Find the vector equation of the plane passing through the point \[(2,\,\,0,\,\,-\,1)\] and perpendicular to the line joining the points (1, 2, 3) and \[(3,\,\,-1,\,\,6).\] |
OR |
Find the equation of the line passing through the point (2, 1, 3) and perpendicular to the lines |
\[\frac{x-1}{1}=\frac{y+1}{2}=\frac{z-2}{3}\] |
and \[\frac{x-4}{-\,3}=\frac{y+1}{2}=\frac{z-1}{5}.\] |
Answer:
Since, the plane passes through the point\[A\,(2,\,\,2,\,\,-\,1)\]. \[\therefore \]Position vector of point\[A,\,\,\,\overrightarrow{a}=2\,\hat{i}-\hat{k}\] \[\therefore \]The given plane is perpendicular to the line joining the points (1, 2, 3) and\[(3,\,\,-\,1,\,\,6)\]. Whose DR?s is\[(3-1),\]\[(-1-2),\]\[(6-3)\]i.e. \[2,\,-3,\,\,3\]. Now, normal vector \[\overrightarrow{n}\] perpendicular to the plane is \[\overrightarrow{n}=2\,\hat{i}-3\hat{j}+3\hat{k}\] \[\therefore \]The vector equation of the plane is \[\overrightarrow{r}\cdot \overrightarrow{n}=\overrightarrow{a}\cdot \overrightarrow{n}\] \[\Rightarrow \] \[\overrightarrow{r}\cdot (2\,\hat{i}-3\hat{j}+3\hat{k})=(2\,\hat{i}-\hat{k})\cdot (2\,\hat{i}-3\hat{j}+3\hat{k})\] \[=(2)\,\,(2)+(0)\,\,(-\,3)+(-\,1)\,\,(3)=1\] Hence, the required vector equation of the plane \[\overrightarrow{r}\cdot (2\,\hat{i}-3\hat{j}+3\hat{k})=1\] OR Given lines are \[\frac{x-1}{1}=\frac{y+1}{2}=\frac{z-2}{3}\] ?(i) and \[\frac{x-4}{-\,3}=\frac{y+1}{2}=\frac{z-1}{5}\] ?(ii) Let a, b, c be the direction ratios of the required line. \[\therefore \]Required line is perpendicular to line (i) and (ii) \[\therefore \] \[(1)\,a+2b+3c=0\] ?(iii) and \[(-\,3)\,\,a+2b+5c=0\] ?(iv) [\[\because \] these two lines with direction ratios \[{{a}_{1}},\,\,{{b}_{1}},\,\,{{c}_{1}}\]and \[{{a}_{2}},\,\,{{b}_{2}},\,\,{{c}_{2}}\]are perpendicular] \[\therefore \]\[{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0\] On solving Eqs. (iii) and (iv), by cross multiplication, we get \[\frac{a}{2}=\frac{b}{-\,7}=\frac{c}{4}\] \[\therefore \]The desired line has direction ratios\[2,\,\,-\,7,\,\,4\]. Now, required equation of the line which passes through the point (2, 1, 3) and \[-\,2,\,\,7,\,\,-\,4\] as its direction ratios, will be \[\frac{x-2}{2}=\frac{y-1}{-\,7}=\frac{z-3}{4}\].
You need to login to perform this action.
You will be redirected in
3 sec