CAIE A-Level Further Math 1.6 Vectors Question Bank

$\overrightarrow{A B}=-2 \mathbf{i}+\mathbf{j}+4 \mathbf{k} \quad \overrightarrow{A C}=-3 \mathbf{i}+3 \mathbf{k} \quad \overrightarrow{B C}=-\mathbf{i}-\mathbf{j}-\mathbf{k}$

B1

Finds direction vectors of two lines in the plane.

$\left|\begin{array}{ccc}\mathbf{i} \mathbf{j} \mathbf{k} -2 1 -3 0 3\end{array}\right|=\left(\begin{array}{c}3 -6 3\end{array}\right) \sim\left(\begin{array}{c}1 -2 1\end{array}\right)$

M1 A1FT

Finds normal to the plane ABC.

$1(1)-2(1)+1(0)=-1 \Rightarrow x-2 y+z=-1$

M1 A1

Substitutes point. CAO

5

$\frac{1}{\sqrt{1^{2}+2^{2}+1^{2}}}=\frac{1}{\sqrt{6}}=0.408$

M1 A1FT

Divides by magnitude of normal vector.
FT their (a).

2

$\overrightarrow{O P}=\left(\begin{array}{c}-2 \lambda \lambda 3 \lambda\end{array}\right), \overrightarrow{O Q}=\left(\begin{array}{c}1-2 \mu 1+\mu 4 \mu\end{array}\right) \Rightarrow \overrightarrow{P Q}=\left(\begin{array}{c}1-2 \mu+2 \lambda 1+\mu-\lambda 4 \mu-3 \lambda\end{array}\right)$

M1 A1

Finds $\overrightarrow{P Q}$, where P is a point on OC and Q is a point on
AB.

$\left(\begin{array}{c}1-2 \mu+2 \lambda 1+\mu-\lambda 4 \mu-3 \lambda\end{array}\right) \cdot\left(\begin{array}{c}-2 1 3\end{array}\right)=0$

M1*

Uses that dot product of $\overrightarrow{P Q}$ with line direction is zero.

$17 \mu-14 \lambda=1$

dM1

Deduces one equation.

$\left(\begin{array}{c}1-2 \mu+2 \lambda 1+\mu-\lambda 4 \mu-3 \lambda\end{array}\right) \cdot\left(\begin{array}{c}-2 1 4\end{array}\right)=0 \Rightarrow 21 \mu-17 \lambda=1$

dM1

Deduces second equation.

$\lambda=-\frac{4}{5} \Rightarrow \overrightarrow{O P}=-\frac{4}{5}\left(\begin{array}{c}-2 1 3\end{array}\right)$

dM1 A1

Solves for $\lambda$ or $\mu$ and substitutes into $\overrightarrow{O P}$.

$\mathbf{r}=-\frac{4}{5}\left(\begin{array}{c}-2 1 3\end{array}\right)+k\left(\begin{array}{l}1 2 0\end{array}\right)$

A1

OE

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