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) =
$\begin{array}{l}
=[(0 / 5)(8 / 5),(-5 / 5)(8 / 5),(0 / 5)(8 / 5)] \\
=0,-8 / 5,0
\end{array}$