A particle moves in space according to the eq | vedclass

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(D) The position vector is given by $\vec{r} = \sin t \hat{i} + \cos t \hat{j} + t \hat{k}$.The velocity vector is $\vec{v} = \frac{d\vec{r}}{dt} = \c

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Never

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(D) The position vector is given by $\vec{r} = \sin t \hat{i} + \cos t \hat{j} + t \hat{k}$.The velocity vector is $\vec{v} = \frac{d\vec{r}}{dt} = \cos t \hat{i} - \sin t \hat{j} + 1 \hat{k}$.The acceleration vector is $\vec{a} = \frac{d\vec{v}}{dt} = -\sin t \hat{i} - \cos t \hat{j} + 0 \hat{k}$.For the position vector and acceleration vector to be perpendicular,their dot product must be zero: $\vec{r} \cdot \vec{a} = 0$.Calculating the dot product: $(\sin t)(-\sin t) + (\cos t)(-\cos t) + (t)(0) = 0$.$-\sin^2 t - \cos^2 t = 0$.$-(\sin^2 t + \cos^2 t) = 0$.$-1 = 0$.Since $-1$ can never be $0$,there is no time $t$ for which the vectors are perpendicular. Therefore,they are never perpendicular.

$A$ particle moves in space according to the equation $\vec{r} = (\sin t \hat{i} + \cos t \hat{j} + t \hat{k}) \text{ m}$. Find the time $t$ when the position vector and acceleration vector are perpendicular to each other.

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