Explanation
We have,
$$ AB=3\overrightarrow{i}+4\overrightarrow{k} $$
$$ AC=5\overrightarrow{i}-2\overrightarrow{j}+4\overrightarrow{k} $$
So, the coordinate of
$$ B\,is\,\left( 3,0,4 \right) $$
$$ C\,is\,\left( 5,-2,4 \right) $$
The mid point of $$BC$$ can be easily found by midpoint formula of two points as
$$\begin{align}
$$ D\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( \dfrac{3+5}{2},\dfrac{0-2}{2},\dfrac{4+4}{2} \right) $$
$$ D\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( 5,-2,4 \right) $$
So, the length of median through A is
$$ AD=\sqrt{{{\left( 4-0 \right)}^{2}}+{{\left( -1-0 \right)}^{2}}+{{\left( 4-0 \right)}^{2}}} $$
$$ AD=\sqrt{33} $$
A particle is projected with velocity 'u' at an angle $$\theta $$ with an inclined plane of inclination $$\theta <45$$ with the horizontal.The time taken when velocity of projectile becomes parallel to the plane
Vector $$\overrightarrow {\text{A}} $$ lies in xy plane and makes an angle will positive y direction. The x components of $$\overrightarrow {\text{A}} $$ is
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