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Mathematics Question on Vector Algebra

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

A

i^+j^\hat {i}+\hat {j}

B

i^−k^\hat {i}-\hat {k}

C

i^\hat {i}

D

i^+j^−k^\hat {i}+\hat {j}-\hat{k}

Answer

i^\hat {i}

Explanation

Solution

Let a⃗=a1i^+a2j^+a3k^\vec{a} =a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}
Given, a⃗.i^=a⃗.(i^+j^)=a⃗.(i^+j^+k^)=1\vec{a} .\hat{i} =\vec{a} . \left(\hat{i} + \hat{j}\right) =\vec{a} .\left(\hat{i} + \hat{j} + \hat{k}\right) = 1
∴     a1=a1+a2=a1+a2+a3=1\therefore\,\,\,\,\, a_{1} =a_{1} + a_{2} =a_{1}+a_{2} + a_{3} = 1
⇒    a1=1,a2=0,a3=0\Rightarrow\,\,\,\, a_{1} = 1 , a_{2} = 0 , a_{3} = 0
∴    a⃗=i^\therefore\,\,\,\, \vec{a} = \hat{i}