Question

find the foot of the perpendicular from p(1 2 -3) to the line x+1/2=y-3/-2=z/-1.

Answer 1

Question :

Find the foot of the perpendicular from A(1,2, -3) to the line

$\sf \frac{x + 1}{2} = \frac{y - 3}{ - 2} = \frac{z }{ - 1}$

Formula's :

If θ is the angle between the lines

$\sf\:\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$

and $\sf\:\frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}$

⇒If two lines are perpendicular then ,

$\bf\:a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0$

Solution :

Let A be the foot of perpendicular from Point P (1,2,-3) to the given line l whose equation is

$\sf \frac{x + 1}{2} = \frac{y - 3}{ - 2} = \frac{z}{ - 1} = k(say)$

Therefore;

x =2k-1 ,y = -2k+3 , z = -k

As point A lies on l , coordinates of A are ( 2k-1,-2k+3,-k) for some value of k .

The direction ratios of PA are

( 2k-2,-2k+1,-k+3k)

Also ,the direction ratios of l are 2,-2 and -1

Since PA ⊥ l

⇒2×(2k-2)+(-2)(-2k+1) + (-1)(-k+3k) = 0

⇒4k-4+4k-2+k-3 = 0

⇒9k=9

⇒k = 1

∴Coordinates of A are ( 1,1,-1)

Hence , (1,1,-1) be the foot of perpendicular from P( 1,2,-3) on the line l