Question

Let a and b be non zero , non collinear vectors . If |a+b| = |a-b| . find the angle between a and b​

Answer 1

$\textbf{Given:}$

$\mathsf{\overrightarrow{a}\;and\;\overrightarrow{b}\;are\;two\;non-zero\;vectors\;and}$

$\mathsf{|\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}-\overrightarrow{b}|}$

$\textbf{To find:}$

$\textsf{The angle between}\;\mathsf{\overrightarrow{a}\;and\;\overrightarrow{b}}$

$\textbf{Solution:}$

$\mathsf{Consider,}$

$\mathsf{|\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}-\overrightarrow{b}|}$

$\textsf{Squaring on bothsides we get}$

$\mathsf{|\overrightarrow{a}+\overrightarrow{b}|^2=|\overrightarrow{a}-\overrightarrow{b}|^2}$

$\mathsf{|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+2\,\overrightarrow{a}\,.\,\overrightarrow{b}=|\overrightarrow{a}|^2+|\overrightarrow{b}|^2-2\,\overrightarrow{a}\,.\,\overrightarrow{b}}$

$\mathsf{2\,\overrightarrow{a}\,.\,\overrightarrow{b}+2\,\overrightarrow{a}\,.\,\overrightarrow{b}=0}$

$\mathsf{4\,\overrightarrow{a}\,.\,\overrightarrow{b}=0}$

$\mathsf{\overrightarrow{a}\,.\,\overrightarrow{b}=0}$

$\implies\mathsf{\overrightarrow{a}\perp\overrightarrow{b}}$

$\therefore\textsf{Angle between the given vectors is}\;\mathsf{90^\circ}$

$\textbf{Find more:}$

If vector A=2i+2j+3k and vector B=3i-2j-4k, find cross product and angle b/w A and B?​

https://brainly.in/question/22534241

For what values of c are the vectors 3i-2j and 2i+3j + ck are perpendicular​

https://brainly.in/question/33019924

Answer 2

$\textbf{Given \: :} \\ \mathsf{ \purple{\overrightarrow{a}\;and\;\overrightarrow{b}\;are\;two\;non-zero\;vectors\;and}}</p><p> \\ \mathsf{ \purple{|\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}-\overrightarrow{b}|}} \\ \\ </p><p>\textbf{To find:} \\ </p><p>\textsf{ \color{orange}{The angle between}}\;\mathsf{ \color{orange}{{\overrightarrow{a}\;and\;\overrightarrow{b}}}} \\ \\ </p><p>\textbf{Solution:} \\ </p><p>\mathsf{ \color{green}{Consider,}} \\ </p><p>\mathsf{ \color{green}{|\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}-\overrightarrow{b}|}} \\ \\ </p><p>\textsf{ \color{royalblue}{Squaring on both \: sides we get}} \\ </p><p>\mathsf{ \color{royalblue}{|\overrightarrow{a}+\overrightarrow{b}|^2=|\overrightarrow{a}-\overrightarrow{b}|^2}} \\ </p><p>\mathsf{ \color{royalblue}{|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+2\,\overrightarrow{a}\,.\,\overrightarrow{b}=|\overrightarrow{a}|^2+|\overrightarrow{b}|^2-2\,\overrightarrow{a}\,.\,\overrightarrow{b}}}\\ </p><p>\mathsf{ \color{royalblue}{2\,\overrightarrow{a}\,.\,\overrightarrow{b}+2\,\overrightarrow{a}\,.\,\overrightarrow{b}=0}} \\ </p><p>\mathsf{ \color{royalblue}{4\,\overrightarrow{a}\,.\,\overrightarrow{b}=0}} \\ </p><p>\mathsf{ \color{royalblue}{\overrightarrow{a}\,.\,\overrightarrow{b}=0}} \\ </p><p>\implies\mathsf{ \color{red} {\overrightarrow{a}\perp\overrightarrow{b}}} \\ \\ \therefore\textsf{\pink{Angle between the given vectors is}}\;\mathsf{ \pink{90^\circ}}$