Question

A uni-modular tangent vector on the curve $x=t^2+2,y=4t-5,z=2t^2-6t$ at $t=2$ is

• 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. $\frac{2}{3}(\hat{i}+\hat{j}+\hat{k})$

Answer 1

The correct option is A $\frac{1}{3}(2\hat{i}+2\hat{j}+\hat{k})$

The position vector of any point at $t$ is
$\overrightarrow{r}=(t^2+2)\hat{i}+(4t-5)\hat{j}+(2t^2-6)\hat{k}$

$\Rightarrow \frac{d\overrightarrow{r}}{dt}=2t\hat{i}+4\hat{j}+(4t-6)\hat{k}$

$\Rightarrow {\begin{array}{c}\frac{d\overrightarrow{r}}{dt}\end{array}|}_{t=2}=4\hat{i}+4\hat{j}+2\hat{k}$
${\begin{array}{c}\left|\frac{d\overrightarrow{r}}{dt}\right|\end{array}|}_{t=2}=\sqrt{16+16+4}=6$

Hence, the required unit tangent vector at $t=2$ is
$\frac{1}{3}(2\hat{i}+2\hat{j}+\hat{k})$