CAIE A-Level Mathematics A2 3.7 Vectors Questions

Perform scalar product of direction vectors and set result equal to zero
2 c+6+4=0.

Use P to find the value of $\lambda$$3-2 \lambda=7 \Rightarrow \lambda=-2[a+\lambda \mathrm{c}=4, b+4 \lambda=-2]$.

Equation for line l may contain $-\lambda$ instead of $+\lambda$ leading to $\lambda=2$ all marks available.

Obtain c=-5 or b=6
a=-6, b=6 and c=-5 all correct
4

SC1: Use P to find the value of $\lambda$ M1

Substitute $\lambda=-2$ into point P, so a-2 c=4, and put $\mu=$ -1 and $\lambda=-1$ into l so a-c=-1, then solve to obtain a=-6, b=6 and c=-5.

All 3 values correct A1.

Marking guidance:

Max 2/4.

Find $\overrightarrow{P Q}$ (or $\overrightarrow{Q P}$ ) for a general point Q on m
$$= \pm((1+2 \mu, 2-3 \mu, 3+\mu)-(a+\lambda c, 3-2 \lambda, b+4 \lambda))$$$$\left[\overrightarrow{P Q} \text { or } \overrightarrow{Q P}= \pm\left(\begin{array}{c} -3+2 \mu -5-3 \mu 5+\mu \end{array}\right)\right]$$

Could be their a, b, c and $\lambda$ values provided M1 M1 gained in (a). Allow expression in answer column.

Equate the scalar product of $\overrightarrow{P Q}$ (or $\overrightarrow{Q P}$ ) and a direction vector for m to
zero and obtain an equation in $\mu$

M1*

$$\begin{aligned} (2(-3+2 \mu)-3(-5-3 \mu)+(5+\mu))=0 \text { Allow } \overrightarrow{P Q}=\overrightarrow{O Q}+\overrightarrow{O P} \text { sign problem. } \end{aligned}$$

Solve and obtain $\mu=-1$$$\begin{aligned} P Q^{2}=(-3+2 \mu)^{2}+(-5-3 \mu)^{2}+(5+\mu)^{2} {\left[=14(\mu+1)^{2}+45\right] .} \text { Min when } \mu=-1 \text { or by differentiation. } \end{aligned}$$

Obtain $\overrightarrow{O Q}=-\mathbf{i}+5 \mathbf{j}+2 \mathbf{k}$ or $\overrightarrow{P Q}=-5 \mathbf{i}-2 \mathbf{j}+4 \mathbf{k}$

Must be labelled correctly
The working may be in (a) provided at least this result is used in (b).

Carry out a method to find the position vector of R

Alternative method for DM1
$$\begin{aligned} & \overrightarrow{OR}=(4,7,-2)+t(-5,-2,4) \overrightarrow{Q R}=\overrightarrow{OR}-\overrightarrow{O Q} & \text { Solve }|\boldsymbol{Q R}|^{2}=\frac{9}{4}|\boldsymbol{P Q}|^{2} \text { or }|\boldsymbol{Q R}|=\frac{3}{2}|\boldsymbol{P Q}| t=2.5 \end{aligned}$$
e.g. Use $\overrightarrow{OR}=\overrightarrow{O P}+\frac{5}{2} \overrightarrow{P Q}$ or $\overrightarrow{OR}=\overrightarrow{O Q}+\frac{3}{2} \overrightarrow{P Q}$ or $\overrightarrow{OR}=\frac{5}{2} \overrightarrow{O Q}-\frac{3}{2} \overrightarrow{O P}$ or $2 \overrightarrow{Q R}=2(\overrightarrow{OR}-\overrightarrow{O Q})=3 \overrightarrow{P Q}$ where $\overrightarrow{OR}=(x, y, z)$.

$\overrightarrow{P Q}$ used in all these approaches, may be incorrect, must be in the correct direction, i.e. not using $\overrightarrow{Q P}$ for $\overrightarrow{P Q}$.

Obtain $-\frac{17}{2} \boldsymbol{i}+2 \boldsymbol{j}+8 \boldsymbol{k}$ from correct working
Accept coordinates.
$$\text { Don't accept }-\frac{17}{2} \mathbf{i}+\frac{4}{2} \mathbf{j}+\frac{16}{2} \mathbf{k} \text {. }$$

SC2 Equate lines, attempt to find $\mu=-1$ or $\lambda=-1$ M1*
$\overrightarrow{O Q}=-\mathbf{i}+5 \mathbf{j}+2 \mathbf{k} \quad \mathrm{~A} 1$.
Attempt to find $\overrightarrow{O Q}$ using other parameter value DM1.
$\overrightarrow{O Q}=-\mathbf{i}+5 \mathbf{j}+2 \mathbf{k}$ therefore intersect A1.
Then use main scheme for the final DM1 A1.

First DM1 A1 are available if they show the 3
coordinates are consistent for the 2 parameter values
instead of attempting to find $\overrightarrow{O Q}$ using the other
parameter value and then showing intersection

Correct direction vector seen or implied ( $\overrightarrow{B C}=3 \mathbf{i}+3 \mathbf{j}-3 \mathbf{k}$ )
Condone $\overrightarrow{B C}=-3 \mathbf{i}-3 \mathbf{j}+3 \mathbf{k}$.

Use a correct method to form a vector equation
Allow for the RHS with no LHS.

Obtain $\mathbf{r}=5 \mathbf{i}+2 \mathbf{j}+\lambda(\mathbf{i}+\mathbf{j}-\mathbf{k})$
ISW

Must have $\mathbf{r}=\ldots$ or $\left(\begin{array}{l}x y z\end{array}\right)=\ldots$, not $l_{1}=\ldots$

Or, equivalent vector form, e.g. $\mathbf{r}=8 \mathbf{i}+5 \mathbf{j}-3 \mathbf{k}+\alpha(\mathbf{i}+\mathbf{j}-\mathbf{k})$ or $\mathbf{r}=5 \mathbf{i}+2 \mathbf{j}+\lambda(3 \mathbf{i}+3 \mathbf{j}-3 \mathbf{k})$.

Condone a column vector with i, j, k.

Use components to form two relevant equations in 2 unknowns
For their $l_{1} \mathrm{~B} 0$ if they use the same unknown for both lines.

B1FT

Two components of $\left(\begin{array}{c}5+\lambda 2+\lambda -\lambda\end{array}\right)=\left(\begin{array}{c}-2+3 \mu 1+\mu 4-2 \mu\end{array}\right)$ seen or implied.

Solve 2 relevant equations in 2 unknowns for $\lambda$ or $\mu$
For their $l_{1}$.

Obtain $\lambda=2$ or $\mu=3$
Or equivalent e.g. using $\overrightarrow{B C}$ as direction vector gives $\lambda=\frac{2}{3}$.

Obtain (7,4,-2)

No need to check the third equation - the question implies that the lines intersect.
Accept position vector. Condone a column vector with i, j, k.
SC: B1 M1 A1 A1 if one component of their line is incorrect but they do not use that component.

State $A B=\sqrt{7^{2}+1^{2}+4^{2}}(=\sqrt{66})$
Or $(A B)^{2}=66$ Condone a sign error in $\overrightarrow{A B}$.

State $\overrightarrow{B D}$ in component form
$\left(\begin{array}{c}-7+3 r -1+r 4-2 r\end{array}\right)$ or equivalent.

$$\begin{aligned} A B=B D \Rightarrow(3 r-7)^{2}+(r-1)^{2}+(-2 r+4)^{2}=66 \left(14 r^{2}-60 r=0\right) \end{aligned}$$
Or equivalent equation in one unknown for their A B and their $\overrightarrow{B D} \neq \overrightarrow{O D}$.

If you never see a correct form and they go direct to $9 r^{2}+49+r^{2}+1 \ldots .$. then M0.

$\Rightarrow r=\frac{30}{7}$
Correct only. Ignore r=0 if seen.

$\overrightarrow{O D}=\frac{76}{7} \mathbf{i}+\frac{37}{7} \mathbf{j}-\frac{32}{7} \mathbf{k}$
Must be a vector.

Condone if also have $\overrightarrow{O D}=\overrightarrow{O A}$.

Carry out correct process for evaluating the scalar product of direction vectors,
equate the result to zero and obtain given value of a=4
E.g. 2(3)+(-1)(-2)+a(-2)=0.

Express general point of at least one line correctly in component form,
i.e. $(1+2 \lambda,-2-\lambda, 3+4 \lambda)$ or $(-1+3 \mu,-1-2 \mu,-1-2 \mu)$
The third component could be implied by a correct final answer.

Equate at least two pairs of corresponding components and solve for $\lambda$ or for $\mu$
Obtain $\lambda=-1$ or $\mu=0$
Obtain position vector of point of intersection is -i-j-k

Equate one component of $l_{1}$ to matching component of A and solve to find $\lambda$
Use $\lambda=-3$ in equation of $l_{1}$ and show this gives position vector of A
AG
Or show $\lambda=-3$ for all three components equated.

Method to find position vector of B
E.g. $\pm 2 \times$ their $(-\mathbf{i}-\mathbf{j}-\mathbf{k}) \pm(-5 \mathbf{i}+\mathbf{j}-9 \mathbf{k})$

Obtain position vector of B is 3 i-3 j+7 k