Question

If ${\displaystyle \overrightarrow{a},\overrightarrow{b},\overrightarrow{c}}$ are three vectors such that ${\displaystyle \overrightarrow{a}.\overrightarrow{b}=\overrightarrow{a}.\overrightarrow{c}}$ and ${\displaystyle \overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{a}\times \overrightarrow{c},\overrightarrow{a}\ne 0}$, then show that ${\displaystyle \overrightarrow{b}=\overrightarrow{c}}$.

Answer 1

Given, ${\displaystyle \overrightarrow{a}.\overrightarrow{b}=\overrightarrow{a}.\overrightarrow{c}}$

⇒ ${\displaystyle \overrightarrow{a}.\overrightarrow{b}-\overrightarrow{a}.\overrightarrow{c}}$ = 0

⇒ ${\displaystyle \overrightarrow{a}.(\overrightarrow{b}-\overrightarrow{c})}$ = 0

⇒ Either ${\displaystyle \overrightarrow{b}=\overrightarrow{c}}$ or ${\displaystyle \overrightarrow{a}\perp (\overrightarrow{b}-\overrightarrow{c})}$

Also, given ${\displaystyle \overrightarrow{a}\times \overrightarrow{b}-\overrightarrow{a}\times \overrightarrow{c}}$

⇒ ${\displaystyle \overrightarrow{a}\times \overrightarrow{b}-\overrightarrow{a}\times \overrightarrow{c}}$ = 0

⇒ ${\displaystyle \overrightarrow{a}\times (\overrightarrow{b}-\overrightarrow{c})}$ = 0

⇒ Either ${\displaystyle \overrightarrow{a}\mid \mid (\overrightarrow{b}-\overrightarrow{c})}$ or ${\displaystyle \overrightarrow{b}=\overrightarrow{c}}$

But vector ${\displaystyle \overrightarrow{a}}$ a cannot be both parallel and perpendicular to vector ${\displaystyle (\overrightarrow{b}-\overrightarrow{c})}$.

Hence, vector ${\displaystyle \overrightarrow{b}=\overrightarrow{c}}$.