| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2012 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Plane containing line and point/vector |
| Difficulty | Standard +0.8 This is a Further Maths question requiring understanding of multiple geometric constraints (parallel to plane, perpendicular to line) and finding a direction vector via cross product. While systematic, it demands careful vector manipulation and understanding of perpendicularity conditions beyond standard A-level, placing it moderately above average difficulty. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04d Angles: between planes and between line and plane |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{b} = [1,-3,4] \times [3,1,2] = [-10,10,10] = k[-1,1,1]\) | M1 | For attempt to find vector product of directions |
| Correct calculation of vector product | M1 | Allow 1 error |
| For correct \(\mathbf{b}\) | A1 | |
| \(\Rightarrow \mathbf{r} = [1,4,2] + t[-1,1,1]\) | B1FT | For correct equation. FT from \(\mathbf{b}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([x,y,z]\cdot[1,-3,4]=0 \Rightarrow x-3y+4z=0\) | M1 | For equation from \(l_2\) perpendicular to normal of plane and equation from \(l_2\) perpendicular to \(l_1\) |
| \([x,y,z]\cdot[3,1,2]=0 \Rightarrow 3x+y+2z=0\) | ||
| Solving \(\Rightarrow [x,y,z] = \mathbf{b} = k[-1,1,1]\) | M1 | |
| A1 | ||
| \(\Rightarrow \mathbf{r} = [1,4,2]+t[-1,1,1]\) | B1FT | For correct equation. FT from \(\mathbf{b}\). Must show "\(\mathbf{r}=\)" |
# Question 1:
**Method 1:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{b} = [1,-3,4] \times [3,1,2] = [-10,10,10] = k[-1,1,1]$ | M1 | For attempt to find vector product of directions |
| Correct calculation of vector product | M1 | Allow 1 error |
| For correct $\mathbf{b}$ | A1 | |
| $\Rightarrow \mathbf{r} = [1,4,2] + t[-1,1,1]$ | B1FT | For correct equation. FT from $\mathbf{b}$ |
**Method 2:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[x,y,z]\cdot[1,-3,4]=0 \Rightarrow x-3y+4z=0$ | M1 | For equation from $l_2$ perpendicular to normal of plane and equation from $l_2$ perpendicular to $l_1$ |
| $[x,y,z]\cdot[3,1,2]=0 \Rightarrow 3x+y+2z=0$ | | |
| Solving $\Rightarrow [x,y,z] = \mathbf{b} = k[-1,1,1]$ | M1 | |
| | A1 | |
| $\Rightarrow \mathbf{r} = [1,4,2]+t[-1,1,1]$ | B1FT | For correct equation. FT from $\mathbf{b}$. Must show "$\mathbf{r}=$" |
---1 The plane $p$ has equation $\mathbf { r } . ( \mathbf { i } - 3 \mathbf { j } + 4 \mathbf { k } ) = 4$ and the line $l _ { 1 }$ has equation $\mathbf { r } = 2 \mathbf { j } - \mathbf { k } + t ( 3 \mathbf { i } + \mathbf { j } + 2 \mathbf { k } )$. The line $l _ { 2 }$ is parallel to $p$ and perpendicular to $l _ { 1 }$, and passes through the point with position vector $\mathbf { i } + 4 \mathbf { j } + 2 \mathbf { k }$. Find the equation of $l _ { 2 }$, giving your answer in the form $\mathbf { r } = \mathbf { a } + t \mathbf { b }$.
\hfill \mbox{\textit{OCR FP3 2012 Q1 [4]}}