Question

Find the length and foot of the perpendicular drawn from the point 2, - 1, 5 on the line x - 11 by 10 is equals to y + 2 upon - 4 is equals to z + 8 upon - 11

Answer 1

Foot of perpendicular = (1,2,3) and length =$\frac{1}{2}\sqrt{\frac{15}{7} }$ units

Step-by-step explanation:

Given equation of straight line is

$\frac{x-11}{10} =\frac{y+2}{-4} = \frac{z+8}{-11} = r (say)$

and given  point is A(2,-1,5)

Any point on the line be B(10r+11,-4r-2,-11r-8)

the d.r of line is (10,-4,-11)

[$\frac{x-11}{10} = r$⇒x-11=10r⇒x=10r + 11,  similarly y = -4r -2 and z= -11r -8]

Therefore d.r of AB is (10r+11-2,-4r-2+1,-11r-8-5)=(10r+9,-4r-1,-11r-13)   [ d.r of line joining by two points X($x_{1}, y_{1}, z_{1}$) and Y($x_{2}, y_{2} ,z_{2}$) is ($x_{2} -x_{1} ,y_{2} -y_{1} ,z_{2} -z_{1}$)]

since the given line and AB are  perpendicular each other ,then

$l_{1} l_{2}+m_{1} m_{2} +n_{1} n_{2} =0$

⇒10(10r+9)+(-4)(-4r-1)+(-11)(-11r-13)=0

⇒100r+90+16r+4+121r+143=0

⇒237r=-237

⇒r=-1

Therefore the coordinate of B is (-10+11,4-2,11-8) =(1,2,3)

The coordinate of foot of perpendicular is (1,2,3).

The length of AB is =$|\frac{2.1+2.(-1)+3.5}{\sqrt{1^{2}+2^{2} +3^{2} }\sqrt{2^{2}+(-1)^{2}+5^{2} } }|$

=$\frac{15}{\sqrt{14} \sqrt{30} }$  units= $\frac{1}{2}\sqrt{\frac{15}{7} }$ units

The length of perpendicular is$\frac{1}{2}\sqrt{\frac{15}{7} }$ units