Answer
$x=2+ \dfrac{t}{2},y=ln 4+ \dfrac{t}{2},z=1+t$
Work Step by Step
Equation can be written as:$r(t)= \sqrt {t^2+3} i+ln(t^2+3) j+t k $
Tangent vector $r'(t)$ can be calculated by taking the derivative of $r(t)$ with respect to $t$.
This yields, $r'(t)=\dfrac{t}{ \sqrt {t^2+3}}i+\dfrac{2t}{ t^2+3}j+1 k $
or, $r'(t)=\lt \dfrac{t}{ \sqrt {t^2+3}},\dfrac{2t}{ t^2+3},1 \gt $
$r'(1)=\lt \dfrac{1}{ \sqrt {(1)^2+3}},\dfrac{2(1)}{ (1)^2+3},1 \gt=\lt \dfrac{1}{2}, \dfrac{1}{2},1 \gt$
As we are given that the equation of tangent line passing through the points $(2,ln 4,1)$, thus,
$r(t)=\lt (2+ \dfrac{t}{2}),(ln 4+ \dfrac{t}{2}), (1+t) \gt$
Hence, the desired parametric equations of the tangent line are:
$x=2+ \dfrac{t}{2},y=ln 4+ \dfrac{t}{2},z=1+t$
