Question

Assuming the plane $4x-3y+7z=0$ to be horizontal, the equation of the line of greatest slope through the point $(2,1,1)$ in the plane $2x+y-5z=0$ is

• A. $\frac{x-2}{3}=\frac{y+1}{1}=\frac{z-1}{1}$
• B. $\frac{x-2}{-3}=\frac{y-1}{-1}=\frac{z-1}{1}$
• C. $\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-1}{1}$
• D. $\frac{x-2}{3}=\frac{y-1}{-1}=\frac{z-1}{1}$

Answer 1

The correct option is D $\frac{x-2}{3}=\frac{y-1}{-1}=\frac{z-1}{1}$

The required line passing throught the point $(2,1,1)$ in the plane $2x+y-5z=0$ and is having greatest slope, so it must be perpendicular to the line of intersection of the planes $2x+y-5z=0\cdots \left(i\right)$ and
$4x-3y+7z=0\cdots \left(ii\right)$
Let the $D.R^{\prime }s$ of the line of intersection of equations $\left(i\right)$ and $\left(ii\right)$ are $(a,b,c)$
$\Rightarrow 2a+b-5c=0$ and $4a-3b+7c=0$
(as $D.R^{\prime }s$ of the straight line $(a,b,c)$ is perpendicular to the $D.R^{\prime }s$ of normal of both the planes)
$\frac{a}{4}=\frac{b}{17}=\frac{c}{5}$
Now, let the direction ratio of required line be proportional to $l,m,n$ then its equation be $\frac{x-2}{l}=\frac{y-1}{m}=\frac{z-1}{n}$
where $2l+m-5n=0$ and $4l+17m+5n=0$
So, $\frac{l}{3}=\frac{m}{-1}=\frac{n}{1}$
Thus the required equation of the line is $\frac{x-2}{3}=\frac{y-1}{-1}=\frac{z-1}{1}$