Question

If $\vec{a}\cdot\hat {i}= \vec{a} \cdot(\hat {i}+\hat {j})= \vec{a}\cdot (\hat {i}+\hat {j}+\hat {k})=1$ then $\vec{a}$=

• A. $\hat {i}+\hat {j}$
• B. $\hat {i}-\hat {k}$
• C. $\hat {i}$
• D. $\hat {i}+\hat {j}-\hat{k} $

Answer 1

Answer: (C)

Let $\vec{a} =a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} $
Given, $ \vec{a} .\hat{i} =\vec{a} . \left(\hat{i} + \hat{j}\right) =\vec{a} .\left(\hat{i} + \hat{j} + \hat{k}\right) = 1 $
$\therefore \,\,\,\,\, a_{1} =a_{1} + a_{2} =a_{1}+a_{2} + a_{3} = 1$
$\Rightarrow \,\,\,\, a_{1} = 1 , a_{2} = 0 , a_{3} = 0 $
$\therefore \,\,\,\, \vec{a} = \hat{i} $