Question

find the angle between two vector​

Answer 1

AnswEr :

Given,

• $\sf \vec{a} = \hat{i} + \hat{j} - \hat{k}$

• $\sf \vec{b} = 2\hat{i} + 3 \hat{j} + \hat{k}$

The Dot Product between any two vectors is given as :

$\sf \: \vec{a} \: . \: \vec{b} = | \vec{a}| | \vec{b}| cos( \alpha )$

For the angle between the two vectors :

$\implies \: \sf \: \cos( \alpha ) = \dfrac{ \vec{a}. \vec{b}}{ | \vec{a} || \vec{b}| }$

The magnitude of a vector r = xi + yj + zk is calculated as :

$\sf | \vec{r}| = \sqrt{ {x}^{2} + {y}^{2} + {z}^{2} }$

Thus,

$\sf \: | \vec{a}| = \sqrt{1 {}^{2} + {1}^{2} + ( - {1})^{2} } = \sqrt{3}$

Also,

$\sf \: | \vec{b}| = \sqrt{2 {}^{2} + {3}^{2} + {1}^{2} } = \sqrt{14}$

Now,

$\sf \vec{a} . \vec{b}= (\hat{i} + \hat{j} - \hat{k})(2 \hat{i} + 3 \hat{j} + \hat{k}) \\ \\ \longrightarrow \: \sf \vec{a}. \vec{b} = 2 { \hat{i}}^{2} + 3 \hat{j} {}^{2} - \hat{k} {}^{2} \\ \\ \longrightarrow \: \sf \: \vec{a}. \vec{b} = 2 + 3 - 1 \\ \\ \longrightarrow \sf \: \vec{a}. \vec{b} = 4$

Therefore,

$\sf \: cos( \alpha ) = \dfrac{4}{ \sqrt{14} \times \sqrt{3} } \\ \\ \implies \: \sf cos( \alpha ) = \dfrac{4}{ \sqrt{42} } \\ \\ \implies \: \sf \: cos( \alpha ) = \dfrac{2 \sqrt{2} }{ \sqrt{21} } \\ \\ \implies \boxed{ \boxed{\sf \: \alpha = cos {}^{ - 1} ( \dfrac{2 \sqrt{2} }{ \sqrt{21} } ) }}$

The angle between the two vectors is arccos(2√2/√21)