Question

Let $F$ be the foot of perpendicular from $P(1,2,-3)$ on the line $L:\frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1}.$ Given $Q(x_1,y_1,z_1)$ and $R(x_2,y_2,z_2)$ are points at a distance of $3$ units from $F$ on line $L.$ Let ${\pi }_1$ and ${\pi }_2$ be two planes passing through $Q$ and $R$ respectively and perpendicular to the line of intersection of the planes $9x-7y+6z+48=0$ and $x+y-z=7.$ Then which of the following statements is (are) CORRECT?

• A. The coordinates of $F$ are $(1,1,-1)$
• B. The value of $\sum _{i=1}^2(x_i^2+y_i^2+z_i^2)$ equals $24$
  
• C. The distance between planes ${\pi }_1$ and ${\pi }_2$ equals $\frac{88}{\sqrt{482}}$
• D. The normal vector of planes ${\pi }_1$ and ${\pi }_2$ is parallel to the vector $\hat{i}+15\hat{j}+16\hat{k}.$

Answer 1

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

The correct options are
A The coordinates of $F$ are $(1,1,-1)$
B The value of $\sum _{i=1}^2(x_i^2+y_i^2+z_i^2)$ equals $24$

C The distance between planes ${\pi }_1$ and ${\pi }_2$ equals $\frac{88}{\sqrt{482}}$
D The normal vector of planes ${\pi }_1$ and ${\pi }_2$ is parallel to the vector $\hat{i}+15\hat{j}+16\hat{k}.$

$L:\frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1}=\lambda$ (let)
$\Rightarrow x=2\lambda -1,y=3-2\lambda ,z=-\lambda$

Clearly, $2(2\lambda -1-1)-2(3-2\lambda -2)-1(-\lambda +3)=0$
$\Rightarrow 2(2\lambda -2)-2(1-2\lambda )-1(3-\lambda )=0$
$\Rightarrow 4\lambda -4-2+4\lambda -3+\lambda =0$ $\Rightarrow 9\lambda =9\Rightarrow \lambda =1$
$\Rightarrow F\equiv (1,1,-1)$

Co-ordinates of a point on $L$ at a distance $r$ from $(x_1,y_1,z_1)$ is $(x_1+lr,y_1+mr,z_1+nr)$
Here, $l=\frac{2}{3},m=\frac{-2}{3},n=\frac{-1}{3},r=\pm 3$ $\{\because \sqrt{a^2+b^2+c^2}=\sqrt{2^2+2^2+1^2}=3\}$
So, $Q\equiv (3,-1,-2)\equiv (x_1,y_1,z_1)$ and $R\equiv (-1,3,0)\equiv (x_2,y_2,z_2)$
$\therefore \sum _{i=1}^2(x_i^2+y_i^2+z_i^2)$ $=x_1^2+y_1^2+z_1^2+x_2^2+y_2^2+z_2^2$ $=(3{)}^2+(-1{)}^2+(-2{)}^2+(-1{)}^2+(3{)}^2+0$
$=24$

Now, normal vector to planes ${\pi }_1$ and ${\pi }_2$ is ${\overrightarrow{n}}_1\times {\overrightarrow{n}}_2=\overrightarrow{n}$ (let)
Given, $\overrightarrow{n_1}=9i-7j+6k,\overrightarrow{n_2}=i+j-k$
So, $\overrightarrow{n}={\overrightarrow{n}}_1\times {\overrightarrow{n}}_2=\left|\begin{array}{ccc}i& j& k\\ 9& -7& 6\\ 1& 1& -1\end{array}\right|$
$=\hat{i}(7-6)-\hat{j}(-9-6)+\hat{k}(9+7)$ $=\hat{i}+15\hat{j}+16\hat{k}$
$\Rightarrow$ Equation of planes ${\pi }_1$ and ${\pi }_2$ are $1(x-3)+15(y+1)+16(z+2)=0$ and $1(x+1)+15(y-3)+16(z-0)=0$
i.e., $x+15y+16z+44=0$ and $x+15y+16z-44=0$

Distance between planes ${\pi }_1$ and ${\pi }_2$ $=\left|\frac{44+44}{\sqrt{1^2+{15}^2+{16}^2}}\right|=\left|\frac{88}{\sqrt{482}}\right|$