Solve the vector $\mathbf{x}$: $$\mathbf{x}\times \mathbf{\beta}=\mathbf{r},$$ $$\mathbf{x}.\mathbf{\alpha}=3,$$ where $\mathbf{\alpha}=\mathbf{i}+2\mathbf{j}+\mathbf{k}$, $\mathbf{\beta}=2\mathbf{i}-\mathbf{j}+\mathbf{k}$, $\mathbf{r}=-4(\mathbf{j}+\mathbf{k})$.
Please help me to solve the vector eqns.
Multiplying vectorially, by $\mathbf{\alpha}$
$(\mathbf{x}.\mathbf{\alpha})\mathbf{\beta}-(\mathbf{x}.\mathbf{\beta})\mathbf{\alpha}=\mathbf{r}\times\mathbf{\alpha}$ implies $(\mathbf{x}.\mathbf{\alpha})\mathbf{\beta}=2i+j-k$ Please help.
Let $\textbf{x}=a\textbf{i}+b\textbf{j}+c\textbf{k}$ $$(a\textbf{i}+b\textbf{j}+c\textbf{k})\times (2\textbf{i}-\textbf{j}+\textbf{k})=-4(\textbf{j}+\textbf{k})$$ $$(b+c)\textbf{i}-(a-2c)\textbf{j}+(-a-2b)\textbf{k}=-4(\textbf{j}+\textbf{k})$$
So we have $$b+c=0$$ $$a-2c=4$$ $$a+2b=4$$
Which gives $$b=-c$$ $$a=4+2c$$
The dot product gives $$(a\textbf{i}+b\textbf{j}+c\textbf{k})\cdot (\textbf{i}+2\textbf{j}+\textbf{k})=3$$ $$a+2b+c=3$$ $$4+2c-2c+c=3$$ $$c=-1$$ $$b=1$$ $$a=2$$
Alternative method: $$\alpha \times (x \times \beta)=\alpha\times r$$ $$(\alpha \cdot \beta)x-(\alpha\cdot x)\beta=\alpha \times r$$ $$1x-3\beta=\alpha\times r$$ $$x=\alpha \times r + 3\beta$$ $$=(1,2,1)\times (0,-4,-4)+3(2,-1,1)$$ $$=(-4,4,-4)+(6,-3,3)$$ $$=(2,1,-1)$$