Question

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

Answer 1

Step-by-step explanation:

Let,

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