The distance of the point (1, 2, 1) from the line \(\rm \dfrac{x-1}{2} = \dfrac{y-2}{1} = \dfrac{z-3}{2}\) is

Approach:

Consider A = (x, y, z) be any point on the line. 

Find the direction ratios of the line and AB.

The sum of the product of direction ratios is zero.

Use the distance formula to find the distance between two-point.

Calculations:

Given, the equation of the line is \(\rm \dfrac{x-1}{2} = \dfrac{y-2}{1} = \dfrac{z-3}{2}\)

⇒\(\rm \dfrac{x-1}{2} = \dfrac{y-2}{1} = \dfrac{z-3}{2}\) = k

Consider A = (x, y, z) be any point on the line.

⇒\(\rm \dfrac{x-1}{2} = k ,\;\dfrac{y-2}{1} = k , \;\dfrac{z-3}{2} = k\) 

⇒ 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 \(\rm \dfrac{x-1}{2} = \dfrac{y-2}{1} = \dfrac{z-3}{2}\) = 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 

⇒ k = \( \rm \dfrac {-4} 9\)

The point A becomes A = (\( \rm \dfrac 1 9\), \( \rm \dfrac {14} 9\), \( \rm \dfrac {19} 9\))

Direction ratios of AB = 2k, k, 2k + 2

\(AB=2 \times \frac{-4}{9}, \frac{-4}{9}, 2\times \frac{-4}{9}+2\)

\(AB=\frac{-8}{9}, \frac{-4}{9}, \frac{10}{9}\)

By distance formula, we have 

\(AB=\rm\sqrt {\dfrac {64}{81}+\dfrac {16}{81}+ \dfrac {100}{81}}\)

AB = \(\dfrac{2\sqrt{5}}{3}\)

Hence, The distance of the point (1, 2, 1) from the line \(\rm \dfrac{x-1}{2} = \dfrac{y-2}{1} = \dfrac{z-3}{2}\) is \(\dfrac{2\sqrt{5}}{3}\)