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. $\frac{1}{3}\left(2\hat{i}+2\hat{j}+\hat{k}\right)$
• B. $\frac{1}{3}\left(\hat{i}-\hat{j}-\hat{k}\right)$
• C. $\frac{1}{6}\left(2\hat{i}+\hat{j}+\hat{k}\right)$
• D. $\frac{12}{3}\left(\hat{i}+\hat{j}+\hat{k}\right)$

Answer 1

Answer: (A)

The position vector of any point at $t$ is
$\vec{r}=\left(2+t^2\right)\hat{i}+\left(4t-5\right)\hat{j}+\left(2t^2-6\right)\hat{k}$
$\Rightarrow \frac{d\vec{r}}{dt}=2t\hat{i}+4\hat{j}+\left(4t-6\right)\hat{k}$
$\Rightarrow \frac{d\vec{r}}{dt}{|}_{t=2}=4\hat{i}+4\hat{j}+2\hat{k}$
and ${\left|\frac{d\vec{r}}{dt}\right|}_{{}_{{}_{t=2}}}=\sqrt{16+16+4}=4$
Hence, the required unit tangent vector at $t=2$ is
$\frac{1}{3}\left(2\hat{i}+2\hat{j}+\hat{k}\right)$