Question

Find the foot of the perpendicular from the point $(2,-1,5)$ on the line $\frac{x-14}{10}=\frac{y+2}{-4}=\frac{z+8}{-11}$.

Answer 1

R.E.F image

$E{q}^n$ of line $CD$

$\frac{x-14}{10}=\frac{y+2}{-4}=\frac{z+8}{-11}=\lambda$

$\begin{array}{c}x=10\lambda +14\\ y=-4\lambda -2\\ z=-11\lambda -8\end{array}\}$ co-ordinates of $B$ on line $CD.$

Now, direction ratio of $AB(x_2-x_1),(y_2-y_1),(z_2-z_1)$

$(2-(10\lambda +14\left)\right),(-1-(4\lambda -2\left)\right),(5-(-11\lambda -8\left)\right)$

$(-10\lambda -12),(-4\lambda +1),(11\lambda +13)$

Direction ration of CD are $10,-4,-11$

Since $AB\perp CD$

$\therefore$ Product of direction ratios = 0

$10(-10\lambda -12)-4(1-4\lambda )-11(11\lambda +13)=0$

$-100\lambda -120-4+16\lambda -121\lambda -143=0$

$-205\lambda -267=0$

$\lambda =\frac{-267}{205}$

$\therefore$ co-ordination of $B.$

$10\left(\frac{-267}{205}\right)+14,-4\left(\frac{-267}{205}\right)-2,-11\left(\frac{-267}{205}\right)-8$

$B\equiv (\frac{40}{41},\frac{658}{205},\frac{1297}{205})$