22M.1.AHL.TZ1.11

METHOD 1

attempt to eliminate a variable                 M1

obtain a pair of equations in two variables

EITHER

$-3x+z=-3$ and          A1

$-3x+z=44$          A1

OR

$-5x+y=-7$ and          A1

$-5x+y=40$          A1

OR

$3x-z=3$ and          A1

$3x-z=-\frac{79}{5}$          A1

THEN

the two lines are parallel ($-3\ne 44$ or $-7\ne 40$ or $3\ne -\frac{79}{5}$)          R1

 

Note: There are other possible pairs of equations in two variables.
To obtain the final R1, at least the initial M1 must have been awarded.

 

hence the three planes do not intersect          AG

 

METHOD 2

vector product of the two normals $=\left(\begin{array}{c}-1\\ -5\\ -3\end{array}\right)$  (or equivalent)          A1

$\mathit{r}=\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)+\lambda \left(\begin{array}{c}1\\ 5\\ 3\end{array}\right)$  (or equivalent)          A1

 

Note: Award A0 if “$r=$” is missing. Subsequent marks may still be awarded.

 

Attempt to substitute $\left(1+\lambda ,-2+5\lambda ,3\lambda \right)$ in $\prod _3$                 M1

$-9\left(1+\lambda \right)+3\left(-2+5\lambda \right)-2\left(3\lambda \right)=32$

$-15=32$, a contradiction          R1

hence the three planes do not intersect          AG

 

METHOD 3

attempt to eliminate a variable                M1

$-3y+5z=6$          A1

$-3y+5z=100$          A1

$0=94$, a contradiction           R1

 

Note: Accept other equivalent alternatives. Accept other valid methods.
To obtain the final R1, at least the initial M1 must have been awarded.

 

hence the three planes do not intersect          AG

 

[4 marks]

$\prod _1:2+2+0=4$  and  $\prod _2:1+4+0=5$          A1

 

[1 mark]

METHOD 1

attempt to find the vector product of the two normals          M1

$\left(\begin{array}{c}2\\ -1\\ 1\end{array}\right)\times \left(\begin{array}{c}1\\ -2\\ 3\end{array}\right)$

$=\left(\begin{array}{c}-1\\ -5\\ -3\end{array}\right)$          A1

$\mathit{r}=\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)+\lambda \left(\begin{array}{c}1\\ 5\\ 3\end{array}\right)$          A1A1

 

Note: Award A1A0 if “$\mathit{r}=$” is missing.
Accept any multiple of the direction vector.
Working for (b)(ii) may be seen in part (a) Method 2. In this case penalize lack of “$\mathit{r}=$” only once.

 

METHOD 2

attempt to eliminate a variable from $\prod _1$ and $\prod _2$          M1

$3x-z=3$  OR  $3y-5z=-6$  OR  $5x-y=7$

Let $x=t$

substituting $x=t$ in $3x-z=3$ to obtain

$z=-3+3t$  and  $y=5t-7$ (for all three variables in parametric form)          A1

$\mathit{r}=\left(\begin{array}{c}0\\ -7\\ -3\end{array}\right)+\lambda \left(\begin{array}{c}1\\ 5\\ 3\end{array}\right)$          A1A1

 

Note: Award A1A0 if “$\mathit{r}=$” is missing.
Accept any multiple of the direction vector. Accept other position vectors which satisfy both the planes $\prod _1$ and $\prod _2$ .

 

[4 marks]

METHOD 1

the line connecting $L$ and $\prod _3$ is given by $L_1$

attempt to substitute position and direction vector to form $L_1$           (M1)

$\mathit{s}=\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)+t\left(\begin{array}{c}-9\\ 3\\ -2\end{array}\right)$          A1

substitute $\left(1-9t,-2+3t,-2t\right)$ in $\prod _3$             M1

$-9\left(1-9t\right)+3\left(-2+3t\right)-2\left(-2t\right)=32$

$94t=47\Rightarrow t=\frac{1}{2}$          A1

attempt to find distance between $\left(1,-2,0\right)$ and their point $\left(-\frac{7}{2},-\frac{1}{2},-1\right)$           (M1)

$=\left|\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)+\frac{1}{2}\left(\begin{array}{c}-9\\ 3\\ -2\end{array}\right)-\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)\right|=\frac{1}{2}\sqrt{{\left(-9\right)}^2+3^2+{\left(-2\right)}^2}$

$=\frac{\sqrt{94}}{2}$          A1

 

METHOD 2

unit normal vector equation of $\prod _3$ is given by $\frac{\left(\begin{array}{c}-9\\ 3\\ 2\end{array}\right)·\left(\begin{array}{c}x\\ y\\ z\end{array}\right)}{\sqrt{81+9+4}}$           (M1)

$=\frac{32}{\sqrt{94}}$          A1

let $\prod _4$ be the plane parallel to $\prod _3$ and passing through $\text{P}$, 
then the normal vector equation of $\prod _4$ is given by

$\left(\begin{array}{c}-9\\ 3\\ 2\end{array}\right)·\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}-9\\ 3\\ 2\end{array}\right)·\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)=-15$             M1

 

unit normal vector equation of $\prod _4$ is given by

$\frac{\left(\begin{array}{c}-9\\ 3\\ 2\end{array}\right)·\left(\begin{array}{c}x\\ y\\ z\end{array}\right)}{\sqrt{81+9+4}}=\frac{-15}{\sqrt{94}}$          A1

distance between the planes is $\frac{32}{\sqrt{94}}-\frac{-15}{\sqrt{94}}$           (M1)

$=\frac{47}{\sqrt{94}}\left(=\frac{\sqrt{94}}{2}\right)$          A1

 

[6 marks]

Part (a) was well attempted using a variety of approaches. Most candidates were able to gain marks for part (a) through attempts to eliminate a variable with many subsequently making algebraic errors. Part (b)(i) was well done. For part (b)(ii) few successful attempts were noted, many candidates failed to use an appropriate notation "r =" while giving the vector equation of a line. Part (c) proved to be challenging for most candidates with very few correct answers seen. Many candidates did not attempt part (c).

[N/A]

[N/A]

[N/A]