Find the vector equation of the plane passing through the three points with position vectors \[\hat{i}+\hat{j}-2\hat{k},\] \[2\hat{i}-\hat{j}+\hat{k}\] and \[\hat{i}+2\hat{j}+\hat{k}.\] Also, find the coordinates of the point of intersection of this plane and the line \[\vec{r}=(3\hat{i}-\hat{j}-\hat{k})+\lambda (2\hat{i}-2\hat{j}+\hat{k}).\] |
OR |
Find the vector equation of the plane through the points \[(2,\,\,1,\,\,-\,1)\] and \[(-\,1,\,\,3,\,\,4)\] and perpendicular to the plane \[x-2y+4z=10.\] |
Answer:
Equation of the plane passing through the three points with position vector\[\hat{i}+\hat{j}-2\hat{k},\]\[2\hat{i}-\hat{j}+\hat{k},\]and \[\hat{i}+2\hat{j}+\hat{k}\]is \[\Rightarrow \] \[\Rightarrow \] \[(x-1)\,\,(-\,6-3)-(y-1)\,\,(3-0)+(z+2)\,\,(1+0)=0\] \[\Rightarrow \] \[(x-1)\,\,(-\,9)-(y-1)\,\,(3)+(z+2)\,\,(1)=0\] \[\Rightarrow \] \[-\,9x+9-3y+3+z+2=0\] \[\Rightarrow \] \[9x+3y-z=14\] ?(i) Now, equation of the plane in vector form is \[\overrightarrow{r}\cdot (9\hat{i}+3\hat{j}-\hat{k})=14\] Vector equation of given line is \[\overrightarrow{r}=(3\hat{i}-\hat{j}-\hat{k})+\lambda \,\,(2\hat{i}-2\hat{j}+\hat{k})\] Equation of line in cartesian form is \[\frac{x-3}{2}=\frac{y+1}{-\,2}=\frac{z+1}{-\,2}=\lambda \] and any point on the given line is \[(2\lambda +3,\,\,-2\lambda -1,\,\,\lambda -1)\] For point of intersection, this point must satisfy the plane (i). \[\therefore \]\[9\,\,(2\lambda +3)+3(-\,2\lambda -1)-(\lambda -1)=14\] \[\Rightarrow \] \[18\lambda +27-6\lambda -3\lambda +1=14\] \[\Rightarrow \] \[11\lambda +25=14\Rightarrow 11\lambda =-\,11\Rightarrow \lambda =-\,1\] \[\therefore \]Point of intersection is \[(2\,(-\,1)+3,\,\,-2\,(-\,1)-1,\,\,-1-1)=(1,\,\,1-2)\] OR General equation of a plane passing through the point\[(2,\,\,1,\,\,-\,1)\]is \[a\,(x-2)+by\,(y-1)+c\,(z+1)=0\] ?.(i) If this plane passes through the point\[(-\,1,\,\,3,\,\,4)\]then \[a\,(-\,3)+b\,(2)+c\,(5)=0\] ?(ii) Plane (i) is perpendicular to the plane \[x-2y+4z=10\] \[\therefore \] \[a\,(1)+b\,(-\,2)+c\,(4)=0\] ?(iii) Eliminating a, b, c from Eqs. (i), (ii) and (iii), we get \[\Rightarrow \]\[(x-2)\,\,(18)-(y-1)\,\,(-\,17)+(z+1)\,\,(4)=0\] \[\Rightarrow \] \[18x-36+17y-17+4z+4=0\] \[\Rightarrow \]\[18x+17y+4z-49=0\]is the equation of the plane. or \[\overrightarrow{r}\cdot (18\,\hat{i}+17\hat{j}+4\hat{k})-49=0\]is vector equation of the plane.
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