question_answer

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

\[\frac{x-2}{2}=\frac{y-1}{-\,7}=\frac{z-3}{4}\].