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Let \(\vec{a}=\hat{j}-\hat{k}\) and\(\vec{c}=\hat{i}-\hat{j}-\hat{k} \) . Then the vector \(\vec{b}\) satisfying \(\vec{a} \times \vec{b}+\vec{c}=\overrightarrow{0} \ and \ \vec{a} . \vec{b}=3\) is :

(1)\(-\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}\)

(2) \(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}\) 

(3)\(\hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}}\)

(4) \(\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}\) 

from [AIEEE-2010]
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1 Answer

 
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Option 1 is correct \(-\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}\)

Explaination::

a x b + c = 0

a x (a x b) + a x c = 0

(a.b)a - (a.a)b + a x c = 0

3a - 2b + (-2i - j - k) = 0

2b = 3(j - k) + (-2i - j - k)

2b=3j-3k-2i-j-k

b=\(-2i+2j-4k\over2\)

b=-i+j-2k

b=\(-\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}\)

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