mathbf{a\cdot c}} \mathbf{c-d} \\ &= \frac{(\mathbf{a\cdot d})\mathbf{c}-(\mathbf{a\cdot c})\mathbf{d}} {\mathbf{a\cdot c}} \\ &= -\frac{\mathbf{a} \times (\mathbf{d\times c})} {\mathbf{a\cdot c}} \\ \end{align*}

Can you proceed?