चित्र में सदिशों overrightarrow{ OA }, overri | vedclass

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Question

Let magnitude be equal to $\lambda$.

$\overrightarrow{ OA }=\lambda\left[\cos 30^{\circ} \hat{ i }+\sin 30 \hat{ j }\right]=\lambda\left[\frac{\sqr

Vedclass · Accepted Answer

$	an ^{-1} \frac{(1-\sqrt{3}-\sqrt{2})}{(1+\sqrt{3}+\sqrt{2})}$

Vedclass · Answer

Let magnitude be equal to $\lambda$.

$\overrightarrow{ OA }=\lambda\left[\cos 30^{\circ} \hat{ i }+\sin 30 \hat{ j }ight]=\lambda\left[\frac{\sqrt{3}}{2} \hat{ i }+\frac{1}{2} \hat{ j }ight]$

$\overrightarrow{ OB }=\lambda\left[\cos 60^{\circ} \hat{ i }-\sin 60 \hat{ j }ight]=\lambda\left[\frac{1}{2} \hat{ i }-\frac{\sqrt{3}}{2} \hat{ j }ight]$

$\overrightarrow{ OC }=\lambda\left[\cos 45^{\circ}(-\hat{ i })+\sin 45 \hat{ j }ight]=\lambda\left[-\frac{1}{\sqrt{2}} \hat{ i }+\frac{1}{\sqrt{2}} \hat{ j }ight]$

$	herefore \overrightarrow{ OA }+\overrightarrow{ OB }-\overrightarrow{ OC }$

$=\lambda\left[\left(\frac{\sqrt{3}+1}{2}+\frac{1}{\sqrt{2}}ight) \hat{ i }+\left(\frac{1}{2}-\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}ight) \hat{ j }ight]$

$	herefore$ Angle with $x$-axis

$	an ^{-1}\left[\frac{\frac{1}{2}-\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2}+\frac{1}{2}+\frac{1}{\sqrt{2}}}ight]=	an ^{-1}\left[\frac{\sqrt{2}-\sqrt{6}-2}{\sqrt{6}+\sqrt{2}+2}ight]$

$=	an ^{-1}\left[\frac{1-\sqrt{3}-\sqrt{2}}{\sqrt{3}+1+\sqrt{2}}ight]$

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यदि $| A + B |=| A |+| B |$ तब $\mathop A\limits^ \to $व $\mathop B\limits^ \to $ के बीच कोण है