Pearson Edexcel IAL Mathematics FP3.5.2 Use of vectors in problems

Practise applying vectors to lines, planes, distances, intersections and angles, choosing forms that make the 3D geometry efficient.

Exam points

substitute a line into a plane to find an intersection point or parameter value
find a line of intersection of two planes using a point and a direction vector
calculate shortest distances or acute angles using normals and direction vectors

[Maximum number: 5]

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.

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

Give your answer in simplest form.

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

M1 A1 M1 M1 A1

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FP3.5.2 - Use of vectors in problems question 1