Question

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 origin.

OR

Find the distance of the point (2, 12, 5) from the point of intersection of the line $\overrightarrow{r}=2\hat{i}-4\hat{j}+2\hat{k}+\lambda (3\hat{i}+4\hat{j}+2\hat{k})$ and the plane $\overrightarrow{r}.(\hat{i}-2\hat{j}+\hat{k})=0$.

Answer 1

The equation of plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 is given as: $x+y+z-1+\lambda (2x+3y+4z-5)=0$

i.e., $(1+2\lambda )\left(1\right)+(1+3\lambda )y+(1+4\lambda )z-1-5\lambda =0\dots \left(i\right)$

Since (i) is perpendicular to the plane $x-y+z=0$,

So $(1+2\lambda )\left(1\right)+(1+3\lambda )(-1)+(1+4\lambda (1)=0\Rightarrow \lambda =\frac{1}{3}$ [Using $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$]

Substituting the value of $\lambda =-\frac{1}{3}$ in (i), we have:

$(1+2(-\frac{1}{3}))x+(1+3(-\frac{1}{3}))y+(1+4(-\frac{1}{3}))z-1-5(-\frac{1}{3})=0$

i.e., $x-z+2=0$

Also, the distance of the plane x=z+2=0 from origin (0,0) is = $\frac{|1.0+0.0-1.0+2|}{\sqrt{1^2+0^2+(-1{)}^3}}=\sqrt{2}$ Units.

OR

Given line is $\overrightarrow{r}=2\hat{i}-4\hat{j}+2\hat{k}+\lambda (3\hat{i}+4\hat{j}+2\hat{k})$ so, the position vector of coordinates of any random point on this line is $\overline{OP}=(2+3\lambda )\hat{i}+(-4+4\lambda )+(2+2\lambda )\hat{k}$

For the given line to intersect the plane $\overrightarrow{r}.(\hat{i}-2\hat{j}+\hat{k})=0$, the position vector of coordinates of any random point $\overline{OP}$ must satisfy the equation of plane for some value of $\lambda$ .

i.e., $\left[\right(2+3\lambda )\hat{i}+(-4+4\lambda )\hat{j}+(2+2\lambda \left)\hat{k}\right].(\hat{i}-2\hat{j}+\hat{k})=0$

$\Rightarrow (2+3\lambda )+(-4+4\lambda )(-2)+(2+2\lambda )=0\Rightarrow \lambda =4$.

$\therefore$ the point of intersection id $\overline{OP}=14\hat{i}+12\hat{j}+10\hat{k}$

i.e., $P(14,12,10)$

So, the required distance of P from (2, 12, 5) is =$\sqrt{(14-2{)}^2+(12-12{)}^2+(10-5{)}^2}=13$ Units.