Standard +0.3 This is a straightforward line intersection problem requiring students to equate parametric forms, solve a simple system of linear equations, and verify consistency. It's slightly easier than average as it's a standard C4 technique with clear methodology and minimal algebraic complexity.
\((s,t)=(-1,4)\) or \((-1,-3)\) or \((-\frac{10}{3},-\frac{2}{3})\)
*A1
or \(s=-1\) & \(-\frac{10}{3}\) or \(t=\) two of \((4,-3,-\frac{2}{3})\)
Show clear contradiction e.g. \(3\neq-4, 4\neq-3, -6\neq1\)
dep*A1 3
Allow \(\checkmark\) unsimplified contradictions. No ISW
SC: If \(s=-\frac{10}{3}\) found from 2nd & 3rd eqns and contradiction shown in 1st eqn, all 3 marks may be awarded
Part (ii):
Answer
Marks
Guidance
Answer/Working
Marks
Guidance
Work with \(\begin{pmatrix}3\\2\\1\end{pmatrix}\) and \(\begin{pmatrix}0\\1\\-1\end{pmatrix}\)
M1
Clear method for scalar product of any 2 vectors
M1
Clear method for modulus of any vector
M1
\(79.1°\) or better (79.1066..) 1.38 (rad) (1.38067..) ISW
A1 4
From \(\frac{1}{\sqrt{14}\cdot\sqrt{2}}\)
Part (iii):
Answer
Marks
Guidance
Answer/Working
Marks
Guidance
Use \(\begin{pmatrix}4+3s\\6+2s\\4+s\end{pmatrix}\cdot\begin{pmatrix}3\\2\\1\end{pmatrix}=0\)
M1
Obtain \(s=-2\)
A1
from \(12+9s+12+4s+4+s=0\)
\(A\) is \(\begin{pmatrix}-2\\2\\2\end{pmatrix}\) or \(-2\mathbf{i}+2\mathbf{j}+2\mathbf{k}\) final answer
B1 3
Accept \((-2,2,2)\)
# Question 5:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempt to set up 3 equations | M1 | of type $4+3s=1, 6+2s=t, 4+s=-t$ |
| $(s,t)=(-1,4)$ or $(-1,-3)$ or $(-\frac{10}{3},-\frac{2}{3})$ | *A1 | or $s=-1$ & $-\frac{10}{3}$ or $t=$ two of $(4,-3,-\frac{2}{3})$ |
| Show clear contradiction e.g. $3\neq-4, 4\neq-3, -6\neq1$ | dep*A1 **3** | Allow $\checkmark$ unsimplified contradictions. No ISW |
| SC: If $s=-\frac{10}{3}$ found from 2nd & 3rd eqns and contradiction shown in 1st eqn, all 3 marks may be awarded | | |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Work with $\begin{pmatrix}3\\2\\1\end{pmatrix}$ and $\begin{pmatrix}0\\1\\-1\end{pmatrix}$ | M1 | |
| Clear method for scalar product of any 2 vectors | M1 | |
| Clear method for modulus of any vector | M1 | |
| $79.1°$ or better (79.1066..) 1.38 (rad) (1.38067..) ISW | A1 **4** | From $\frac{1}{\sqrt{14}\cdot\sqrt{2}}$ |
## Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use $\begin{pmatrix}4+3s\\6+2s\\4+s\end{pmatrix}\cdot\begin{pmatrix}3\\2\\1\end{pmatrix}=0$ | M1 | |
| Obtain $s=-2$ | A1 | from $12+9s+12+4s+4+s=0$ |
| $A$ is $\begin{pmatrix}-2\\2\\2\end{pmatrix}$ or $-2\mathbf{i}+2\mathbf{j}+2\mathbf{k}$ final answer | B1 **3** | Accept $(-2,2,2)$ |
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