Question

Consider a plane $P:x-2y+3z=7$ and a line $L:\frac{x-1}{7}=\frac{y}{2}=\frac{z+1}{-1}$. $P_1$ is the plane containing the line $L$ and perpendicular to the plane $P$. If $A(a,b,0)$ is a point on the line of intersection of the planes $P$ and $P_1,$ and two points $B(1,0,–1)$ and $C(8,2,–2)$ lie on the line $L$, then

$[$ Here, $Ar\left(△ABC\right)$ denotes area of $△ABC.]$

• A. $a-2b=9$
• B. $a-2b=7$
• C. $Ar\left(△ABC\right)=\frac{27}{2}\sqrt{\frac{3}{7}}$ sq. units
• D. $Ar\left(△ABC\right)=9$ sq. units

Answer 1

Answer: (B), (C)

$Ar\left(△ABC\right)=\frac{27}{2}\sqrt{\frac{3}{7}}$ sq. units



Given, plane $P:x-2y+3z=7[n_1:(1,-2,3\left)\right]$
and line $L:\frac{x-1}{7}=\frac{y}{2}=\frac{z+1}{-1}[n_2:(7,2,-1\left)\right]$
Equation of plane $P_1$ is
$\left|\begin{array}{ccc}x-1& y& z+1\\ 7& 2& -1\\ 1& -2& 3\end{array}\right|=0$
$\Rightarrow 2x-11y-8z=10$

Point of intersection of the planes
$P$ and $P_1$ with $z-$coordinate is equal to $0.$
$\begin{array}{c}2x-11y=10\\ x-2y=7\end{array}\}\Rightarrow x=\frac{57}{7},y=\frac{4}{7}$
$\therefore (a,b,0)\equiv (\frac{57}{7},\frac{4}{7},0)$
$\Rightarrow a-2b=\frac{57}{7}-2\times \frac{4}{7}=7$

$h=\frac{|1-3-7|}{\sqrt{1^2+(-2{)}^2+3^2}}=\frac{9}{\sqrt{14}}$
Required area of $△ABC$ is $\frac{1}{2}\times BC\times h$
$=\frac{1}{2}\sqrt{54}\times \frac{9}{\sqrt{14}}=\frac{27}{2}\sqrt{\frac{3}{7}}$