Question

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

Answer 1

The required line passing through the point $P(2,1,1)$ in the plane $2x+y-5z=0$ and is having greates slope, thus it must be perpendicular to the line of intersection of the planes, i.e.,

$2x+y-5z=0$ and $4x-3y+7z=0$

Let the direction ratio's of the line of intersection of plane $2x+y-5z=0$ and $4x-3y+7z=0$ are $a,b,c$.

Therefore,

$2a+b-5c=0$

$4a-3b+7c=0$

As we know that direction ratio's of any straight line is perpendicular to the direction ratio's of its noral to the plane.

Therefore,

$\frac{a}{4}=\frac{b}{17}=\frac{c}{5}$

Now let the direction ratios of required line be proportional to $l,m,n$ and the line passes through the point $(2,1,1)$.

Therefore, the equation of line will be-

$\frac{x-2}{l}=\frac{y-1}{m}=\frac{z-1}{n}.....\left(1\right)$

Whereas,

$2l+m-5n=0$

$4l+17m+5n=0$

Therefore,

$\frac{l}{3}=\frac{m}{-1}=\frac{n}{1}.....\left(2\right)$

From ${eq}^n\left(1\right)\&\left(2\right)$, we get

$\frac{x-2}{3}=\frac{y-1}{-1}=\frac{z-1}{1}$

Hence the required line is $\frac{x-2}{3}=\frac{y-1}{-1}=\frac{z-1}{1}$