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# The point of intersection of the lines $\overrightarrow{r}=(-\overrightarrow{i}+\overrightarrow{2j}+\overrightarrow{3k})+\overrightarrow{t}=(-\overrightarrow{2i}+\overrightarrow{j}+\overrightarrow{k})$ and $\overrightarrow{r}=(\overrightarrow{2i}+\overrightarrow{3j}+\overrightarrow{5k})+\overrightarrow{s}=(\overrightarrow{i}+\overrightarrow{2j}+\overrightarrow{3k})$ is

$\begin{array}{1 1}(1)(2 , 2 , 1 )&(2)(1 , 2 , 1)\\(3)(1 , 1 ,2 )&(4)(1 , 1 , 1)\end{array}$

Can you answer this question?

$\overrightarrow{r}=(-\overrightarrow{i}+\overrightarrow{2j}+\overrightarrow{3k})+\overrightarrow{t}(-\overrightarrow{2i}+\overrightarrow{j}+\overrightarrow{k})$
$x\overrightarrow{i}+ y\overrightarrow{j} + z \overrightarrow{k}=(-\overrightarrow{i}+\overrightarrow{2j}+\overrightarrow{3k})+\overrightarrow{t}(-\overrightarrow{2i}+\overrightarrow{j}+\overrightarrow{k})$
$(x+1)\overrightarrow{i} +(y-2) \overrightarrow{j} +(z-3)\overrightarrow{k}=-2t \overrightarrow{i}+t \overrightarrow{j}+ t\overrightarrow{k}$
$x+1=-2t,y-2=t,z-3=t$
$\large\frac{x+1}{-2} =-t,\;$$\large\frac{y-2}{1} =t\;,$$\large\frac{y-2}{1} =t$
The cartesian equation of the line
$\large\frac{x+1}{2}=\frac{y-2}{1} =\large\frac{z-3}{1}$------(1)
Similarly the cartesian equation of the line
$\overrightarrow{r}=(\overrightarrow{2i}+\overrightarrow{3j}+\overrightarrow{5k})+\overrightarrow{s}=(\overrightarrow{i}+\overrightarrow{2j}+\overrightarrow{3k})$
$\large\frac{x-2}{1}=\frac{y-3}{2} =\large\frac{z-5}{3}$------(2)
(1,1,2) is the point satisfying equations (1) and (2)
Hence 3 is correct answer.
answered May 13, 2014 by