Browse Questions

# $\overrightarrow{AB}=3 \hat i-\hat j+\hat k \;and\; \overrightarrow{CD}=-3\hat i+2 \hat j+4\hat k$ are two vectors. The position vectors of the 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 vector of a point $P$ on the line $AB$ and a point $Q$ on the line $CD$ such that $\overrightarrow{PQ}$ is perpendicular to $\overrightarrow{AB}\;and\;\overrightarrow{CD}$ both.

$\begin{array}{1 1} P(3\hat i + 8\hat j + 3\hat k) \: and \: Q(-3\hat i - 7\hat j + 6\hat k ) \\ P(2\hat i + 8\hat j + 3\hat k) \: and \: Q(-2\hat i - 7\hat j + 6\hat k )\\ P(3\hat i + 7\hat j + 3\hat k) \: and \: Q(-3\hat i +7\hat j + 6\hat k ) \\P(3\hat i + 8\hat j - 3\hat k) \: and \: Q(3\hat i - 7\hat j + 6\hat k ) \end{array}$

Toolbox:
• Write the cartesian equations of the lines $\overline{AB}\: and \: \overline{CD}$
• Two lines $\perp$ means dot product of thier d.r = 0
• Position vector of point is coordinates of the point.
Step 1
Write the equation of AB and CD in cartesian form.
$AB \rightarrow \large\frac{x-6}{3}=\large\frac{y-7}{-1}=\large\frac{z-4}{1}=\lambda$( say)
$CD \rightarrow \large\frac{x+0}{-3}=\large\frac{y+9}{2}=\large\frac{z-2}{4}=\mu$( say)
Step 2
Point p on $\overline{AB}$ and point Q on CD are given by
$p(3\lambda+6, -\lambda+7, \lambda+4)$
$Q(-3\mu, 2\mu-9, 4\mu+2)$
Step 3
$\overline{PQ}=(3\lambda+3\mu+6,-\lambda-24+16, \lambda-4\mu+2)$
Step 4
given $\overline{PQ}$ is perpendicular to
$AB\: and \: CD \: \Rightarrow \overrightarrow{PQ}.\overrightarrow{AB}=0\: and \: \overrightarrow{PQ}.\overrightarrow{CD}=0$
$\Rightarrow 11\lambda+7\mu+4=0\: and \: -7\mu-29\mu+22=0$
Step 5
Solving which $\lambda$=
$\mu =$
Step 6
Put the values of $\lambda \: and \: \mu$ in P and Q and get the required vector
$P(3\hat i + 8\hat j + 3\hat k) \: and \: Q(-3\hat i - 7\hat j + 6\hat k )$

edited Apr 5, 2013