Figure 1 shows a methane molecule consisting of a carbon atom with four hydrogen atoms symmetrically

Question

Figure 1 shows a methane molecule consisting of a carbon atom with four hydrogen atoms symmetrically placed around it. Figure 2 shows the tetrahedron structure of the methane molecule with the centre of the hydrogen atoms represented by points $Q$, $R$, $S$ and $T$ and the centre of the carbon atom represented by point $V$. The points $Q$, $R$ and $S$ has coordinates $\left( 8,1,8 \right)$, $\left( 8,7,2 \right)$ and $\left( 2,1,2 \right)$ respectively and form an equilateral triangle. [4] Find a cartesian equation of the plane ${{p}_{1}}$ which passes through the midpoint of $QR$ and is perpendicular to $QR$. [2] Plane ${{p}_{2}}$ which passes through the midpoint of $RS$and is perpendicular to $RS$ has equation $x+y=9$. Find a vector equation of the line $l$ where ${{p}_{1}}$ and ${{p}_{2}}$ meet. [1] The point $T$ is on the line $l$ such that $QRST$ is a regular tetrahedron with $QR=QT$. Show that the possible coordinates for the point $T$ is $\left( 2,7,8 \right)$. Hence, or otherwise, find the coordinates of a point on plane $p$ that is the closest to point $T$. [5] Given that $TV$ is $\frac{3}{4}$ of the vertical height of tetrahedron $QRST$. Find the coordinates of point $V$ and hence show that the bonding angle $TVQ$ of the methane molecule is $109.5{}^\circ $ (correct to 1 decimal place). [4]

Accepted Answer

(i) x - y - z = -1; (ii) p₁: y - z = -1; (iii) l: r = (10,-1,0) + λ(-1,1,1); (iv) T(2,7,8), closest point (6,3,4); (v) V(5,4,5), ∠TVQ = 109.5°

2018 ACJC P1 Q11

2018 ACJC P1 Q11

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