# Vectors $\vec{AB}=3\hat{i}-\hat{j}+\hat{k}$ and $\vec{CD}=-3\hat{i}+2\hat{j}+4\hat{k}$ are not coplanar.

Vectors $\vec{AB}=3\hat{i}-\hat{j}+\hat{k}$ and $\vec{CD}=-3\hat{i}+2\hat{j}+4\hat{k}$ are not coplanar.The position vectors of points $A$ and $C$ are $6\hat{i}+7\hat{j}+4\hat{k}$ and $-9\hat{j}+2\hat{k}$ respectively.Find the position vectors of a point $P$ on the line $AB$ and a point $Q$ on the line $CD$ such that $\vec{PQ}$ is perpendicular to $\vec{AB}$ and $\vec{CD}$ both.

Let the position vectors of $P$ and $Q$ be $a\hat{i}+b\hat{j}+c\hat{k}$ and $p\hat{i}+q\hat{j}+r\hat{k}$ respectively.Then $\vec{PQ}=(p-a)\hat{i}+(q-b)\hat{j}+(r-c)\hat{k}$ and as $\vec{PQ}$ is perpendicular to $\vec{AB}$ and $\vec{CD}.$

So $3(p-a)-(q-b)+(r-c)=0$
$-3(p-a)+2(q-b)+4(r-c)=0$

• I was also thinking that two vectors are always coplanar.But it is written in the question.Answer given is $P(3,8,3)$ and $Q(-3,-7,6)$. – Vinod Kumar Punia Nov 8 '15 at 13:57
$P=(6,7,4)+a(3,-1,1)$ and $Q=(0,-9,2)+c(-3,2,4)$
$PQ=(-6,-16,-2)+c(-3,2,4)-a(3,-1,1)$
$PQ.AB=3(-6-3c-3a)-1(-16+2c+a)+1(-2+4c-a)=0$
$PQ.CD=...$
You now have two equations that involve $a$ and $c$. Solve them simultaneously.