Question

The angle θ between 5i-j+k & 2i-j+k is​

Answer 1

Explanation:

Let,

`\sf \vec{a} = 5 \hat{i} - \hat{j} + \hat{k} \\ \therefore \: \sf \: | \vec{a}| = \sqrt{ {5}^(2) + {( - 1)}^(2) + {1}^(2) } \\ = √(25 + 1 + 1) \\ = √(27) \\ \\ \sf \vec{b} = 2\hat{i} - \hat{j} + \hat{k} \\ \therefore \: \sf \: | \vec{b}| = \sqrt{ {2}^(2) + {( - 1)}^(2) + {1}^(2) } \\ = √(4 + 1 + 1) \\ = √(6) \\ \\\sf \: \vec{a}. \vec{b} = (5 \hat{i} - \hat{j} + \hat{k}).(2\hat{i} - \hat{j} + \hat{k}) \\ = 5 * 2 + ( - 1) * ( - 1) + 1 * 1 \\ = 10 + 1 + 1 \\ = 12 \\ \\ \sf \: angle \: between \: \vec{a} \: and \: \vec{b} \: = \theta \\ \\ \: so \\ \sf \vec{a}. \vec{b} = | \vec{a}| . | \vec{b}| cos\theta \\ = > \sf \: cos \theta \: = \frac{ \vec{a}. \vec{b}}{ | \vec{a}| . | \vec{b}| } \\ = > cos \theta = (12)/( √(27) * √(6) ) = 0.94 \\ = > \theta = {cos}^( - 1) (0.94) \\ = > \green{\theta = 19.47 ^( \circ) }`