Question

Let $p_1:x-2y+3z=5$ and $p_2:2x+3y+z+4=0$ be two planes. If $P$ is the foot of the perpendicular dropped from the origin $O$ to the line of intersection of the planes, then

• A. $\overrightarrow{OP}\cdot ((\hat{i}-2\hat{j}+3\hat{k})\times (2\hat{i}+3\hat{j}+\hat{k}))=0$
• B. if $Q$ is another point on the line of intersection of planes, then equation of the plane containing a triangle $OPQ$ is $14x+7y+17z=0.$
• C. equation of the plane perpendicular to the line of intersection of the planes and passing through $(1,1,1)$ is $11x-5y-7z+1=0.$
• D. If $N_1$ and $N_2$ are foot of the perpendicular from the origin $O$ to the planes $p_1$ and $p_2$ respectively, then $O{N}_1+O{N}_2$ is equal to $\frac{9}{\sqrt{14}}.$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\overrightarrow{OP}\cdot ((\hat{i}-2\hat{j}+3\hat{k})\times (2\hat{i}+3\hat{j}+\hat{k}))=0$
B if $Q$ is another point on the line of intersection of planes, then equation of the plane containing a triangle $OPQ$ is $14x+7y+17z=0.$
C equation of the plane perpendicular to the line of intersection of the planes and passing through $(1,1,1)$ is $11x-5y-7z+1=0.$
D If $N_1$ and $N_2$ are foot of the perpendicular from the origin $O$ to the planes $p_1$ and $p_2$ respectively, then $O{N}_1+O{N}_2$ is equal to $\frac{9}{\sqrt{14}}.$

Let $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$
$p_1\equiv \overrightarrow{r}\cdot (\hat{i}-2\hat{j}+3\hat{k})=5p_2\equiv \overrightarrow{r}\cdot (2\hat{i}+3\hat{j}+\hat{k})=-4$


DR's of intersection line $\left(L\right)$ of $p_1$ and $p_2$ is
$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -2& 3\\ 2& 3& 1\end{array}\right|=-11\hat{i}+5\hat{j}+7\hat{k}$

Now, $OP$ is $\perp$ to $L$
and $\hat{i}-2\hat{j}+3\hat{k},2\hat{i}+3\hat{j}+\hat{k}$ are also perpendicular to $L.$
This means $\overrightarrow{OP}\cdot ((\hat{i}-2\hat{j}+3\hat{k})\times (2\hat{i}+3\hat{j}+\hat{k}))=0$

Equation of planes passing through intersection of $p_1$ and $p_2$ is
$p_1+\lambda p_2=0(x-2y+3z-5)+\lambda (2x+3y+z+4)=0x(1+2\lambda )+y(3\lambda -2)+z(3+\lambda )-5+4\lambda =0\cdots \left(1\right)$
Put $(0,0,0)$
$0+0+0+4\lambda -5=0\Rightarrow \lambda =\frac{5}{4}$
Putting this value in equation $\left(1\right),$
$x(1+2\left(\frac{5}{4}\right))+y(3\left(\frac{5}{4}\right)-2)+z(3+\frac{5}{4})+0=0\Rightarrow 14x+7y+17z=0$

Vector along the line of intersection of planes is
$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -2& 3\\ 2& 3& 1\end{array}\right|=-11\hat{i}+5\hat{j}+7\hat{k}$

$\therefore -11(x-1)+5(y-1)+7(z-1)=0$
or, $11x-5y-7z+1=0$

$O{N}_1+O{N}_2\left|\frac{5}{\sqrt{1^2+2^2+3^2}}\right|+\left|\frac{4}{\sqrt{2^2+3^2+1^2}}\right|=\frac{5}{\sqrt{14}}+\frac{4}{\sqrt{14}}=\frac{9}{\sqrt{14}}$