Question

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

$3y+4z-6=0$

• A. $(0,\frac{24}{25},\frac{18}{25})$
• B. $(0,\frac{24}{25},\frac{24}{25})$
• C. $(0,\frac{18}{25},\frac{24}{25})$
• D. None of these
  

Answer 1

Answer: (C)

$(0,\frac{18}{25},\frac{24}{25})$

Let the coordinates of the foot of perpendicular $P$ from the origin to the plane be $(x_1,y_1,z_1)$
$3y+4z-6=0$
$\Rightarrow$ $0x+3y+4z=6=\mathrm{6....}\left(1\right)$
The direction ratios of the normal are $0,3$ and $4$
$\therefore$ $\sqrt{0+{\left(3\right)}^2+{\left(4\right)}^2}=5$
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
$(0,\frac{3}{5}.\frac{6}{5},\frac{4}{5}.\frac{6}{5})$ i.e., $(0,\frac{18}{25},\frac{24}{25})$