# Find the angle between​​ ​​vector A = i + j - 2k and vector B = i + j - k

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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 = $\frac{4}{\sqrt{18}}$