| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Angle between line and plane |
| Difficulty | Standard +0.3 This is a straightforward FP3 vectors question testing standard techniques: finding the angle between a line and plane using dot product of direction vectors, and using the perpendicular distance formula. Both parts are direct applications of memorized formulas with minimal problem-solving required, though slightly above average difficulty due to being Further Maths content. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04d Angles: between planes and between line and plane4.04j Shortest distance: between a point and a plane |
| Answer | Marks | Guidance |
|---|---|---|
| \(\theta = \sin^{-1}\frac{[5,6,-7] \cdot [1,2,-1]}{\sqrt{5^2+6^2+(-7)^2}\sqrt{1^2+2^2+(-1)^2}}\) | M1* | For using scalar product of line and plane vectors |
| \(\theta = \sin^{-1}\frac{24}{\sqrt{110}\sqrt{6}}= 69.1°\) (69.099...°, 1.206) | M1 A1 | For both moduli seen; For correct scalar product; For correct angle |
| \(\phi = \sin^{-1}\frac{[5,6,-7] \times [1,2,-1]}{\sqrt{5^2+6^2+(-7)^2}\sqrt{1^2+2^2+(-1)^2}}\) | SR M1* M1 | For vector product of line and plane vectors AND finding modulus of result; For moduli of line and plane vectors seen |
| \(\phi = \sin^{-1}\frac{\sqrt{84}}{\sqrt{110}\sqrt{6}} = 20.9° \Rightarrow \theta = 69.1°\) | A1 A1 | For correct modulus \(\sqrt{84}\); For correct angle |
| Answer | Marks | Guidance |
|---|---|---|
| \(d = \frac{ | 1+12+3-40 | }{\sqrt{1^2+2^2+(-1)^2}} = \frac{24}{\sqrt{6}} = 4\sqrt{6} = 9.80\) |
| Answer | Marks | Guidance |
|---|---|---|
| \((1+\lambda) + 2(6+2\lambda) - (-3-\lambda) = 40\) | M1 | For substituting parametric form into plane |
| \(\Rightarrow \lambda = 4 \Rightarrow d = 4\sqrt{6}\) | A1 | For correct distance |
| OR distance from (1, 6, −3) to (5, 14, −7) \(= \sqrt{4^2+8^2+(-4)^2} = \sqrt{96}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Plane through (1, 6, −3) parallel to \(p\) is | M1 | For finding parallel plane through (1, 6, −3) |
| \(x + 2y - z = 16 \Rightarrow d = \frac{40-16}{\sqrt{6}} = \frac{24}{\sqrt{6}}\) | A1 | For correct distance |
| Answer | Marks | Guidance |
|---|---|---|
| e.g. (0, 0, −40) on \(p\) | M1 | For using any point on \(p\) to find vector and scalar product seen |
| \(\Rightarrow\) vector to (1, 6, −3) is \(\pm(1, 6, 37)\) | e.g. [1, 6, 37], [1, 2, −1] | |
| \(d = \frac{ | [1,6,37] \cdot [1,2,-1] | }{\sqrt{6}} = \frac{24}{\sqrt{6}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(l\) meets \(p\) where \((1+5t)+2(6+6t)-(-3-7t) = 40\) | For finding \(l\) where \(l\) meets \(p\) and linking \(d\) with triangle | |
| \(\Rightarrow t = 1 \Rightarrow d = [5,6,-7]\sin\theta\) | M1 | |
| \(\Rightarrow d = \sqrt{110} \cdot \frac{24}{\sqrt{110}\sqrt{6}} = \frac{24}{\sqrt{6}}\) | A1 | For correct distance |
## (i)
$\theta = \sin^{-1}\frac{[5,6,-7] \cdot [1,2,-1]}{\sqrt{5^2+6^2+(-7)^2}\sqrt{1^2+2^2+(-1)^2}}$ | M1* | For using scalar product of line and plane vectors
$\theta = \sin^{-1}\frac{24}{\sqrt{110}\sqrt{6}}= 69.1°$ (69.099...°, 1.206) | M1 A1 | For both moduli seen; For correct scalar product; For correct angle
$\phi = \sin^{-1}\frac{[5,6,-7] \times [1,2,-1]}{\sqrt{5^2+6^2+(-7)^2}\sqrt{1^2+2^2+(-1)^2}}$ | SR M1* M1 | For vector product of line and plane vectors AND finding modulus of result; For moduli of line and plane vectors seen
$\phi = \sin^{-1}\frac{\sqrt{84}}{\sqrt{110}\sqrt{6}} = 20.9° \Rightarrow \theta = 69.1°$ | A1 A1 | For correct modulus $\sqrt{84}$; For correct angle
## (ii) METHOD 1
$d = \frac{|1+12+3-40|}{\sqrt{1^2+2^2+(-1)^2}} = \frac{24}{\sqrt{6}} = 4\sqrt{6} = 9.80$ | M1 A1 2 | For use of correct formula; For correct distance
## METHOD 2
$(1+\lambda) + 2(6+2\lambda) - (-3-\lambda) = 40$ | M1 | For substituting parametric form into plane
$\Rightarrow \lambda = 4 \Rightarrow d = 4\sqrt{6}$ | A1 | For correct distance
OR distance from (1, 6, −3) to (5, 14, −7) $= \sqrt{4^2+8^2+(-4)^2} = \sqrt{96}$ | |
## METHOD 3
Plane through (1, 6, −3) parallel to $p$ is | M1 | For finding parallel plane through (1, 6, −3)
$x + 2y - z = 16 \Rightarrow d = \frac{40-16}{\sqrt{6}} = \frac{24}{\sqrt{6}}$ | A1 | For correct distance
## METHOD 4
e.g. (0, 0, −40) on $p$ | M1 | For using any point on $p$ to find vector and scalar product seen
$\Rightarrow$ vector to (1, 6, −3) is $\pm(1, 6, 37)$ | | e.g. [1, 6, 37], [1, 2, −1]
$d = \frac{|[1,6,37] \cdot [1,2,-1]|}{\sqrt{6}} = \frac{24}{\sqrt{6}}$ | A1 | For correct distance
## METHOD 5
$l$ meets $p$ where $(1+5t)+2(6+6t)-(-3-7t) = 40$ | | For finding $l$ where $l$ meets $p$ and linking $d$ with triangle
$\Rightarrow t = 1 \Rightarrow d = [5,6,-7]\sin\theta$ | M1 |
$\Rightarrow d = \sqrt{110} \cdot \frac{24}{\sqrt{110}\sqrt{6}} = \frac{24}{\sqrt{6}}$ | A1 | For correct distance
---A line $l$ has equation $\frac{x-1}{5} = \frac{y-6}{6} = \frac{z+3}{-7}$ and a plane $p$ has equation $x + 2y - z = 40$.
\begin{enumerate}[label=(\roman*)]
\item Find the acute angle between $l$ and $p$. [4]
\item Find the perpendicular distance from the point $(1, 6, -3)$ to $p$. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR FP3 2011 Q1 [6]}}