Question

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

Answer 1

(a) Let the coordinates of the foot of the perpendicular P from the origin to the plane be (x1,y1,z1).

2x+3y+4z-12=0
$\Rightarrow$ 2x+3y+4z=12...(1)

The direction ratios of normal are 2, 3, and 4.

∴ $\sqrt{(2)^2+(3)^2+(4)^2}=\sqrt{29}$

Dividing both sides of equation (1) by $\sqrt{29}$, we obtain
$\frac{2}{\sqrt{29}}x+\frac{3}{\sqrt{29}}y+\frac{4}{\sqrt{29}}z=\frac{12}{\sqrt{29}}$

This equation is of the form lx+my+nz=d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.
The coordinates of the foot of the perpendicular are given by (ld, md, nd).

Therefore, the coordinates of the foot of the perpendicular are
$\bigg(\frac{2}{\sqrt{29}}.\frac{12}{\sqrt{29}}.\frac{3}{\sqrt{29}}.\frac{12}{\sqrt{29}}.\frac{4}{\sqrt{29}}.\frac{12}{\sqrt{29}}\bigg)$i.e., $\bigg(\frac{24}{29}.\frac{36}{49}.\frac{48}{29}\bigg)$.

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(b) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x1,y1,z1).

3y+4z-6=0
$\Rightarrow$ 0x+3y+4z=6...(1)

The direction ratios of the normal are 0, 3, and 4.

∴ $\sqrt{0+3^2+4^2}=5$

Dividing both sides of equation (1) by 5, we obtain

$0x+\frac{3}{5}y+\frac{4}{5}z=\frac{6}{5}$

This equation is of the form lx+my+nz=d, where l,m,n are the direction cosines of normal to the plane and d is the distance of normal from the origin.
The coordinates of the foot of the perpendicular are given by (ld,md,nd).

Therefore, the coordinates of the foot of the perpendicular are
$\bigg(0,\frac{3}{5},\frac{6}{5},\frac{4}{5},\frac{6}{5}\bigg)$i.e., $\bigg(0,\frac{18}{25},\frac{24}{25}\bigg).$.

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(c) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x1,y1,z1).

x+y+z=1...(1)

The direction ratios of the normal are 1, 1, and 1.

∴$\sqrt{1^2+1^2+1^2}=\sqrt3$

Dividing both sides of equation(1) by $\sqrt 3$, we obtain

$\frac{1}{\sqrt 3}x+\frac{1}{\sqrt 3}y+\frac{1}{\sqrt 3}z=\frac{1}{\sqrt 3}$

This equation is of the form lx+my+nz=d,where l,m,n are the direction cosines of normal to the plane and d is the distance of normal from the origin.
The coordinates of the foot of the perpendicular are given by (ld,md,nd)

Therefore, the coordinates of the foot of the perpendicular are
$\bigg(\frac{1}{\sqrt 3}.\frac{1}{\sqrt 3}.\frac{1}{\sqrt 3}.\frac{1}{\sqrt 3}.\frac{1}{\sqrt 3}.\frac{1}{\sqrt 3}\bigg)$i.e., $\bigg(\frac{1}{3}.\frac{1}{3}.\frac{1}{3}.\bigg)$.

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(d) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x1,y1,z1).

5y+8=0
$\Rightarrow$ 0x-5y+0z=8...(1)

The direction ratios of the normal are 0,-5,and 0.
∴$\sqrt{0+(-5)^2+0}$ = 5

Dividing both sides of equation(1) by 5,we obtain
-y= $\frac{8}{5}$

This equation is of the form lx+my+nz=d, where l,m,n are the direction cosines of normal to the plane and d is the distance of normal from the origin.
The coordinates of the foot of the perpendicular are given by (ld,md,nd).

Therefore, the coordinates of the foot of the perpendicular are
$\bigg(0,-1\bigg(\frac{8}{5}\bigg),0\bigg)$i.e., $\bigg(0,-\frac{8}{5},0\bigg)$.