Find the coordinates of point on line \[\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{6},\] which are at a distance of 3 units from the point \[(1,\,\,-2,\,\,3).\] |
OR |
Show that the lines \[\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda (3\hat{i}-\hat{j})\] and \[\vec{r}-=(4\hat{i}-\hat{k})+\mu (2\hat{i}+3\hat{k})\] are coplanar. Also, find the equation of the plane containing them. |
Answer:
Let \[\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{6}=\lambda \] Then, arbitrary point on the line is \[(2\lambda +1,\,\,3\lambda -2,\,\,6\lambda +3)\] Suppose \[P\,(2\lambda +1,\,\,3\lambda -2,\,\,6\lambda +3)\] be the point at a distance 3 units from the point \[Q\,(1,\,\,-2,\,\,3)\] \[\therefore \] \[PQ=3\] \[{{(PQ)}^{2}}={{(3)}^{2}}=9\] \[\Rightarrow \]\[{{(2\lambda +1-1)}^{2}}+{{(3\lambda -2+2)}^{2}}+{{(6\lambda +3-3)}^{2}}=9\] \[\Rightarrow \] \[{{(2\lambda )}^{2}}+{{(3\lambda )}^{2}}+{{(6\lambda )}^{2}}=9\Rightarrow 49{{\lambda }^{2}}=9\] \[\Rightarrow \] \[{{\lambda }^{2}}=\frac{9}{49}\] \[\Rightarrow \] \[\lambda =\pm \frac{3}{7}\] \[\therefore \]Points are \[\left( 2\left( \frac{3}{7} \right)+1,\,\,3\left( \frac{3}{7} \right)-2,\,\,6\left( \frac{3}{7} \right)+3 \right)\] \[\left( 2\left( \frac{-\,3}{7} \right)+1,\,\,3\left( \frac{-\,3}{7} \right)-2,\,\,6\left( \frac{-\,3}{7} \right)+3 \right)\] \[=\left( \frac{13}{7},\,\,\frac{-\,5}{7},\,\,\frac{39}{7} \right)\] or \[\left( \frac{1}{7},\,\,\frac{-\,23}{7},\,\,\frac{3}{7} \right)\] OR Given lines are \[\overrightarrow{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda \,(3\,\hat{i}-\hat{j})\] and \[\overrightarrow{r}=(4\,\hat{i}-\hat{k})+\mu \,(2\,\hat{i}+3\hat{k})\] On comparing both equations of lines with \[\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}\]respectively, we get \[\overrightarrow{{{a}_{1}}}=\hat{i}+\hat{j}-\hat{k},\,\,\overrightarrow{{{b}_{1}}}=3\hat{i}-\hat{j}\] And \[\overrightarrow{{{a}_{2}}}=4\,\hat{i}-\hat{k},\,\,\overrightarrow{{{b}_{2}}}=2\hat{i}+3\hat{k}\] \[\therefore \] \[=\hat{i}\,(-\,3-0)-\hat{j}\,(9-0)+\hat{k}\,(0+2)\] \[=-\,3\,\hat{i}-9\hat{j}+2\hat{k}\] and \[{{\overrightarrow{a}}_{2}}-{{\overrightarrow{a}}_{1}}=(4\hat{i}-\hat{k})-(\hat{i}+\hat{j}-\hat{k})=3\hat{i}-\hat{j}\] Now, \[({{\overrightarrow{a}}_{2}}-{{\overrightarrow{a}}_{1}})\cdot ({{\overrightarrow{b}}_{1}}\times {{\overrightarrow{b}}_{2}})=(3\,\hat{i}-\hat{j})\cdot (-\,3\,\hat{i}-9\hat{j}+2\hat{k})\] \[=-\,9+9=0\] Hence, the given lines are coplanar. Now, Cartesian equations of given lines are \[\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}\]and \[\frac{x-4}{2}=\frac{y-0}{0}=\frac{z+1}{3}\]. Then, equation of plane containing them is \Rightarrow \left| \begin{matrix} x-1 & y-1 & z-1 \\ 3 & -1 & 0 \\ 2 & 0 & 3 \\ \end{matrix} \right|=0\]\[\Rightarrow \]\[(x-1)(-3-0)-(y-1)(9-0)+(z+1)(0+2)=0\] \[\Rightarrow \] \[-\,3x+3-9y+9+2z+2=0\] \[\therefore \] \[3x+9y-2z=14\]
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