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If \( \vec{a}=3 \hat{\imath}-2 \hat{\jmath}+\hat{k}, \vec{b}=2 \hat{\imath}-4 \hat{\jmath}+\hat{k} \), then find \( |\vec{a}-2 \vec{b}| \).

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\(\vec a=3\hat i-2\hat j+\hat k\)

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

\(\vec a.\vec b\) = \((3\hat i-2\hat j+\hat k).(2\hat i-4\hat j+\hat k)\)

 = 6 + 8 + 1 = 15

\(|\vec a|=\sqrt{9+4+1}=\sqrt{14}\)

\(|\vec b|=\sqrt{4+16+1}=\sqrt{21}\) 

\(|\vec a-2\hat b|^2=(\vec a-2\vec b).(\vec a-2\vec b)\) 

 = \(|\vec a|^2-4\vec a \vec b+4|\vec b|^2\)

 = 14 - 4 x 15 + 4 x 21 = 14 - 60 + 84 = 38

∴ |\(\vec a-2\vec b\)| = \(\sqrt{38}\)

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