A pair of skew lines ${{l}_{1}}$ and ${{l}_{2}}$ have equations
$\mathbf{r}=\left( \begin{matrix}
-1 \\
1 \\
3 \\
\end{matrix} \right)+\lambda \left( \begin{matrix}
-2 \\
2 \\
1 \\
\end{matrix} \right)$, $\lambda \in \mathbb{R}$ and $\mathbf{r}=\left( \begin{matrix}
1 \\
3 \\
2 \\
\end{matrix} \right)+\mu \left( \begin{matrix}
3 \\
1 \\
6 \\
\end{matrix} \right)$, $\mu \in \mathbb{R}$
respectively. Suppose the points $P$ and $Q$ lie on ${{l}_{1}}$ and ${{l}_{2}}$ respectively such that $PQ$ is the shortest distance between ${{l}_{1}}$ and ${{l}_{2}}$ , find the length of $PQ$.[4]
(Hint: What relationship(s) does $\overrightarrow{PQ}$ have with ${{l}_{1}}$ and ${{l}_{2}}$?)

A pair of skew lines ${{l}_{1}}$ and ${{l}_{2}}$ have equations

$\mathbf{r}=\left( \begin{matrix}
-1 \\
1 \\
3 \\
\end{matrix} \right)+\lambda \left( \begin{matrix}
-2 \\
2 \\
1 \\
\end{matrix} \right)$, $\lambda \in \mathbb{R}$ and $\mathbf{r}=\left( \begin{matrix}
1 \\
3 \\
2 \\
\end{matrix} \right)+\mu \left( \begin{matrix}
3 \\
1 \\
6 \\
\end{matrix} \right)$, $\mu \in \mathbb{R}$

respectively. Suppose the points $P$ and $Q$ lie on ${{l}_{1}}$ and ${{l}_{2}}$ respectively such that $PQ$ is the shortest distance between ${{l}_{1}}$ and ${{l}_{2}}$ , find the length of $PQ$.

(Hint: What relationship(s) does $\overrightarrow{PQ}$ have with ${{l}_{1}}$ and ${{l}_{2}}$?)

Method I
(Use of Magic $\vartriangle $ without finding $P$

$\left| \overrightarrow{AB}\cdot \frac{\overrightarrow{PQ}}{\left| \overrightarrow{PQ} \right|} \right|$ $\overrightarrow{AB}=\left( \begin{matrix}
1 \\
3 \\
2 \\
\end{matrix} \right)-\left( \begin{matrix}
-1 \\
1 \\
3 \\
\end{matrix} \right)$

Shortest Distance
$=$Projection length of $AB$ onto $\overrightarrow{PQ}$
$=\left| \overrightarrow{AB}\cdot \frac{\overrightarrow{PQ}}{\left| \overrightarrow{PQ} \right|} \right|$; $\overrightarrow{PQ}=\left( \begin{matrix}
-2 \\
2 \\
1 \\
\end{matrix} \right)\times \left( \begin{matrix}
3 \\
1 \\
6 \\
\end{matrix} \right)=\left( \begin{matrix}
11 \\
15 \\
-8 \\
\end{matrix} \right)$
$=\frac{\left| \left( \begin{matrix}
2 \\
2 \\
-1 \\
\end{matrix} \right)\cdot \left( \begin{matrix}
11 \\
15 \\
-8 \\
\end{matrix} \right) \right|}{\sqrt{{{11}^{2}}+{{15}^{2}}+{{\left( -8 \right)}^{2}}}}=\frac{\left| 22+30+8 \right|}{\sqrt{410}}$
$=\frac{60}{\sqrt{410}}\cdot \frac{\sqrt{410}}{\sqrt{410}}=\frac{60}{410}\cdot \sqrt{410}$

Method II (Find $P$ and $Q$ first)

For some $\lambda \in \mathbb{R}$: $\overrightarrow{OP}=\left( \begin{matrix}
-1 \\
1 \\
3 \\
\end{matrix} \right)+\lambda \left( \begin{matrix}
-2 \\
2 \\
1 \\
\end{matrix} \right)$

For some $\mu \in \mathbb{R}$: $\overrightarrow{OQ}=\left( \begin{matrix}
1 \\
3 \\
2 \\
\end{matrix} \right)+\mu \left( \begin{matrix}
3 \\
1 \\
6 \\
\end{matrix} \right)$
$\overrightarrow{PQ}=\left[ \left( \begin{matrix}
1 \\
3 \\
2 \\
\end{matrix} \right)+\mu \left( \begin{matrix}
3 \\
1 \\
6 \\
\end{matrix} \right) \right]$

$-\left[ \left( \begin{matrix}
-1 \\
1 \\
3 \\
\end{matrix} \right)+\lambda \left( \begin{matrix}
-2 \\
2 \\
1 \\
\end{matrix} \right) \right]$
=$\left( \begin{matrix}
2 \\
2 \\
-1 \\
\end{matrix} \right)+\mu \left( \begin{matrix}
3 \\
1 \\
6 \\
\end{matrix} \right)+\lambda \left( \begin{matrix}
2 \\
-2 \\
-1 \\
\end{matrix} \right)$

$\overrightarrow{PQ}\cdot {{\mathbf{d}}_{1}}=0$
$\left[ \left( \begin{matrix}
2 \\
2 \\
-1 \\
\end{matrix} \right)+\mu \left( \begin{matrix}
3 \\
1 \\
6 \\
\end{matrix} \right)+\lambda \left( \begin{matrix}
2 \\
-2 \\
-1 \\
\end{matrix} \right)\cdot \left( \begin{matrix}
-2 \\
2 \\
1 \\
\end{matrix} \right) \right]=0$
$\left( -4+4-1 \right)+\mu \left( -6+2+6 \right)+\lambda \left( -4-4-1 \right)=0$
$2\mu -9\lambda =1$------- $1$.

$\overrightarrow{PQ}\cdot {{\mathbf{d}}_{2}}=0$
$\left[ \left( \begin{matrix}
2 \\
2 \\
-1 \\
\end{matrix} \right)+\mu \left( \begin{matrix}
3 \\
1 \\
6 \\
\end{matrix} \right)+\lambda \left( \begin{matrix}
2 \\
-2 \\
-1 \\
\end{matrix} \right)\cdot \left( \begin{matrix}
3 \\
1 \\
6 \\
\end{matrix} \right) \right]=0$
$\left( \left( 6+2-6 \right)+\mu \left( 9+1+36 \right)+\lambda \left( 6-2-6 \right) \right)=0$
$46\mu -2\lambda =-2$--------$2$.

Solve $1\And 2$:
$\mu =-\frac{2}{41}$, $\lambda =-\frac{5}{41}$

$\overrightarrow{OP}=\left( \begin{matrix}
-1 \\
1 \\
3 \\
\end{matrix} \right)-\frac{5}{41}\left( \begin{matrix}
-2 \\
2 \\
1 \\
\end{matrix} \right)$
$=\left( \begin{matrix}
-\frac{31}{41} \\
\frac{31}{41} \\
\frac{118}{41} \\
\end{matrix} \right)=\frac{1}{41}\left( \begin{matrix}
-31 \\
31 \\
118 \\
\end{matrix} \right)$

$\overrightarrow{OQ}=\left( \begin{matrix}
1 \\
3 \\
2 \\
\end{matrix} \right)-\frac{2}{41}\left( \begin{matrix}
3 \\
1 \\
6 \\
\end{matrix} \right)$
$\left( \begin{matrix}
\frac{35}{41} \\
\frac{121}{41} \\
\frac{70}{41} \\
\end{matrix} \right)=\frac{1}{41}\left( \begin{matrix}
35 \\
121 \\
70 \\
\end{matrix} \right)$

$\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}$
$ =\frac{1}{41}\left( \begin{matrix}
35 \\
121 \\
70 \\
\end{matrix} \right)-\frac{1}{41}\left( \begin{matrix}
-31 \\
31 \\
118 \\
\end{matrix} \right)$
$=\frac{1}{41}\left( \begin{matrix}
66 \\
90 \\
-48 \\
\end{matrix} \right)$

$\left| \overrightarrow{PQ} \right|$
$ =\sqrt{{{\left( \frac{1}{41} \right)}^{2}}\left( {{66}^{2}}+{{90}^{2}}+{{48}^{2}} \right)}$
$ =\sqrt{\frac{14760}{{{41}^{2}}}}=\sqrt{\frac{360}{41}}$
$=\sqrt{\frac{36\times 10}{41}}$
$=\frac{6\sqrt{10}}{\sqrt{41}}\cdot \frac{\sqrt{41}}{\sqrt{41}}$

Vectors - General

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H2 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

Shortest Distance Between 2 Skew Lines

**Method I (Magic Triangle - without finding P and Q)**

Let $A$ and $B$ be points $\begin{pmatrix} -1 \\ 1 \\ 3 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 3 \\ 2 \end{pmatrix}$ respectively.

$\overrightarrow{PQ}$ is perpendicular to both $\mathbf{d}_1$ and $\mathbf{d}_2$

$\overrightarrow{PQ} = \mathbf{d}_1 \times \mathbf{d}_2$

$\overrightarrow{AB} = \begin{pmatrix} 1 \\ 3 \\ 2 \end{pmatrix} - \begin{pmatrix} -1 \\ 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}$

$\overrightarrow{PQ} = \begin{pmatrix} -2 \\ 2 \\ 1 \end{pmatrix} \times \begin{pmatrix} 3 \\ 1 \\ 6 \end{pmatrix} = \begin{pmatrix} 11 \\ 15 \\ -8 \end{pmatrix}$

Shortest distance = Projection length of $\overrightarrow{AB}$ onto $\overrightarrow{PQ}$

$= \left| \overrightarrow{AB} \cdot \frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|} \right|$

$= \frac{\left| \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 11 \\ 15 \\ -8 \end{pmatrix} \right|}{\sqrt{11^2 + 15^2 + (-8)^2}}$

$= \frac{|22 + 30 + 8|}{\sqrt{410}}$

$= \frac{60}{\sqrt{410}} = \frac{6}{41}\sqrt{410}$ units

**Method II (Find P and Q first)**

For some $\lambda \in \mathbb{R}$: $\overrightarrow{OP} = \begin{pmatrix} -1 \\ 1 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} -2 \\ 2 \\ 1 \end{pmatrix}$

For some $\mu \in \mathbb{R}$: $\overrightarrow{OQ} = \begin{pmatrix} 1 \\ 3 \\ 2 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ 1 \\ 6 \end{pmatrix}$

$\overrightarrow{PQ} = \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ 1 \\ 6 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -2 \\ -1 \end{pmatrix}$

Using $\overrightarrow{PQ} \cdot \mathbf{d}_1 = 0$:
$2\mu - 9\lambda = 1$ --- (1)

Using $\overrightarrow{PQ} \cdot \mathbf{d}_2 = 0$:
$46\mu - 2\lambda = -2$ --- (2)

Solving (1) and (2):
$\mu = -\frac{2}{41}$, $\lambda = -\frac{5}{41}$

$\overrightarrow{OP} = \frac{1}{41}\begin{pmatrix} -31 \\ 31 \\ 118 \end{pmatrix}$

$\overrightarrow{OQ} = \frac{1}{41}\begin{pmatrix} 35 \\ 121 \\ 70 \end{pmatrix}$

$\overrightarrow{PQ} = \frac{1}{41}\begin{pmatrix} 66 \\ 90 \\ -48 \end{pmatrix}$

$|\overrightarrow{PQ}| = \sqrt{\frac{1}{41^2}(66^2 + 90^2 + 48^2)} = \sqrt{\frac{14760}{1681}} = \frac{6}{41}\sqrt{410}$ units

Shortest Distance Between 2 Skew Lines

Shortest Distance Between 2 Skew Lines

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