The angle made by the vector left( {hat | vedclass

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Question

$a = \sqrt {{a_x}^2\,\, + \;\,{a_y}^2} = \sqrt {{1^2} + \;\,{1^2}} = \sqrt 2$

$\cos \alpha = \frac{{{a_x}}}{a} = \frac{1}{{\sqrt 2 }}$

$	herefo

Vedclass · Accepted Answer

$45^o, 45^o$

Vedclass · Answer

$a = \sqrt {{a_x}^2\,\, + \;\,{a_y}^2} = \sqrt {{1^2} + \;\,{1^2}} = \sqrt 2$

$\cos \alpha = \frac{{{a_x}}}{a} = \frac{1}{{\sqrt 2 }}$

$	herefore \alpha = 45^\circ ;\,\cos \,\beta = \frac{{{a_y}}}{a} = \frac{1}{{\sqrt 2 }}$

$	herefore \beta = 45^\circ $

The angle made by the vector $\left( {\hat i\,\, + \;\,\hat j} \right)$ with $x-$ axis and $y$ axis is

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