Answer
(a) See the figure below.
(b)(i) $D_{E}$=Distance of Earth from the Sun=$0.9857AU$
(ii) $D_{M}$=Distance of Mars from the Sun=$1.3820AU$
(iii) $|R|$=Distance of Mars form the Earth=$1.6953AU$
(c) The angle between the direction to the Sun and the direction to Mars on December 3,1999 is $54.6^{\circ}$.
(d)On December 3,1999 Mars was not visible from our location at midnight because the angle between the direction from Earth to Sun and from Earth to Mars was less than $90^{\circ}$.
Work Step by Step
(a) See the figure.
(b) We assume that the Sun is at the origin of our coordinate system with the Earth's orbit in the xy-plane.
Let $E$ and $M$ be the position vector of Earth and Mars from the Sun.
Let $R$ be the displacement vector from Earth to Mars.
We are also given the coordinates of Earth and Mars from the sun on the table.
Form the table,
(i) $D_{E}$=Distance of Earth from the Sun=$\sqrt (0.3182^2+0.9329^2)=0.9857AU$
(ii) $D_{M}$=Distance of Mars from the Sun=$\sqrt (1.3087^2+(-0.4423)^2+(-0.0414)^2)=1.3820AU$
(iii) Since , $M=E+R$
$R=M-E=(1.3087-0.3182)i+(-0.4423-0.9329)j+(-0.0414)k=0.9905i+(-1.3752)j+(-0.0414)k$
Now, $|R|$=Distance of Mars form the Earth=$\sqrt (0.9905^2+(-1.3752)^2+(-0.0414)^2)=1.6953AU$
(c) Let the displacement vector of the Sun form Earth be $S$ and $S=-E$
Using the scalar product of $S$ and $R$ ,we get
$S.R=|S||R|cos\theta$ , where $\theta$ is the angle between $S$ and $R$
$=>-0.3182\times0.9905+0.9329\times1.3752=\sqrt ((-0.3182)^2+(-0.9329)^2)\times\sqrt (0.9905^2+(-1.3752)^2+(-0.0414)^2)cos\theta$
$=>\theta=cos^{-1}(\frac{-0.3182\times0.9905+0.9329\times1.3752}{\sqrt ((-0.3182)^2+(-0.9329)^2)\times\sqrt (0.9905^2+(-1.3752)^2+(-0.0414)^2)})=54.6^{\circ}$
Thus ,the angle between the direction to the Sun and the direction to Mars on December 3,1999 is $54.6^{\circ}$.
(d) At midnight, the Sun is on the opposite side of the Earth from our location.
On December 3,1999 Mars was not visible from our location at midnight because the angle between the direction from Earth to Sun and from Earth to Mars was $54.6^{\circ}$.
For Mars to be visible at midnight ,this angle should be greater than $90^{\circ}$
