Vector Algebra Question 217
Question: A uni-modular tangent vector on the curve $ x=t^{2}+2 $ , $ y=4t-5 $ , $ z=2t^{2}-6t $ at t=2 is
Options:
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}) $
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Answer:
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}+(2t^{2}-6)\hat{k} $
$ \Rightarrow \frac{d,\vec{r}}{dt}=2t,\hat{i}+4\hat{j}+(4t-6)\hat{k} $
$ \Rightarrow {{. \frac{d,\vec{r}}{dt} |}{t=2}}=4\hat{i}+4\hat{j}+2\hat{k} $ and $ {{. | \frac{d,\vec{r}}{dt} | |}{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}). $
BETA
