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Let a⃗\vec{a} be any vector such that a⃗.i^=a⃗.(i^+j^)=a⃗.(i^+j^+k^)=1\vec{a}.\hat{i} = \vec{a}.(\hat{i}+\hat{j}) = \vec{a}.(\hat{i}+\hat{j}+\hat{k}) = 1, then a⃗\vec{a} is

Solution

✅ Correct Option: 2

From a⃗⋅i^=1\vec a\cdot\hat i=1 and a⃗⋅(i^+j^)=1\vec a\cdot(\hat i+\hat j)=1 we get a⃗⋅j^=0\vec a\cdot\hat j=0; comparing the second and third conditions gives a⃗⋅k^=0\vec a\cdot\hat k=0. So the components are 1,0,01,0,0 and a⃗=i^\vec a=\hat i.

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