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Q.
A uni-modular tangent vector on the curve $x = t^{2} + 2, y - 4t - 5, z = 2t^{2} -6t$ at $t = 2$ is
Vector Algebra
Solution:
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
$\frac{1}{3}\left(2\hat{i}+2\hat{j}+\hat{k}\right)$