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The line $L_1$ is parallel to the vector $i-2j-3k$ and passes through $A$, whose position vector is $3i+3j-4k$.

The line $L_2$ is parallel to the vector $-2i+j+3k$ and passes through the point $B$, whose position vector is $-3i-j+2k$.

The point $P$ on $L_1$ and the point $Q$ on $L_2$ are such that $PQ$ is perpendicular to both $L_1$ and $L_2$.

Find:

  1. The length of $PQ$
  2. The cartesian equation of the plane $PI$ contaning $PQ$ and $L_2$
  3. The perpendocular distance of $A$ from $PI$
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  • $\begingroup$ What is the question, exactly? $\endgroup$
    – mlc
    Mar 22 '17 at 5:53
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    $\begingroup$ Welcome to math.SE: you may find it useful to know a few things. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Proper formatting is expected; for some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$
    – mlc
    Mar 22 '17 at 5:54
  • $\begingroup$ Find i) the length of PQ $\endgroup$
    – Rubber
    Mar 22 '17 at 6:14
  • $\begingroup$ ii) The cartesian equation of the Plane PI containing PQ and L2 $\endgroup$
    – Rubber
    Mar 22 '17 at 6:15
  • $\begingroup$ iii) The perpendicular distance of A from PI $\endgroup$
    – Rubber
    Mar 22 '17 at 6:15
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Write equations of the lines in parametric form: $x_1=m_1t_1+b_1$ and $x_2=m_2t_2+b_2$ where $m_1=[1 \,\,-2\,\,-3]^T$, $m_2=[-2\,\,1\,\,3]^T$, $b_1=[3 \,\,3\,\,-4]^T$ and $b_2=[-3\,\,-1\,\,2]^T$. $t_1$ and $t_2$ are scalars. The distance $d$ between the lines is $$d^2=(x_1-x_2) \bullet (x_1-x_2) = (x_1-x_2)^T(x_1-x_2)$$ which, if it is minimized, will give you $P$ and $Q$.

\begin{align} {\partial \over \partial t_1}d^2&=(x_1-x_2)^Tm_1+m_1^T(x_1-x_2)=2m_1^T(x_1-x_2) \\ -{\partial \over \partial t_2}d^2&=(x_1-x_2)^Tm_2+m_2^T(x_1-x_2)=2m_2^T(x_1-x_2) \end{align}

Now set these partials equal zero and solve for $t_1$ and $t_2$. If $t_1'$ and $t_2'$ are the solution, then $P=m_1t_1'+b_1$ and $Q=m_2t_2'+b_2$. I get $t_1'=4.2222$ and $t_2'=-5.7778$.

You should be able to complete the answer with a few web searches, e.g. "plane from 3 points", etc.

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