Q.
The coordinates of the foot perpendicular from the point $(1,0,0)$ to the line
$\frac{x-1}{2} = \frac{y+1}{-3} = \frac{z+10}{8}$ are
AIEEEAIEEE 2012Three Dimensional Geometry
Solution:
Let the equation of AB is
$\frac{x-1}{2} = \frac{y-\left(-1\right)}{-3} = \frac{z-\left(-10\right)}{8} = k$
Let L be the foot of the perpendicular drawn form P$\left(1, 0,0\right)$.
$\therefore \quad L=\left(2k+ 1,-3k-1,8k-10\right)$.
Now, direction ratio of PL $= \left(2k, - 3k-1,8k -10\right)$ and direction ratio of AB $= \left(2, - 3,8\right)$ Since, PL is perpendicular to AB
$\therefore \quad2\left(2k\right) - 3\left(-3k-1\right) + 8 \left(8k-10\right) = 0$
Now, $k = \frac{2\left(1 -1\right) + \left(-3\right)\left(0 +1\right) + 8\left(0 + 10\right)}{\left(2\right)^{2} + \left(-3\right)^{2} + \left(8\right)^{2}}$
$= \frac{0-3+80}{4+9+64} = \frac{77}{77} = 1$
$\therefore \quad$ Required co-ordinate
$= Z = \left(2+ 1,-3-1, 8-10\right) = \left(3,-4,-2\right).$
