A line passes through A(4, –6, –2) and B(16, –2, 4). the point P(a, b, c), where a, b, c are non-negative integers, on the line AB lies at a distance of 21 units, from the point A. The distance between the points P(a, b, c) and Q(4, –12, 3) is equal to __

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Solution

Q.
A line passes through A(4, –6, –2) and B(16, –2, 4). the point P(a, b, c), where a, b, c are non-negative integers, on the line AB lies at a distance of 21 units, from the point A. The distance between the points P(a, b, c) and Q(4, –12, 3) is equal to __________. [2024]

Ans.
(22)

Equation of the line through A(4, –6, –2) and B(16, –2, 4) is

$\frac{x - 4}{16 - 4} = \frac{y + 6}{- 2 + 6} = \frac{z + 2}{4 + 2} = λ, λ \in R$

$\Rightarrow \frac{x - 4}{12} = \frac{y + 6}{4} = \frac{z + 2}{6} = λ$

$∴ x = 12 λ + 4, y = 4 λ - 6, z = 6 λ - 2$

Let $a = 12 λ + 4, b = 4 λ - 6, c = 6 λ - 2$ .

Now ${(12 λ + 4 - 4)}^{2} + {(4 λ - 6 + 6)}^{2} + {(6 λ - 2 + 2)}^{2} = {(21)}^{2}$

$144 λ^{2} + 16 λ^{2} + 36 λ^{2} = 441$

$\Rightarrow 196 λ^{2} = 441$

$λ^{2} = \frac{441}{196} \Rightarrow λ = \pm \frac{21}{14} = \pm \frac{3}{2}$

When $λ = \frac{3}{2}$

$a = 12 \times \frac{3}{2} + 4 = 22; b = 4 \times \frac{3}{2} - 6 = 0; c = 6 \times \frac{3}{2} - 2 = 7$

When $λ = \frac{- 3}{2}$

$a = 12 \times \frac{- 3}{2} + 4 = - 14; b = 4 \times \frac{- 3}{2} - 6 = - 12; c = 6 \times \frac{- 3}{2} - 2 = - 11$

Distance between (22, 0, 7) and (4, –12, 3)

$= \sqrt{18^{2} + 12^{2} + 4^{2}} = \sqrt{484} = 22$ .

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