In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

(a) $2x + 3y + 4z - 12 = 0$
(b) $3y + 4z - 6 = 0$
(c)$x + y + z = 1$
(d) $5y + 8 = 0$

.: (a) Let the coordinates of the foot of the perpendicular. P from the origin to the plane are$({x_1},{y_1},{z_1})$.

Then, the direction ratios of the line OP are ${x_1},{y_1},{z_1}$

Writing the equation of the plane in the normal form, we have
$\cfrac{2}{{\sqrt {29} }}x + \cfrac{3}{{\sqrt {29} }}y + \cfrac{4}{{\sqrt {29} }}z = \cfrac{{12}}{{\sqrt {29} }}$

where, $\cfrac{2}{{\sqrt {29} }},\cfrac{3}{{\sqrt {29} }},\cfrac{4}{{29}}$

are the direction cosines of OP.

Since direction cosines and direction ratios of a line are proportional, then
$\cfrac{{{x_1}}}{{\cfrac{2}{{\sqrt {29} }}}} = \cfrac{{{y_1}}}{{\cfrac{3}{{\sqrt {29} }}}} = \cfrac{{{z_1}}}{{\cfrac{4}{{\sqrt {29} }}}} = k$ i.e., ${x_1}$

$= \cfrac{{2k}}{{\sqrt {29} }},{y_1} = \cfrac{{3k}}{{\sqrt {29} }},{z_1} = \cfrac{{4k}}{{\sqrt {29} }}$

Substituting these in the equation of plane, we get $k = \cfrac{{12}}{{\sqrt {29} }}$

Hence, foot of perpendicular is $\left( {\cfrac{{24}}{{29}},\cfrac{{36}}{{29}},\cfrac{{49}}{{29}}} \right)$ .

(b) Let the coordinates of the foot of the perpendicular P from the origin to the plane are $({x_1},{y_1},{z_1})$.

Then, the direction ratios of the line OP are ${x_1},{y_1},{z_1}.$

Writing the equation of the plane in the normal form, we have $\cfrac{3}{5}y + \cfrac{4}{5}z = \cfrac{6}{5}$

where, $\cfrac{3}{5},\cfrac{4}{5}$ are the direction cosines of OP.

Since direction cosines and direction ratios of the line are proportional,

then
$\cfrac{{{x_1}}}{0} = \cfrac{{{y_1}}}{{\cfrac{3}{5}}} = \cfrac{{{z_1}}}{{\cfrac{4}{5}}} = k\;\;i.e.,{x_1} = 0,{y_1} = \cfrac{{3k}}{5},{z_1} = \cfrac{{4k}}{5}$

Substituting these in the equation of the plane, we get $k = \cfrac{6}{5}$

Hence, the foot of perpendicular is $\left( {0,\cfrac{{18}}{{25}},\cfrac{{24}}{{25}}} \right)$

(c) Let the coordinates of the foot of the perpendicular P from the origin to the plane are $({x_1},{y_1},{z_1})$

Then, the direction ratios of the line OP are ${x_1},{y_1},{z_1}$.

Writing the equation of the plane in the normal form, we have $\cfrac{1}{{\sqrt 3 }}x + \cfrac{1}{{\sqrt 3 }}y + \cfrac{1}{{\sqrt 3 }}z = \cfrac{1}{{\sqrt 3 }}$

where ,$\cfrac{1}{{\sqrt 3 }},\cfrac{1}{{\sqrt 3 }},\cfrac{1}{{\sqrt 3 }}$ are the direction cosines of OP.

Since direction cosines and direction ratios of a line are proportional,

then
$\cfrac{{{x_1}}}{{\cfrac{1}{{\sqrt 3 }}}} = \cfrac{{{y_1}}}{{\cfrac{1}{{\sqrt 3 }}}} = \cfrac{{{z_1}}}{{\cfrac{1}{{\sqrt 3 }}}} = k$

i.e., ${x_1} = \cfrac{k}{{\sqrt 3 }},{y_1} = \cfrac{k}{{\sqrt 3 }},{z_1} = \cfrac{k}{{\sqrt 3 }}$

Substituting these in the equation of the plane,

we get $k = \cfrac{1}{{\sqrt 3 }}$
Hence, the foot of perpendicular is $\left( {\cfrac{1}{3},\cfrac{1}{3},\cfrac{1}{3}} \right)$

Let the coordinates of the foot of the perpendicular P from the origin to the plane are $({x_1},{y_1},{z_1})$

Then, the direction ratios of the line OP are ${x_1},{y_1},{z_1}$

Writing the equation of the plane in the normal form, we have $0x + \cfrac{5}{5}y + 0z = - \cfrac{8}{5}$

where 0, 1, 0 are the direction cosines of OP.

Since direction cosines and direction ratios of a line are proportional,

then
$\cfrac{{{x_1}}}{0} = \cfrac{{{y_1}}}{1} = \cfrac{{{z_1}}}{0} = k$ i.e., ${x_1} = 0,{y_1} = k,{z_1} = 0$

Substituting these in the equation of the plane, we get $k = - \cfrac{8}{5}$

Hence, the foot of perpendicular is $\left( {0, - \cfrac{8}{5},0} \right)$