Question

The position vectors of two points A and C are $9\hat{i}-\hat{j}+7\hat{k}$ and $7\hat{i}-2\hat{j}+7\hat{k}$ respectively. The point of intersection of the lines containing vectors $\overrightarrow{AB}=4\hat{i}-\hat{j}+3\hat{k}\text{and}\overrightarrow{CD}=2\hat{i}-\hat{j}+2\hat{k}$ is P. If a vector $\overrightarrow{PQ}$ is perpendicular to $\overrightarrow{AB}$~\text {and }\overrightarrow{CD} \text { and }PQ = 15\) units, the possible position vectors of Q are $x_1\hat{i}+x_2\hat{j}+x_3\hat{k}\text{and}{y}_1\hat{i}+y_2\hat{j}+y_3\hat{k}$. Then the value of $\sum _{i=1}^3(x_i+y_i)$ is equal to

Answer 1

Cut equation of two lines AB, CD then their point of intersection P.
Equations of lines AB and CD are:
$AB:\overrightarrow{r}=(9\hat{i}-\hat{j}+7\hat{k})+t(4\hat{i}-\hat{j}+3\hat{k})$ ...(1)
$CD:\overrightarrow{r}=(7\hat{i}-2\hat{j}+7\hat{k})+s(2\hat{i}-\hat{j}+2\hat{k})$ ...(2)
Co-ordinates of point of intersection p of lines
(1) and (2)
$\frac{x-9}{4}=\frac{y+1}{-1}=\frac{z-7}{3}=t$

$\Rightarrow$ Point on line AB is $(9+4t,–1–t,7+3t)$ and
point on line CD are $(7+2s,–2–s,7+2s)$
$\therefore 9+4t=7+2s$
$\Rightarrow 2s–4t=2$
$\Rightarrow s–2t=1$ ...(1)
$–1–t=–2–s$
$\Rightarrow s–t=–1$ ...(2)
and $7+3t=7+2s$
$\Rightarrow 2s–3t=0$ ...(3)
From (3), $t=\frac{2s}{3}$
$\therefore$ From (2), $s-\frac{2s}{3}=-1\Rightarrow s=-3$
$\therefore$
$\Rightarrow t=–2$
$\therefore P\equiv (9–8),–1+2,7–6)P\equiv (1,1,1)$
Let $Q\equiv (x,y,z)$
$\therefore \overrightarrow{PQ}=(x-1)\hat{i}+(y+1)\hat{j}+(2-1)\hat{k}$
Given, $\left|\overrightarrow{PQ}\right|=15$
$\Rightarrow \sqrt{{(x-1)}^2+{(y-1)}^2+{(2-1)}^2}=15$
$\Rightarrow (x–1{)}^2+(y–1{)}^2+(z–1{)}^2=225$ ...(4)
given, $\overrightarrow{PQ}\perp \overrightarrow{AB}\text{and}\overrightarrow{PQ}\perp \overrightarrow{CD}$
$\Rightarrow \overrightarrow{PQ}\Vert \overrightarrow{AB}\times \overrightarrow{CD}$
Where $\overrightarrow{AB}\times \overrightarrow{CD}=\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& -1& 3\\ 2& -1& 2\end{array}$

$=\hat{i}(-2+3)-\hat{j}(8-6)+\hat{k}(-4+2)$
$=\hat{i}-2\hat{j}-2\hat{k}$
and so $\frac{x-1}{1}=\frac{y-1}{-2}=\frac{z-1}{-2}=\lambda \left(\text{say}\right)$ ...(5)
$\Rightarrow x-1=\lambda ,y-1=-2\lambda ,z-1=-2\lambda$
Putting these values in (4)
${\lambda }^2+(-2\lambda {)}^2+(-2\lambda {)}^2=225$
$\Rightarrow 9{\lambda }^2=225\Rightarrow \lambda =\pm 5$
when $\lambda =5$, then from (5)
$x=6,y=–9,z=–9$
when $\lambda =–5,$ then from (5),
$x=–4,y=11,z=11$
 $\therefore$Coordinates of Q can be $(6,–9,–9)\text{and}(–4,11,11)$
So that $x_1=6,x_2=–6,x_3=–9\text{and}{y}_1=–4,y_2=11,$
$y_3=11$
$\therefore \sum _{i=1}^3(x_i+y_i)=(x_1+y_1)+(x_2+y_2)+(x_3+y_3)$
$=(x_1+x_2+x_3)+(y_1+y_2+y_3)$
$=(6–9–9)+(–4+11+11)$
$=–12+18=6$