Let {a}={j}-{k} and {c}={i}-{j}-{k}. Then the vector {b} satisfying {a}× {b}+{c}={0} and {a}· {b}=3 [AIEEE 2010]

Correct Answer is : −i^+j^−2k^-{i}+{j}-2 {k}−i^+j^−2k^

Correct Answer is : −i^+j^−2k^-{i}+{j}-2 {k}−i^+j^−2k^

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AIEEE 2010 Solved Paper

Section: Mathematics

Question : 85 of 89

Marks: +1, -0

Let

\vec{a}=\hat{j}-\hat{k}

\vec{c}=\hat{i}-\hat{j}-\hat{k}

\vec{b}

\vec{a}\times \vec{b}+\vec{c}=\vec{0}

\vec{a}\cdot \vec{b}=3

Solution:

\vec{c}=\vec{b}\times \vec{a}

\Rightarrow \vec{b}\cdot \vec{c}=\vec{b}\cdot(\vec{b}\times \vec{a}) \Rightarrow \vec{b}\cdot \vec{c}=0

\Rightarrow (b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}) \cdot(\hat{i}-\hat{j}-\hat{k})=0

where

\vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}

b_1-b_2-b_3=0 \;\; \dots (i)

\;\;\text{ and }\; \vec{a}\cdot \vec{b}=3

\;\Rightarrow(\hat{j}-\hat{k}) \cdot (b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}) =3

\;\Rightarrow b_2-b_3=3

From equation

(i)

\;b_1=b_2+b_3= (3+b_3) +b_3=3+2 b_3

\;\vec{b}= (3+2 b_3) \hat{i}+ (3+b_3) \hat{j}+b_3 \hat{k}

From the option given, it is clear that

b_3

equal to either 2 or

-2

b_3=2

then

\vec{b}=7 \hat{i}+5 \hat{j}+2 \hat{k}

which is not possible

b_3=-2

, then

\vec{b}=-\hat{i}+\hat{j}-2 \hat{k}

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