MCQ
If $\vec{\text{a}}.\vec{\text{b}}=\vec{\text{a}}.\vec{\text{c}}$ and $\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{a}}\times\vec{\text{c}}.\vec{\text{a}}\neq0,$ then:
  • $\vec{\text{b}}=\vec{\text{c}}$
  • B
    $\vec{\text{b}}=\vec{0}$
  • C
    $\vec{\text{b}}+\vec{\text{c}}=\vec{0}$
  • D
    $\text{None of these}$

Answer

Correct option: A.
$\vec{\text{b}}=\vec{\text{c}}$
$\vec{\text{a}}.\vec{\text{b}}=\vec{\text{a}}.\vec{\text{c}}$
$\Rightarrow\vec{\text{a}}.\vec{\text{b}}-\vec{\text{a}}.\vec{\text{c}}=0$
$\Rightarrow\vec{\text{a}}.\big(\vec{\text{b}}-\vec{\text{c}}\big)=0$
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\big(\vec{\text{b}}-\vec{\text{c}}\big)$
$|\vec{\text{a}}|\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|\cos\theta\dots(1)$
and $\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{a}}\times\vec{\text{c}}$
$\Rightarrow\vec{\text{a}}\times\vec{\text{b}}-\vec{\text{a}}\times\vec{\text{c}}=0$
$\Rightarrow\vec{\text{a}}\times\big(\vec{\text{b}}-\vec{\text{c}}\big)=0$
Then, $|\vec{\text{a}}|\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|\sin\theta=0\dots(2)$
Here, it is given that $\vec{\text{a}}\neq0$
Therefore, for eq. (1) and eq. (2) to be 0
We have,
$\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|\cos\theta=0$
For $\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|\cos\theta=0,$ one of $\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|$ or $\cos\theta$ must be 0
Case 1 :
Let $\cos\theta=0$
$\Rightarrow\theta=90^\circ$
$\Rightarrow\sin\theta=1$
& if $\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|\sin\theta=0$ and $\sin\theta=1$
Then $\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|=0$
$\Rightarrow\vec{\text{b}}=\vec{\text{c}}$
Case 2 :
Let $\big|\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|=0$
$\Rightarrow\vec{\text{b}}=\vec{\text{c}}$
Hence, $\vec{\text{b}}=\vec{\text{c}}$

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