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Consider the following parallel lines.
\(L_1:\ {\underset{\sim}{r}}_1=\underset{\sim}{ i }+3 \underset{\sim}{ j }+\underset{\sim}{k}+s(\underset{\sim}{ i }+\underset{\sim}{ j }+\underset{\sim}{ k })\) and \(L_2:\ {\underset{\sim}{r}}_2=-2 \underset{\sim}{ i }+\underset{\sim}{ j }+3 \underset{\sim}{k}+t(\underset{\sim}{ i }+\underset{\sim}{ j }+\underset{\sim}{k })\) where \(s, t \in R\).
The shortest distance between \(L_1\) and \(L_2\) is
\(B\)
\({\underset{\sim}{r}}_1=\underset{\sim}{a}+s \underset{\sim}{d}, \quad{\underset{\sim}{r}}_2=\underset{\sim}{b}+t \underset{\sim}{d}\)
\(\underset{\sim}{a}-\underset{\sim}{b}=\left(\begin{array}{c}1 \\ 3 \\ 1\end{array}\right)-\left(\begin{array}{c}-2 \\ 1 \\ 3\end{array}\right)=\left(\begin{array}{c}3 \\ 2 \\ -2\end{array}\right)\)
\(d=\left(\begin{array}{l}1 \\ 1 \\ 1\end{array}\right) \Rightarrow \abs{d}=\sqrt{3}\)
| \(\text{Shortest distance}\) | \(=\abs{\hat{\underset{\sim}{d}} \times \left(\underset{\sim}{a}-\underset{\sim}{b}\right)}\) |
| \(=\abs{\dfrac{1}{\sqrt{3}}\left(\underset{\sim}{i}+\underset{\sim}{j}+\underset{\sim}{k}\right) \times \left(3 \underset{\sim}{i}+2\underset{\sim}{j}-2 \underset{\sim}{k}\right)}\) | |
| \(=\abs{\dfrac{1}{\sqrt{3}}\left(-4\underset{\sim}{i}+5\underset{\sim}{j}-\underset{\sim}{k}\right)}\) | |
| \(=\sqrt{14}\) |
\(\Rightarrow B\)
Let the lines \(l_1\) and \(l_2\) be defined by \(l_1:{\underset{\sim}{r}}_1(\lambda)=\underset{\sim}{i}+m \underset{\sim}{k}+\lambda(\underset{\sim}{i}+2\underset{\sim}{j}+\underset{\sim}{k})\) and \(l_2:{\underset{\sim}{r}}_2(\mu)=2 \underset{\sim}{ i }-\underset{\sim}{ k }+\mu(-\underset{\sim}{ i }+3 \underset{\sim}{ j }+2 \underset{\sim}{ k })\), where \(m \in R \backslash\left\{-\dfrac{4}{5}\right\}\) and \(\lambda, \mu \in R\). If the shortest distance between the two skew lines \(l_1\) and \(l_2\) is \(\dfrac{14}{\sqrt{35}}\), find the values of \(m\). (3 marks) --- 10 WORK AREA LINES (style=lined) --- \(m=\dfrac{-18}{5}\ \text{or}\ 2\) \begin{array}{lll} \(\underset{\sim}{n}={\underset{\sim}{d}}_1 \times{\underset{\sim}{d}}_2=\left|\begin{array}{ccc}\underset{\sim}{i} & \underset{\sim}{j} & \underset{\sim}{k} \\ 1 & 2 & 1 \\ -1 & 3 & 2\end{array}\right|=\underset{\sim}{i}+3 \underset{\sim}{j}+5 \underset{\sim}{k}\) \(\hat{\underset{\sim}{n}} = \dfrac{\underset{\sim}{i}+3 \underset{\sim}{j}+5 \underset{\sim}{k}}{\sqrt{1+9+25}}=\dfrac{1}{\sqrt{35}}\left(\underset{\sim}{i}+3\underset{\sim}{j}+5\underset{\sim}{k}\right)\)
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\text{Let} \ & {\underset{\sim}{a}}_1=\underset{\sim}{i}+m\underset{\sim}{k}, &{\underset{\sim}{d}}_1=\underset{\sim}{i}+2 \underset{\sim}{j}+\underset{\sim}{k}\\
\quad &{\underset{\sim}{a}}_2=2\underset{\sim}{i}-\underset{\sim}{k}, & {\underset{\sim}{d}}_2=\underset{\sim}{i}+3\underset{\sim}{j}+2 \underset{\sim}{k}\\
\end{array}
♦♦ Mean mark 27%.
\(\text{Distance}\)
\(=\abs{\left({\underset{\sim}{a}}_2-{\underset{\sim}{a}}_1 \right) \cdot \hat{\underset{\sim}{n}}}\)
\(=\abs{\left( \underset{\sim}{i}-\left(1+m\right)\underset{\sim}{k}\right) \cdot \dfrac{1}{\sqrt{3}} \left(\underset{\sim}{i}+3 \underset{\sim}{j}+5 \underset{\sim}{k}\right) }\)
\(=\abs{\dfrac{1+0-5-5 m}{\sqrt{3}}}\)
\(=\abs{\dfrac{-4-5 m}{\sqrt{35}}}\)
\(\text{Find \(m\) such that:}\)
\(\abs{\dfrac{-4-5 m}{\sqrt{35}}}\)
\(=\dfrac{14}{\sqrt{35}}\)
\(-4-5 m\)
\(=14\)
\(\qquad 4+5 m\)
\(=14\)
\(m\)
\( =\dfrac{-18}{5}\ \ \ \ \ \ \text{or}\)
\(m\)
\(=2\)