Question

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

2x +3y +4z -12=0

3y +4z -6 =0

x +y +z =1

Answer 1

Given plane is 2x+3y+4z-12=0 .....(i)

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

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

Equation of the line through origin and at right angles to Eq. (i) are $\frac{x-0}{2}=\frac{y-0}{3}=\frac{z-0}{4}$

Any point on this is (2t,3t,4t). This point lies in the plane(i), if 2(2t)+3(3t)+4(4t)-12 =0 i.e., $t=\frac{12}{29}$

$\therefore$ The required foot of perpendicular is

$(2\times \frac{12}{29},3\times \frac{12}{29},4\times \frac{12}{29})or(\frac{24}{29},\frac{36}{29},\frac{48}{29}).$

Given plane is 0x + 3y +4z -6 =0 ...(ii)

$\therefore$ Dr's of any line perpendicular to plane (ii) are 0,. 3,4.

Hence, equation of the line through origin and at right angles to plane (ii) are $\frac{x-0}{0}=\frac{y-0}{3}=\frac{z-0}{4}$

Any point on this line is (0,3t,4t). This point lies in the plane (ii)

If $0+3\times \left(3t\right)+4\times \left(4t\right)-6=0i.e.,t=\frac{6}{25}$

Hence, the required foot of the perpendicular is

$(0+3\times \frac{6}{25},4\times \frac{6}{25})$ or $(0,\frac{18}{25},\frac{24}{25})$

Given plane is x+y+z=1 ...(iii)

$\therefore$ Dr's of any line perpendicular to plane (iii) are (1,1,1)

Hence, equation of the line through origin and at right angle to Eq. (iii) are $\frac{x-0}{1}=\frac{y-0}{1}=\frac{z-0}{1}$ ...(iv)

Any point on this line is (t,t,t). This point lies in the plane , if

t+t+t=1 ie., 3t=1 $\Rightarrow t=\frac{1}{3}$

Hence, the reqired foot of the perpendicular is $(\frac{1}{3},\frac{1}{3},\frac{1}{3}).$

Given plane is 0x +5y +0z +8= 0 ....(v)

Dr's of any line perpendicular to plane (v) are 0,5,0.

Hence, equation of the line through origin and at right angle to plane (v) are

$\frac{x-0}{0}=\frac{y-0}{5}=\frac{z-0}{0}$ ...(vi)

Any point on this line is (0,5t,0). This lies in plane (v) if $5\times 5t+8=0$ i.e., $t=\frac{-8}{25}$

Hence, the required foot of perpendicular is $(0,5(-\frac{8}{25}),0)(0,\frac{8}{25},0)$