| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Perpendicular distance point to plane |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on standard vector geometry techniques. Part (i) requires recognizing that the direction vector of the line is the normal to the plane, then using the point to find the equation—a direct application of a standard method. Part (ii) is a routine formula application for distance from origin to plane. Part (iii) requires understanding that parallel planes have the same normal vector and using the distance formula, involving slightly more thought but still procedural. All parts are textbook-standard with no novel insight required, making this slightly easier than average. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04h Shortest distances: between parallel lines and between skew lines |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Obtain \(2x - 3y + 6z\) for LHS of equation | B1 | |
| Obtain \(2x - 3y + 6z = 23\) | B1 | [2] |
| (ii) Either Use correct formula to find perpendicular distance | M1 | |
| Obtain unsimplified value \(\frac{\pm 23}{\sqrt{2^2 + (-3)^2 + 6^2}}\), following answer to (i) | A1✓ | |
| Obtain \(\frac{23}{7}\) or equivalent | A1 | [3] |
| OR 1 Use scalar product of \((4, -1, 2)\) and a vector normal to the plane | M1 | |
| Use unit normal to plane to obtain \(\pm\frac{(8 + 3 + 12)}{\sqrt{49}}\) | A1 | |
| Obtain \(\frac{23}{7}\) or equivalent | A1 | [3] |
| OR 2 Find parameter intersection of \(p\) and \(\mathbf{r} = \mu(2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k})\) | M1 | |
| Obtain \(\mu = \frac{23}{49}\) [and \(\left(\frac{46}{49}, -\frac{69}{49}, \frac{138}{49}\right)\) as foot of perpendicular] | A1 | |
| Obtain distance \(\frac{23}{7}\) or equivalent | A1 | [3] |
| (iii) Either Recognise that plane is \(2x - 3y + 6z = k\) and attempt use of formula for perpendicular distance to plane at least once | M1 | |
| Obtain \(\frac{ | 23 - k | }{7} = 14\) or equivalent |
| Obtain \(2x - 3y + 6z = 121\) and \(2x - 3y + 6z = -75\) | A1 | [3] |
| OR Recognise that plane is \(2x - 3y + 6z = k\) and attempt to find at least one point on \(q\) using \(l\) with \(\lambda = \pm 2\) | M1 | |
| Obtain \(2x - 3y + 6z = 121\) | A1 | |
| Obtain \(2x - 3y + 6z = -75\) | A1 | [3] |
(i) Obtain $2x - 3y + 6z$ for LHS of equation | B1 |
Obtain $2x - 3y + 6z = 23$ | B1 | [2]
(ii) Either **Use correct formula to find perpendicular distance** | M1 |
Obtain unsimplified value $\frac{\pm 23}{\sqrt{2^2 + (-3)^2 + 6^2}}$, following answer to (i) | A1✓ |
Obtain $\frac{23}{7}$ or equivalent | A1 | [3]
OR 1 **Use scalar product of $(4, -1, 2)$ and a vector normal to the plane** | M1 |
Use unit normal to plane to obtain $\pm\frac{(8 + 3 + 12)}{\sqrt{49}}$ | A1 |
Obtain $\frac{23}{7}$ or equivalent | A1 | [3]
OR 2 **Find parameter intersection of $p$ and $\mathbf{r} = \mu(2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k})$** | M1 |
Obtain $\mu = \frac{23}{49}$ [and $\left(\frac{46}{49}, -\frac{69}{49}, \frac{138}{49}\right)$ as foot of perpendicular] | A1 |
Obtain distance $\frac{23}{7}$ or equivalent | A1 | [3]
(iii) Either **Recognise that plane is $2x - 3y + 6z = k$ and attempt use of formula for perpendicular distance to plane at least once** | M1 |
Obtain $\frac{|23 - k|}{7} = 14$ or equivalent | A1 |
Obtain $2x - 3y + 6z = 121$ and $2x - 3y + 6z = -75$ | A1 | [3]
OR **Recognise that plane is $2x - 3y + 6z = k$ and attempt to find at least one point on $q$ using $l$ with $\lambda = \pm 2$** | M1 |
Obtain $2x - 3y + 6z = 121$ | A1 |
Obtain $2x - 3y + 6z = -75$ | A1 | [3]7 The straight line $l$ has equation $\mathbf { r } = 4 \mathbf { i } - \mathbf { j } + 2 \mathbf { k } + \lambda ( 2 \mathbf { i } - 3 \mathbf { j } + 6 \mathbf { k } )$. The plane $p$ passes through the point $( 4 , - 1,2 )$ and is perpendicular to $l$.\\
(i) Find the equation of $p$, giving your answer in the form $a x + b y + c z = d$.\\
(ii) Find the perpendicular distance from the origin to $p$.\\
(iii) A second plane $q$ is parallel to $p$ and the perpendicular distance between $p$ and $q$ is 14 units. Find the possible equations of $q$.
\hfill \mbox{\textit{CAIE P3 2014 Q7 [8]}}