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Find the angle between \(\vec A=\hat i+\hat j-2\hat k \,and\, \vec B=\hat i+\hat j-\hat k \)

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We have

\(\vec A.\vec B\)= | A | | B | cos θ

Here  | A | = \(\sqrt{1^2+1^2+(-2)^2}=\sqrt6\)

| B | =\(\sqrt{1^2+1^2+(-1)^2}=\sqrt3\)

\(\vec A.\vec B=(\hat i+\hat j-\hat k).(\hat i+\hat j-\hat k)\)

= 1 × 1 + 1 × 1 + (-2)(-1)

= 1 + 1 + 2 = 4 

∴   cos θ = \(\frac{\vec A.\vec B}{|A||B|}=\frac{4}{\sqrt6\sqrt3}\)

cos θ = \(\frac{4}{\sqrt{18}}\)

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