If \overrightarrow{a} \cdot \stackrel{\^}{i}=\overrightarrow{a} \cdot(\stackrel{\^}{i}+\stackrel{\^}{j})=\overrightarrow{a} \cdot(\stackrel{\^}{i}+\stackrel{\^}{j}+\stackrel{\^}{k})=1, then \overrightarrow{a} is

\stackrel{\^}{i}+\stackrel{\^}{j}+\stackrel{\^}{k}

CBSE Class 12 Math 2023 Delhi Set 3 Solved Paper - Question 6

Question : 6 of 12

Marks: +1, -0

{a}↖{→} ⋅ {i}↖{\^}={a}↖{→} ⋅({i}↖{\^}+{j}↖{\^})={a}↖{→} ⋅({i}↖{\^}+{j}↖{\^}+{k}↖{\^})=1

, then

{a}↖{→}

Solution:

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