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# help

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Compute the distance between the parallel lines given by $$\begin{pmatrix} 1 \\ 4 \end{pmatrix} + t \begin{pmatrix} 4 \\ 3 \end{pmatrix}$$ and $$\begin{pmatrix} -5 \\ 6 \end{pmatrix} + s \begin{pmatrix} 4 \\ 3 \end{pmatrix}$$

Feb 23, 2020

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The distance is 3/2.

Feb 23, 2020
#2
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Compute the distance between the parallel lines given by  $$\begin{pmatrix} 1 \\ 4 \end{pmatrix} + t \begin{pmatrix} 4 \\ 3 \end{pmatrix}$$
and $$\begin{pmatrix} -5 \\ 6 \end{pmatrix} + s \begin{pmatrix} 4 \\ 3 \end{pmatrix}$$.

$$\text{The orthogonal vector of the lines and normalized is }\\ \binom{-3}{4}*\frac{1}{\sqrt{(-3)^2+4^2}}~\text{ or }~\binom{4}{-3}*\frac{1}{\sqrt{3^2+(-4)^2}} \\ \text{because } \binom{4}{3}\binom{-3}{4} = 0 ~\text{ respectively }~\binom{4}{3}\binom{4}{-3} = 0\\$$

$$\text{Let  \hat{n} = \frac{1}{5}\binom{-3}{4}}$$

The distance between the parallel lines is:

$$\begin{array}{|rcll|} \hline d &=&\dfrac{1}{5}\dbinom{-3}{4}\dbinom{-5}{6}-\frac{1}{5}\dbinom{-3}{4}\dbinom{1}{4} \\\\ d &=&\dfrac{1}{5}\Big(15+24-( -3+16 ) \Big) \\\\ d &=&\dfrac{26}{5} \\\\ \mathbf{d} &=& \mathbf{5.2} \\ \hline \end{array}$$

Feb 24, 2020