Question

$A+B=a,A\times B=b,A\cdot a=1$

Find the vector $A$.

• A. $\frac{(a\times b)+a}{b^2}$
• B. $\frac{(a\times b)+a}{a^2}$
• C. $\frac{(a\times b)+b}{a^2}$
• D. $\frac{(a\times b)+b}{b^2}$

Answer 1

The correct option is B $\frac{(a\times b)+a}{a^2}$

given,

$A+B=a⟶\left(1\right)$, $A\times B=b⟶(2$), $A.a=1⟶\left(3\right)$

$\left(2\right)\Rightarrow A\times B=b$

$\overline{a}\times \left(2\right)\Rightarrow \overline{a}\times (A\times B)=\overline{a}\times \overline{b}$

$(\overline{a}.\overline{B})\overline{A}-(\overline{a}.\overline{A})\overline{B}=\overline{a}\times \overline{b}$ .......... $\left(4\right)$

$\left(1\right)\Rightarrow \overline{A}+\overline{B}=\overline{a}$

Multiply with $\overline{a}.$ on both sides (Dot produces)

$\Rightarrow \overline{A}.\overline{a}+\overline{B}.\overline{a}={\left(a\right)}^2$

$1+(\overline{B}.\overline{a})={\left(a\right)}^2$

$(\overline{B}.\overline{a})={\left(a\right)}^2-1$

$\left(4\right)\Rightarrow ({\left(a\right)}^2-1)\overline{A}-\overline{B}=\overline{a}\times \overline{b}$

Add $\left(1\right)and\left(4\right)$

$\Rightarrow ({\left(a\right)}^2-1)\overline{A}-\overline{B}+\overline{A}+\overline{B}=\overline{a}+\overline{a}\times \overline{b}$

$⟹\overline{A}=\frac{\overline{a}+\overline{a}\times \overline{b}}{a^2-1+1}=\frac{\overline{a}+(\overline{a}\times \overline{b})}{a^2}$