In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
(a) x + y + z = 1
(b) 5y + 8 = 0
In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
(a) x + y + z = 1
(b) 5y + 8 = 0

Solution:

(a) x+y+z=1
Let the coordinate of the foot of \perp \mathrm{P} from the origin to the given plane be P(x, y, z) x+y+z=1
The direction ratio are (1,1,1)
\begin{array}{l} \sqrt\left[(1)^{2}+(1)^{2}+(1)^{2}\right]=\sqrt{(1+1+1)} \\ =\sqrt{3} \end{array}
Now,
On dividing both sides of the eq. (1) by \sqrt{3}, we obtain
1 \mathrm{x} /(\sqrt{3})+1 \mathrm{y} /(\sqrt{3})+1 \mathrm{z} /(\sqrt{3})=1 / \sqrt{3}
So this is of the form I x+m y+n z=d
Where, I, \mathrm{m}, \mathrm{n} are the direction cosines and \mathrm{d} is the distance
\therefore The direction cosines are 1 / \sqrt{3}, 1 / \sqrt{3}, 1 / \sqrt{3}
The coordinate of the foot (Id, md, nd) =
\begin{array}{l} =[(1 / \sqrt{3})(1 / \sqrt{3}),(1 / \sqrt{3})(1 / \sqrt{3}),(1 / \sqrt{3})(1 / \sqrt{3})] \\ =1 / 3,1 / 3,1 / 3 \end{array}

(b) 5 y+8=0
Let the coordinate of the foot of \perp P from the origin to the given plane be P(x, y, z)
0 x-5 y+0 z=8 \ldots
The direction ratio are (0,-5,0)
\begin{array}{l} \sqrt\left[(0)^{2}+(-5)^{2}+(0)^{2}\right]=\sqrt{(0+25+0)} \\ =\sqrt{25} \\ =5 \end{array}
Now,
The dividing both sides of the eq. (1) by 5 , we obtain
0 x /(5)-5 y /(5)+0 z /(5)=8 / 5
So this is of the form I x+m y+n z=d
Where, I, \mathrm{m}, \mathrm{n} are the direction cosines and \mathrm{d} is the distance
As a result, the direction cosines are 0,-1,0
Coordinate of the foot (Id, md, nd) =
\begin{array}{l} =[(0 / 5)(8 / 5),(-5 / 5)(8 / 5),(0 / 5)(8 / 5)] \\ =0,-8 / 5,0 \end{array}