A particle moves in space according to equation
$\vec r = (\sin \,t\,\hat i\, + \,\cos \,t\,\hat j\, + \,t\,\hat k)m$
Find time $'t'$ when position vector and acceleration vector are perpendicular to each other
Medium
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$\overrightarrow{\mathrm{r}}=(\sin t \hat{i}+\cos t \hat{j}+t \hat{k}) m$
$\overrightarrow{\mathrm{v}}=\frac{\mathrm{d} \overrightarrow{\mathrm{r}}}{\mathrm{dt}}=(\cos t \hat{i}-\sin t \hat{j})$
$\overrightarrow{\mathrm{a}}=\frac{\mathrm{d} \overrightarrow{\mathrm{v}}}{\mathrm{dt}}=(-\sin t \hat{\mathrm{i}}-\cos t \hat{\mathrm{j}})$
According to question
$\overrightarrow{r} \cdot \overrightarrow{a}=0$
$\Rightarrow-\sin ^{2} t-\cos ^{2} t=0$
$\Rightarrow 1=0$ which is not possible
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