question_answer

Find the image of the point \[2\hat{i}+3\hat{j}-4\hat{k}\] in the plane \[\vec{r}\cdot (2\hat{i}-\hat{j}+\hat{k})=3.\]

OR

Find the equation of the plane through the line of intersection of the planes \[x+y+z=1\] and \[2x+3y+4z=5,\] which is perpendicular to the plane \[x-y+z=0.\] Also, find the distance of the plane obtained above from the point (1, 1, 1).

Answer:

Let B be the image of the point \[A\,(2\,\hat{i}+3\hat{j}-4\hat{k})\]in the plane \[\overrightarrow{r}\cdot (2\,\hat{i}-\hat{j}+4\hat{k})-3=0\].

\[\therefore \overrightarrow{AB}\]is normal to the plane.

Equation of AB is\[\overrightarrow{r}=(2\,\hat{i}+3\hat{j}-4\hat{k})+\lambda \,(2\,\hat{i}-\hat{j}+\hat{k})\]

Let the position vector of

\[B=(2+2\lambda )\,\hat{i}+(3-\lambda )\hat{j}-(4-\lambda )\hat{k}\]

Let C be the mid-point of AB.

Position vector of C

\[=\frac{[(2+2\lambda )\,\hat{i}+(3-\lambda )\hat{j}-(4-\lambda )\hat{k}]+2\,\hat{i}+3\hat{j}+4\hat{k}}{2}\]

\[=(2+\lambda )\,\hat{i}+\left( 3-\frac{\lambda }{2} \right)\hat{j}-\left( \frac{4-\lambda }{2} \right)\hat{k}\]

Since, C lies on the plane\[\overrightarrow{r}\cdot (2\,\hat{i}-\hat{j}+\hat{k})-3=0\]

\[\therefore \]\[\left\{ (2+\lambda )\,\hat{i}+\left( 3-\frac{\lambda }{2} \right)\hat{j}-\left( 4-\frac{\lambda }{2} \right)\hat{k} \right\}\cdot (2\,\hat{i}-\hat{j}+\hat{k})\]

\[\Rightarrow \]\[2\,(2+\lambda )+\left( 3+\frac{\lambda }{2} \right)\hat{j}-1\left( 4-\frac{\lambda }{2} \right)-3=0\]

\[\Rightarrow \]\[4+2\lambda -3+\frac{\lambda }{2}-4+\frac{\lambda }{2}-3=0\Rightarrow 3\lambda -6=0\]

\[\Rightarrow \]                                       \[\lambda =\frac{6}{3}=2\]

\[\therefore \]Position of vector of \[B=(2+4)\,\hat{i}+(3-2)\hat{j}-(4-2)\hat{k}\] [using Eq. (i)]

\[=6\,\hat{i}+\hat{j}-2\hat{k}\]

Thus, the image of \[2\,\hat{i}+3\hat{j}-4\hat{k}\] in the given plane is\[(6,\,\,1,\,\,-\,2)\]

OR

Equation of plane through the intersection of given two planes is

\[(x+y+z-1)+\lambda \,(2x+3y+4z-5)=0\]

\[\Rightarrow \]\[(1+2\lambda )\,x+(1+3\lambda )\,y+(1+4\lambda )\,z-1-5\lambda =0\]                  ?(i)

\[\therefore \]Plane (i) is perpendicular to the plane

\[x-y+z=0\]

So,

\[1\,(1+2\lambda )-1\,(1+3\lambda )+1\,(1+4\lambda )=0\Rightarrow 3\lambda =-\,1\]

\[\therefore \]                  \[\lambda =-\frac{1}{3}\]

\[\therefore \]Equation of the plane is

\[\left( 1-\frac{2}{3} \right)x+(1-1)y+\left( 1-\frac{4}{3} \right)z-1+\frac{5}{3}=0\]

\[\Rightarrow \]               \[x-z+2=0\]

Distance of above plane from (1, 1, 1),

\[d=\left| \frac{1-1+2}{\sqrt{1+1}} \right|\Rightarrow d=\frac{2}{\sqrt{2}}=\sqrt{2}\,\,units\]