UPSC NDA Math Model Paper 10

Given, the equation of the line is $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{2}$ ⇒ $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{2}=k$ Consider A = (x, y, z) be any point on the line. $\Rightarrow \frac{x-1}{2}=k, \frac{y-2}{1}=k, \frac{z-3}{2}=k$ $\Rightarrow x = 2k + 1, y = k + 2, z = 2k + 3$ Hence, A = (2k + 1, k + 2, 2k + 3 ) be any point on the line. Consider the point B = (1, 2, 1) Direction ratios of AB = 2k, k, 2k + 2 Direction ratios of line $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{2}=2,1,2$ Since, AB is perpendicular to the line The sum of the product of direction ratios is zero. ⇒ 2(2k) + (1)(k) + (2)(2k + 2) = 0 $\Rightarrow k = \frac{-4}{9}$ The point $A$ becomes $A = \left( \frac{1}{9}, \frac{14}{9}, \frac{19}{9} \right)$ By distance formula, we have $AB = \sqrt{ \frac{64}{81} + \frac{16}{81} + \frac{100}{81} }$ $AB = \frac{2 \sqrt{5}}{3}$ Hence, The distance of the point (1, 2, 1) from the line $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{2}$ is $\frac{2 \sqrt{5}}{3}$