question_answer

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.