A) \[\frac{1}{3}(2\hat{i}+2\hat{j}+\hat{k})\]
B) \[\frac{1}{3}(\hat{i}-\hat{j}-\hat{k})\]
C) \[\frac{1}{6}(2\hat{i}+\hat{j}+\hat{k})\]
D) \[\frac{2}{3}(\hat{i}+\hat{j}+\hat{k})\]
Correct Answer: A
Solution :
[a] The position vector of any point at t is \[\vec{r}=(2+{{t}^{2}})\hat{i}+(4t-5)\hat{j}+(2{{t}^{2}}-6)\hat{k}\] \[\Rightarrow \frac{d\,\vec{r}}{dt}=2t\,\hat{i}+4\hat{j}+(4t-6)\hat{k}\] \[\Rightarrow {{\left. \frac{d\,\vec{r}}{dt} \right|}_{t=2}}=4\hat{i}+4\hat{j}+2\hat{k}\] and \[{{\left. \left| \frac{d\,\vec{r}}{dt} \right| \right|}_{t=2}}=\sqrt{16+16+4}=4\] Hence, the required unit tangent vector at t=2 is \[\frac{1}{3}(2\hat{i}+2\hat{j}+\hat{k}).\]
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