Answer: (a) \(\dfrac1{\sqrt{14}}\)
(b) \(7.5^\circ\) approximately.
(c) \(\mathbf r=\dfrac14\begin{pmatrix}6\\-1\\-1\end{pmatrix}+t\begin{pmatrix}1\\1\\1\end{pmatrix}\)
(a)
The perpendicular distance from \((0,0,0)\) to \(x+3y+2z-1=0\) is
\(\frac{|0+0+0-1|}{\sqrt{1^2+3^2+2^2}}=\frac1{\sqrt{14}}.\)
(b)
Use \(\overrightarrow{OA}=(0,-1,2)\) and \(\overrightarrow{OB}=(2,0,-1)\). A normal to plane \(OAB\) is
\(\overrightarrow{OA}\times\overrightarrow{OB}=(1,4,2).\)
A normal to \(\Pi\) is
\((1,3,2).\)
Thus, if \(\alpha\) is the acute angle between the planes,
\(\cos\alpha=\frac{(1,4,2)\cdot(1,3,2)}{\sqrt{1^2+4^2+2^2}\sqrt{1^2+3^2+2^2}}=\frac{17}{\sqrt{21}\sqrt{14}}.\)
Therefore
\(\alpha\approx7.5^\circ.\)
(c)
The line \(OC\) has direction \((2,-1,-1)\), so a point \(P\) on \(OC\) has
\(\overrightarrow{OP}=\lambda\begin{pmatrix}2\\-1\\-1\end{pmatrix}.\)
The line \(AB\) has direction
\(\overrightarrow{AB}=(2,0,-1)-(0,-1,2)=\begin{pmatrix}2\\1\\-3\end{pmatrix}.\)
A point \(Q\) on \(AB\) has
\(\overrightarrow{OQ}=\begin{pmatrix}0\\-1\\2\end{pmatrix}+\mu\begin{pmatrix}2\\1\\-3\end{pmatrix}=\begin{pmatrix}2\mu\\-1+\mu\\2-3\mu\end{pmatrix}.\)
Then
\(\overrightarrow{PQ}=\begin{pmatrix}2\mu-2\lambda\\-1+\mu+\lambda\\2-3\mu+\lambda\end{pmatrix}.\)
For the common perpendicular, \(\overrightarrow{PQ}\) is perpendicular to both line directions.
Using \(\overrightarrow{PQ}\cdot(2,-1,-1)=0\),
\(6\mu-6\lambda=1.\)
Using \(\overrightarrow{PQ}\cdot(2,1,-3)=0\),
\(14\mu-6\lambda=7.\)
Solving gives \(\mu=\frac34\). Hence
\(\overrightarrow{OQ}=\begin{pmatrix}\frac32\\-\frac14\\-\frac14\end{pmatrix}=\frac14\begin{pmatrix}6\\-1\\-1\end{pmatrix}.\)
The common perpendicular direction is
\((2,-1,-1)\times(2,1,-3)=4(1,1,1).\)
Therefore an equation of the common perpendicular is
\(\mathbf r=\frac14\begin{pmatrix}6\\-1\\-1\end{pmatrix}+t\begin{pmatrix}1\\1\\1\end{pmatrix}.\)
Mark scheme
| Part | Marks | Expected result | Notes |
|---|
| (a) | M1 A1 | \(\dfrac1{\sqrt{1^2+3^2+2^2}}=\dfrac1{\sqrt{14}}\) | Allow 0.267 or better. |
| (b) | M1 A1 | \(\overrightarrow{OA}\times\overrightarrow{OB}=(1,4,2)\) | Finds normal to \(OAB\). Allow 1 error for the M mark. |
| (b) | M1 | \((1,4,2)\cdot(1,3,2)=\sqrt{21}\sqrt{14}\cos\alpha\), so \(\cos\alpha=\dfrac{17}{\sqrt{21}\sqrt{14}}\) | Takes dot product of normal vectors. |
| (b) | A1 | \(7.5^\circ\) | Accept 0.131 radians. Mark final answer of 7.5° or better. |
| (c) | M1 A1 | \(\overrightarrow{OP}=\lambda(2,-1,-1)^T,\quad \overrightarrow{OQ}=(2\mu,-1+\mu,2-3\mu)^T\), so \(\overrightarrow{PQ}=(2\mu-2\lambda,-1+\mu+\lambda,2-3\mu+\lambda)^T\) | Finds \(\overrightarrow{PQ}\) or direction of common perpendicular. |
| (c) | M1 | \(\overrightarrow{PQ}\cdot(2,-1,-1)=0\) | Uses that dot product of \(\overrightarrow{PQ}\) with line direction is zero, or that \(\overrightarrow{PQ}\) is a multiple of the common perpendicular \((1,1,1)^T\). |
| (c) | A1 | \(6\mu-6\lambda=1\) | Deduces one equation. |
| (c) | A1 | \(\overrightarrow{PQ}\cdot(2,1,-3)=0\), hence \(14\mu-6\lambda=7\) | Deduces second equation. |
| (c) | M1 A1 | \(\mu=\frac34\), so \(\overrightarrow{OQ}=\frac14(6,-1,-1)^T\); alternatively \(\lambda=\frac7{12}\) and \(\overrightarrow{OP}=\frac7{12}(2,-1,-1)^T\) | Solves for \(\mu\) and substitutes into \(\overrightarrow{OQ}\) and \(\overrightarrow{PQ}\), or uses the alternative. |
| (c) | B1FT | \(\mathbf r=\frac14(6,-1,-1)^T+k(1,1,1)^T\) | OE. Follow through using their common perpendicular. Must have \(\mathbf r=\). |