A regular tetrahedron and the methane angle · source reference

Printed 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 centres of the hydrogen atoms represented by points $Q,R,S,T$ and the centre of the carbon atom represented by point $V$.

The points $Q,R,S$ have coordinates $(8,1,8),(8,7,2),(2,1,2)$ respectively and form an equilateral triangle.

(i) Find a cartesian equation of the plane $p$ which passes through the points $Q,R,S$. [4]

(ii) 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 $x+y=9$.

(iii) Find a vector equation of the line $l$ where $p_1,p_2$ meet. [1]

(iv) The point $T$ is on the line $l$ such that $QRST$ is a regular tetrahedron with $QR=QT$. Show that the possible coordinates for point $T$ is $(2,7,8)$. Hence, or otherwise, find the coordinates of a point on plane $p$ that is closest to point $T$. [5]

(v) Given that $TV$ is $\frac34$ 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]

[2018/ACJC/Prelim/I/11]

Printed working

(i) $\overrightarrow{OQ}=(8,1,8)$, $\overrightarrow{OR}=(8,7,2)$, $\overrightarrow{OS}=(2,1,2)$.

$\overrightarrow{QR}=\overrightarrow{OR}-\overrightarrow{OQ}=(8,7,2)-(8,1,8)=(0,6,-6)=6(0,1,-1)$.

$\overrightarrow{QS}=\overrightarrow{OS}-\overrightarrow{OQ}=(2,1,2)-(8,1,8)=(-6,0,-6)=6(-1,0,-1)$.

$\mathbf n=(0,1,-1)\times(-1,0,-1)=(-1,1,1)$.

$\mathbf r\cdot(-1,1,1)=(2,1,2)\cdot(-1,1,1)\Rightarrow\mathbf r\cdot(-1,1,1)=1$.

The cartesian equation of plane $p$ is $-x+y+z=1$.

(ii) Midpoint of $QR=\frac{\overrightarrow{OR}+\overrightarrow{OQ}}2=\frac{(8,7,2)+(8,1,8)}2=(8,4,5)$.

$\mathbf r\cdot(0,1,-1)=(8,4,5)\cdot(0,1,-1)\Rightarrow\mathbf r\cdot(0,1,-1)=-1$.

The cartesian equation of plane $p_1$ is $y-z=-1$.

(iii) $p_1:y-z=-1$; $p_2:x+y=9$.

Using GC: The vector equation of the line $l$ is $\mathbf r=(10,-1,0)+\lambda(-1,1,1)$, where $\lambda\in\mathbb R$.

(iv) Since point $T$ is on line $l$, $\overrightarrow{OT}=(10-\lambda,-1+\lambda,\lambda)$.

$|\overrightarrow{QR}|=|\overrightarrow{QT}|$.

$|(0,6,-6)|=|(10-\lambda,-1+\lambda,\lambda)-(8,1,8)|$.

$|(0,6,-6)|=|(2-\lambda,-2+\lambda,\lambda-8)|$.

$\sqrt{72}=\sqrt{(2-\lambda)^2+(-2+\lambda)^2+(\lambda-8)^2}$.

$72=3\lambda^2-24\lambda+72$.

$\lambda=0$ or $\lambda=8$.

When $\lambda=8$, $\overrightarrow{OT}=(10-8,-1+8,8)=(2,7,8)$.

When $\lambda=0$, $\overrightarrow{OT'}=(10+0,-1+0,0)=(10,-1,0)$.

Hence possible coordinates for point $T$ are $(2,7,8)$ (Shown).

Method 1

$\frac{(2,7,8)+(10,-1,0)}2=(6,3,4)$.

The coordinates of point on plane $p$ that is closest to point $T$ is $(6,3,4)$.

Method 2

The vector equation of line through point $T$ perpendicular to plane $p$ is $\mathbf r=(2,7,8)+\mu(-1,1,1)$, where $\mu\in\mathbb R$.

$(2-\mu,7+\mu,8+\mu)\cdot(-1,1,1)=1\Rightarrow-2+\mu+7+\mu+8+\mu=1\Rightarrow\mu=-4$.

When $\mu=-4$, $\mathbf r=(2,7,8)+(-4)(-1,1,1)=(6,3,4)$.

The coordinates of point on plane $p$ that is closest to point $T$ is $(6,3,4)$.

(v) Using ratio theorem,

$\overrightarrow{OV}=\frac{3(6,3,4)+(2,7,8)}4=(5,4,5)$.

The coordinates of $V$ is $(5,4,5)$.

$\overrightarrow{OQ}=(8,1,8)$, $\overrightarrow{OV}=(5,4,5)$, $\overrightarrow{OT}=(2,7,8)$.

$\overrightarrow{VQ}=\overrightarrow{OQ}-\overrightarrow{OV}=(8,1,8)-(5,4,5)=(3,-3,3)$.

$\overrightarrow{VT}=\overrightarrow{OT}-\overrightarrow{OV}=(2,7,8)-(5,4,5)=(-3,3,3)$.

$\angle TVQ=\cos^{-1}\frac{(3,-3,3)\cdot(-3,3,3)}{\sqrt{3^2+(-3)^2+3^2}\sqrt{(-3)^2+3^2+3^2}}$.

$=\cos^{-1}\frac{-9-9+9}{27}=\cos^{-1}\frac{-9}{27}=109.471^\circ=109.5^\circ$ (1 d.p.).

Teacher notes

Since plane $p_1$ is perpendicular to $QR$, $(0,1,-1)$ parallel to line $QR$ is normal to plane $p_1$.

Method 1: Pt on plane $p$ closest to $T$ is $M$, midpt of $T,T'$.

Method 2: The point on plane $p$ that is closest to $T$ is the foot of perpendicular, $F$, from $T$ to plane $p$. $F$ is the point of intersection between the line through $T$ perpendicular to $p$ and plane $p$.

$\cos\angle TVQ=\frac{\overrightarrow{VQ}\cdot\overrightarrow{VT}}{|\overrightarrow{VQ}||\overrightarrow{VT}|}$.

To find $\angle TVQ$, the 2 vectors bounding the angle must both be directed towards pt $V$ ($\overrightarrow{TV},\overrightarrow{QV}$) or directed away from pt $V$ ($\overrightarrow{VT},\overrightarrow{VQ}$).

Note: Do not put in the modulus sign for $\overrightarrow{VT}\cdot\overrightarrow{VQ}$.

Printed answer key

(i) $-x+y+z=1$. (ii) $y-z=-1$. (iii) $\mathbf r=(10,-1,0)+\lambda(-1,1,1)$ where $\lambda\in\mathbb R$. (iv) $(6,3,4)$. (v) $(5,4,5)$.

Editorial qualification

There are two tetrahedron apexes: T=(2,7,8) and T′=(10,−1,0). The source shows both in its working, then selects the requested T=(2,7,8). “Vertical height” here is the perpendicular altitude to plane p, not the ambient z-axis. V=(5,4,5) and angle109.5° refer to the selected apex.

Source provenance

2025 TJC H2 Math (9758) JC2 MSM: VECTORS. Question booklet PDF pages 4–12 (printed 80–88); solution booklet PDF pages 17–44. Column vectors are typeset as ordered coordinate triples here; the entries, products, intermediate equations and original parameter/domain restrictions are retained.