Question

If $\overline{x}+\overline{y}=\overline{a},\overline{x}\times \overline{y}=\overline{b},\overline{x}\cdot \overline{a}=1$ find $\overline{x}$ and $\overline{y}$

• A. $x=\frac{\overline{a}+\overline{a}\times \overline{b}}{a^2},y=\overline{a}-\overline{x}$
• B. $x=\frac{\overline{a}-\overline{a}\times \overline{b}}{a^2},y=\overline{a}-\overline{x}$
• C. $x=\frac{\overline{a}+\overline{a}\times \overline{b}}{a^2},y=\overline{a}+\overline{x}$
• D. $x=\frac{\overline{a}-\overline{a}\times \overline{b}}{a^2},y=\overline{a}+\overline{x}$

Answer 1

The correct option is A $x=\frac{\overline{a}+\overline{a}\times \overline{b}}{a^2},y=\overline{a}-\overline{x}$

Consider $\overline{x}=i$ and $\overline{y}=j$

Then

$\overline{a}=i+j$

And

$\overline{b}=k$

Then

$\frac{\overline{a}+(\overline{a}\times \overline{b})}{|a{|}^2}$

$=\frac{i+j+\left(\right(i+j)\times k)}{2}$

$=\frac{i+j-j+i}{2}$

$=\frac{2i}{2}$

$=i$

$=\overline{x}$

And

$\overline{a}-\overline{x}$

$=i+j-i$

$=j$

$=\overline{y}$

Hence the correct is option A.