{a}·{i}={a}·(2{i}+{j})={a}({i}+{j}+3{k})=1, then {a} is equal to :

Correct Answer is : 13(3i^−3j^+k^)1/3(3{i}-3{j}+{k})31(3i^−3j^+k^)

Correct Answer is : 13(3i^−3j^+k^)1/3(3{i}-3{j}+{k})31(3i^−3j^+k^)

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EAMCET Engineering 2006 Solved Paper

Section: Mathematics

Question : 118 of 160

Marks: +1, -0

\vec{a}\cdot\hat{i}=\vec{a}\cdot(2\hat{i}+\hat{j})=\vec{a}(\hat{i}+\hat{j}+3\hat{k})=1

\vec{a}

Solution:

\;\;\text{ Let }\; \vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}

\; \because \;\; \vec{a}\cdot\hat{i}=1

\; \therefore \;\; (a_1\hat{i}+a_2\hat{j}+a_3\hat{k})\cdot\hat{i}=1

\;\Rightarrow \;\; a_1=1

\;\;\text{ Now, }\; \;\; \vec{a}\cdot(2\hat{i}+\hat{j})=1

\;\Rightarrow \;\; (a_1\hat{i}+a_2\hat{j}+a_3\hat{k})\cdot(2\hat{i}+\hat{j})=1

\;\Rightarrow \;\; 2a_1+a_2=1

\;\Rightarrow \;\; a_2=1-2 \;\; (\because a_1=1)

\;\Rightarrow \;\; a_2=-1

\;\;\text{ and }\; \;\; \vec{a}\cdot(\hat{i}+\hat{j}+3\hat{k})=1

\;\Rightarrow \;\; (a_1\hat{i}+a_2\hat{j}+a_3\hat{k})\cdot(\hat{i}+\hat{j}+3\hat{k})=1

\;\Rightarrow \;\; a_1+a_2+3a_3=1

\;\Rightarrow \;\; 1-1+3a_3=1

\;\Rightarrow \;\; a_3=\;\frac{1}{3}

\; \therefore \vec{a}=\hat{i}-\hat{j}+\;\frac{1}{3}\hat{k}=\;\frac{1}{3}(3\hat{i}-3\hat{j}+\hat{k})

\;

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