Question

A=î, B=2î + K now what is the angle between A & B Vector?? I ans: 26.577° (explain how?)​

Answer 1

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Given :- }}}$

$\sf\implies Vector \ A \ = \ \hat{i} \\\sf\implies Vector \ B \ =\ 2\hat{i}+\hat{k} .$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: To \ Find :- }}}$

$\implies\textsf{ The angle between A and B .}$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Answer :- }}}$

Here two vectors are given to us and we need to find the angle between them m.The vectors are A= î and B = 2î + $\sf\hat{k}$ . We know dot product as |A||B| cosθ ( for vectors A & B ) so ,

$\sf:\implies\pink{ \vec{A}. \vec{B}= |A| |B| \cos\theta }\\\\\sf:\implies cos\theta = \dfrac{ \vec{A}. \vec{B}}{|A| |B|}\\\\\sf:\implies cos\theta = \dfrac{ ( \hat{i} + 0\hat{j} + 0\hat{k})(2\hat{i} + 0\hat{j} + 1\hat{k})}{ (\sqrt{ 1^2 + 0^2+0^2})(\sqrt{ 2^2+0^2+1^2})} \\\\\sf:\implies cos\theta = \dfrac{ 1(2) + 0(0)+0(1)}{ (1)(\sqrt{5})} \\\\\sf:\implies cos\theta= \dfrac{ 2}{\sqrt5} \\\\\sf:\implies \boxed{\pink{\sf \theta = cos^{-1}\bigg( \dfrac{ 2}{\sqrt5}\bigg) }}$

$\underline{\blue{\sf \therefore Hence \ the \ angle \ between\ them \ \ is \ \bf{\textsf{\textbf{cos}}^{-\textsf{\textbf{1}}}\dfrac{\textsf{\textbf{2}}}{\sqrt5} }. }}$