# Further Mathematics Vector A level, How to solve it

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 $PI$
• What is the question, exactly?
– mlc
Mar 22 '17 at 5:53
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– mlc
Mar 22 '17 at 5:54
• Find i) the length of PQ Mar 22 '17 at 6:14
• ii) The cartesian equation of the Plane PI containing PQ and L2 Mar 22 '17 at 6:15
• iii) The perpendicular distance of A from PI Mar 22 '17 at 6:15

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 $Q$.
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 $t_2'=-5.7778$.