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

     (b)     3y + 4z - 6 =  0

Answers (1)

Let the coordinates of the foot of perpendicular P from the origin to the plane be (x_{1},y_{1},z_{1})

Given a plane equation 3y+4z-6=0,

Or, 0x+3y+4z=6

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

Therefore \sqrt{(0)^2+(3)^2+(4)^2} = 5

So, now dividing both sides of the equation by 5 we will obtain,

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

This equation is similar to lx+my+nz = d where, l,\ m,\ n are the directions cosines of normal to the plane and d is the distance of normal from the origin.

Then finding the coordinates of the foot of the perpendicular are given by (ld,md,nd)

\therefore The coordinates of the foot of the perpendicular are;

\left (0,\frac{3}{5}.\frac{6}{5},\frac{4}{5}.\frac{6}{5} \right )  or  \left ( 0, \frac{18}{25}, \frac{24}{25} \right ).

 

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