Find the coordinates of the foot of the perpendicular drawn from the origin to the plane $(c)\; x + y + z = 1 $

Question

This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com

Answer 1

Toolbox:

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

Step 1:

The given equation of the plane is $x+y+z=1$------(1)

Hence the direction cosines of the normal are $(1,1,1)$

Therefore $\sqrt{1^2+1^2+1^2}=\sqrt 3$

Divide both sides of the equation by $\sqrt 3$

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

Step 2:

Therefore the coordinates of the foot of the perpendicular are $(ld,md,nd)$

(i.e)$\big(\large\frac{1}{\sqrt 3}.\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}.\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}.\frac{1}{\sqrt 3}\big)$

$\Rightarrow (\large\frac{1}{ 3},\large\frac{1}{ 3},\large\frac{1}{ 3})$

Hence the coordinates are $(\large\frac{1}{ 3},\large\frac{1}{ 3},\large\frac{1}{ 3})$