| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Foot of perpendicular from origin to line |
| Difficulty | Standard +0.3 This is a standard two-part vectors question requiring routine techniques: (i) using perpendicularity condition (dot product = 0) to find a point on a line, and (ii) finding position vectors parallel to a direction with given magnitude. Both parts are textbook exercises with straightforward algebraic manipulation, making this slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{OP} = \begin{pmatrix}-5+2\lambda \\ 1-3\lambda \\ 6+\lambda\end{pmatrix}\) or coordinates of \(P\) are \((-5+2\lambda, 1-3\lambda, 6+\lambda)\) | B1 | May be implied by use in scalar product |
| \(\overrightarrow{OP} \cdot \begin{pmatrix}2\\-3\\1\end{pmatrix} = 0 \Rightarrow 2(-5+2\lambda) - 3(1-3\lambda) + 1(6+\lambda) = 0 \Rightarrow 14\lambda = 7\) | M1 | Attempts scalar product with \(\begin{pmatrix}2\\-3\\1\end{pmatrix} = 0\); may use any multiple of \(2\mathbf{i}-3\mathbf{j}+\mathbf{k}\) |
| \(\lambda = \frac{1}{2}\) | A1 | |
| Substitutes \(\lambda = \frac{1}{2}\) into \(\overrightarrow{OP}\) | dM1 | Dependent on previous M |
| \(P\) has coordinates \((-4, -0.5, 6.5)\) | A1 | Accept column vector or \(-4\mathbf{i} - 0.5\mathbf{j} + 6.5\mathbf{k}\) |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{OA} = k\times\begin{pmatrix}5\\-3\\4\end{pmatrix}\) or coordinates of \(A\) are \((5k,-3k,4k)\) | M1 | May be implied by correct coordinates later |
| \(k\sqrt{5^2+(-3)^2+4^2} = \sqrt{2}\) or \(k^2(5^2+(-3)^2+4^2)=2\) | A1 | Correct equation in \(k\) using 3D Pythagoras |
| Finds at least one value of \(k\) and substitutes into \(\overrightarrow{OA}\) | dM1 | |
| As \(k = (\pm)\frac{1}{5}\), \(A\) has possible positions \(\begin{pmatrix}1\\-\frac{3}{5}\\\frac{4}{5}\end{pmatrix}\) or \(\begin{pmatrix}-1\\\frac{3}{5}\\-\frac{4}{5}\end{pmatrix}\) | A1 | Any notation |
| Both positions stated | A1 | |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{OA} = \begin{pmatrix}x\\-\frac{3}{5}x\\\frac{4}{5}x\end{pmatrix}\) or coordinates of \(A\) are \(\left(x, -\frac{3}{5}x, \frac{4}{5}x\right)\) | M1 | |
| \(x\sqrt{1^2+\left(-\frac{3}{5}\right)^2+\left(\frac{4}{5}\right)^2} = \sqrt{2}\) or equivalent | A1 | |
| Finds \(x\) and substitutes into \(\overrightarrow{OA}\) | dM1 | |
| As \(x=(\pm)1\), \(A\) has coordinates \(\left(1,-\frac{3}{5},\frac{4}{5}\right)\) or \(\left(-1,\frac{3}{5},-\frac{4}{5}\right)\) | A1 | |
| Both positions stated | A1 | |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Writes \(\frac{5}{x} = -\frac{3}{y} = \frac{4}{z}\) | M1 | 2 equations, 3 unknowns |
| Writes \(x^2+y^2+z^2 = 2\) | A1 | Third equation |
| Eliminates two variables to find the other values | dM1 | |
| \(A\) has coordinates \(\left(1,-\frac{3}{5},\frac{4}{5}\right)\) or \(\left(-1,\frac{3}{5},-\frac{4}{5}\right)\) | A1 | |
| Both positions stated | A1 | |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States \(\frac{\sqrt{2}}{\sqrt{50}}\) or \(\frac{\sqrt{50}}{\sqrt{2}}\) | M1A1 | |
| Deduces position vectors \(\mathbf{i}-\frac{3}{5}\mathbf{j}+\frac{4}{5}\mathbf{k}\) or \(-\mathbf{i}+\frac{3}{5}\mathbf{j}-\frac{4}{5}\mathbf{k}\) | M1A1 | |
| Both: \(\mathbf{i}-\frac{3}{5}\mathbf{j}+\frac{4}{5}\mathbf{k}\) and \(-\mathbf{i}+\frac{3}{5}\mathbf{j}-\frac{4}{5}\mathbf{k}\) | A1 | |
| (5) | ||
| (10 marks) |
# Question 12:
**Part (i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{OP} = \begin{pmatrix}-5+2\lambda \\ 1-3\lambda \\ 6+\lambda\end{pmatrix}$ or coordinates of $P$ are $(-5+2\lambda, 1-3\lambda, 6+\lambda)$ | B1 | May be implied by use in scalar product |
| $\overrightarrow{OP} \cdot \begin{pmatrix}2\\-3\\1\end{pmatrix} = 0 \Rightarrow 2(-5+2\lambda) - 3(1-3\lambda) + 1(6+\lambda) = 0 \Rightarrow 14\lambda = 7$ | M1 | Attempts scalar product with $\begin{pmatrix}2\\-3\\1\end{pmatrix} = 0$; may use any multiple of $2\mathbf{i}-3\mathbf{j}+\mathbf{k}$ |
| $\lambda = \frac{1}{2}$ | A1 | |
| Substitutes $\lambda = \frac{1}{2}$ into $\overrightarrow{OP}$ | dM1 | Dependent on previous M |
| $P$ has coordinates $(-4, -0.5, 6.5)$ | A1 | Accept column vector or $-4\mathbf{i} - 0.5\mathbf{j} + 6.5\mathbf{k}$ |
| | **(5)** | |
**Part (ii) Way 1:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{OA} = k\times\begin{pmatrix}5\\-3\\4\end{pmatrix}$ or coordinates of $A$ are $(5k,-3k,4k)$ | M1 | May be implied by correct coordinates later |
| $k\sqrt{5^2+(-3)^2+4^2} = \sqrt{2}$ or $k^2(5^2+(-3)^2+4^2)=2$ | A1 | Correct equation in $k$ using 3D Pythagoras |
| Finds at least one value of $k$ and substitutes into $\overrightarrow{OA}$ | dM1 | |
| As $k = (\pm)\frac{1}{5}$, $A$ has possible positions $\begin{pmatrix}1\\-\frac{3}{5}\\\frac{4}{5}\end{pmatrix}$ or $\begin{pmatrix}-1\\\frac{3}{5}\\-\frac{4}{5}\end{pmatrix}$ | A1 | Any notation |
| Both positions stated | A1 | |
| | **(5)** | |
**Part (ii) Way 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{OA} = \begin{pmatrix}x\\-\frac{3}{5}x\\\frac{4}{5}x\end{pmatrix}$ or coordinates of $A$ are $\left(x, -\frac{3}{5}x, \frac{4}{5}x\right)$ | M1 | |
| $x\sqrt{1^2+\left(-\frac{3}{5}\right)^2+\left(\frac{4}{5}\right)^2} = \sqrt{2}$ or equivalent | A1 | |
| Finds $x$ and substitutes into $\overrightarrow{OA}$ | dM1 | |
| As $x=(\pm)1$, $A$ has coordinates $\left(1,-\frac{3}{5},\frac{4}{5}\right)$ or $\left(-1,\frac{3}{5},-\frac{4}{5}\right)$ | A1 | |
| Both positions stated | A1 | |
| | **(5)** | |
**Part (ii) Way 3:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Writes $\frac{5}{x} = -\frac{3}{y} = \frac{4}{z}$ | M1 | 2 equations, 3 unknowns |
| Writes $x^2+y^2+z^2 = 2$ | A1 | Third equation |
| Eliminates two variables to find the other values | dM1 | |
| $A$ has coordinates $\left(1,-\frac{3}{5},\frac{4}{5}\right)$ or $\left(-1,\frac{3}{5},-\frac{4}{5}\right)$ | A1 | |
| Both positions stated | A1 | |
| | **(5)** | |
**Part (ii) Way 4 (minimal working):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| States $\frac{\sqrt{2}}{\sqrt{50}}$ or $\frac{\sqrt{50}}{\sqrt{2}}$ | M1A1 | |
| Deduces position vectors $\mathbf{i}-\frac{3}{5}\mathbf{j}+\frac{4}{5}\mathbf{k}$ **or** $-\mathbf{i}+\frac{3}{5}\mathbf{j}-\frac{4}{5}\mathbf{k}$ | M1A1 | |
| Both: $\mathbf{i}-\frac{3}{5}\mathbf{j}+\frac{4}{5}\mathbf{k}$ **and** $-\mathbf{i}+\frac{3}{5}\mathbf{j}-\frac{4}{5}\mathbf{k}$ | A1 | |
| | **(5)** | |
| | **(10 marks)** | |\begin{enumerate}
\item (i) Relative to a fixed origin $O$, the line $l _ { 1 }$ is given by the equation
\end{enumerate}
$$l _ { 1 } : \mathbf { r } = \left( \begin{array} { r }
- 5 \\
1 \\
6
\end{array} \right) + \lambda \left( \begin{array} { r }
2 \\
- 3 \\
1
\end{array} \right) \text { where } \lambda \text { is a scalar parameter. }$$
The point $P$ lies on $l _ { 1 }$. Given that $\overrightarrow { O P }$ is perpendicular to $l _ { 1 }$, calculate the coordinates of $P$.\\
(ii) Relative to a fixed origin $O$, the line $l _ { 2 }$ is given by the equation
$$l _ { 2 } : \mathbf { r } = \left( \begin{array} { r }
4 \\
- 3 \\
12
\end{array} \right) + \mu \left( \begin{array} { r }
5 \\
- 3 \\
4
\end{array} \right) \text { where } \mu \text { is a scalar parameter. }$$
The point $A$ does not lie on $l _ { 2 }$. Given that the vector $\overrightarrow { O A }$ is parallel to the line $l _ { 2 }$ and $| \overrightarrow { O A } | = \sqrt { 2 }$ units, calculate the possible position vectors of the point $A$.
\hfill \mbox{\textit{Edexcel C34 2015 Q12 [10]}}