Pearson Edexcel IAL Mathematics A2 FP3.5 Vectors

[Maximum number: 10]

The skew lines $l_{1}$ and $l_{2}$ have equations

l_{1}: \mathbf{r}=(\mathbf{i}+2 \mathbf{j}-5 \mathbf{k})+\lambda(5 \mathbf{i}+\mathbf{j})

and

l_{2}: \mathbf{r}=(2 \mathbf{i}-4 \mathbf{j}+4 \mathbf{k})+\mu(8 \mathbf{i}-2 \mathbf{j}+3 \mathbf{k})

where $\lambda$ and $\mu$ are scalar parameters.

(a)

Determine a vector that is perpendicular to both $l_{1}$ and $l_{2}$

[ 2 ]

$(5\mathbf i+\mathbf j)\times(8\mathbf i-2\mathbf j+3\mathbf k)$
or solve
$(u\mathbf i+v\mathbf j+w\mathbf k)\cdot(5\mathbf i+\mathbf j)=0, \quad (u\mathbf i+v\mathbf j+w\mathbf k)\cdot(8\mathbf i-2\mathbf j+3\mathbf k)=0$ $5u+v=0,\quad 8u-2v+3w=0$ $\mathbf n=3\mathbf i-15\mathbf j-18\mathbf k$
or any non-zero multiple of $\mathbf i-5\mathbf j-6\mathbf k$ .

(b)

Determine an equation of the plane parallel to $l_{1}$ that contains $l_{2}$

[ 3 ]

(i)

in the form r=a+s b+t c

[ 1 ]

$\mathbf r=(2\mathbf i-4\mathbf j+4\mathbf k) +s(8\mathbf i-2\mathbf j+3\mathbf k) +t(5\mathbf i+\mathbf j)$

(ii)

in the form r.n =p

[ 2 ]

$(2\mathbf i-4\mathbf j+4\mathbf k)\cdot(3\mathbf i-15\mathbf j-18\mathbf k)=-6$ $\mathbf r\cdot(3\mathbf i-15\mathbf j-18\mathbf k)=-6$
or
$\mathbf r\cdot(-\mathbf i+5\mathbf j+6\mathbf k)=2$

(c)

Determine the shortest distance between $l_{1}$ and $l_{2}$

Give your answer in simplest form.

[ 5 ]

Way 1:
$\frac{6}{\sqrt{3^2+15^2+18^2}}=\frac{2}{\sqrt{62}}, \quad \frac{21}{\sqrt{62}}$ $d=\frac{2}{\sqrt{62}}+\frac{21}{\sqrt{62}} =\frac{23}{\sqrt{62}} =\frac{23\sqrt{62}}{62}$

OR

Way 2:
$\overrightarrow{AB}=\pm(-\mathbf i+6\mathbf j-9\mathbf k)$ $d=\frac{|\overrightarrow{AB}\cdot\mathbf n|}{|\mathbf n|} =\frac{69}{\sqrt{558}} =\frac{23}{\sqrt{62}}$

Edexcel A-Level Mathematics A2 Fp3 5 Vectors Questions

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (i)

Question (ii)

Question (c)