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# Find the coordinates of the foot of the perpendicular drawn from the origin to the plane $(a)\; 2x + 3y + 4z -12 = 0$

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

$\begin{array}{1 1} (\large\frac{24}{29},\frac{36}{29},\frac{-48}{29}) \\(\large\frac{2}{29},\frac{3}{29},\frac{4}{29}) \\ (\large\frac{24}{29},\frac{36}{29},\frac{48}{29}) \\ (2,3,4) \end{array}$

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 coordinates of the foot of the perpendicular is $(ld,md,nd)$.
Step 1:
Given equation of the plane=$2x+3y+4z-12=0$
(i.e) $2x+3y+4z=12$----(1)
Let the coordinates of the foot of the perpendicular $P$ from the origin to the plane be $(x_1,y_1,z_1)$
The direction cosines of the plane are $(2,3,4)$
Therefore $\sqrt{(2)^2+(3)^2+(4)^2}=\sqrt{29}$
Step 2:
Divide equ(1) on both sides by $\sqrt{29}$
$\large\frac{2}{\sqrt{29}}$$x+\large\frac{3}{\sqrt{29}}$$y+\large\frac{4}{\sqrt{29}}$$z=\large\frac{12}{\sqrt{29}}$
Therefore the coordinates of the foot of the perpendicular are $\big(\large\frac{2}{\sqrt{29}}.\large\frac{12}{\sqrt{29}}\big),\big(\large\frac{3}{\sqrt{29}}.\large\frac{12}{\sqrt{29}}\big),\big(\large\frac{4}{\sqrt{29}}.\large\frac{12}{\sqrt{29}}\big)$
$\Rightarrow \big(\large\frac{24}{29}\big),\big(\large\frac{36}{29}\big),\big(\large\frac{48}{29}\big)$