Question

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 \(\_\_\_\_\_\).

Answer 1

Correct Answer: 22

The direction ratios of line \( AB \) are given by:

\[ (16 - 4, -2 - 6, 4 - (-2)) = (12, -8, 6) \]

The parametric equation of the line passing through point \( A(4, 6, -2) \) in the direction of \( AB \) is:

\[ x = 4 + 12t, \quad y = 6 - 8t, \quad z = -2 + 6t \]

Given that the distance from point \( A \) to point \( P(a, b, c) \) is 21 units, we use the distance formula:

\[ \sqrt{(12t)^2 + (-8t)^2 + (6t)^2} = 21 \]

Squaring both sides:

\[ 144t^2 + 64t^2 + 36t^2 = 441 \] \[ 244t^2 = 441 \implies t^2 = \frac{441}{244} \implies t = \pm \frac{21}{\sqrt{244}} = \pm \frac{21}{2\sqrt{61}} \]

Substituting the value of \( t \) into the parametric equations:

\[ a = 4 + 12 \left( \frac{6}{7} \right) = 22, \quad b = 6 - 8 \left( \frac{6}{7} \right) = 0, \quad c = -2 + 6 \left( \frac{6}{7} \right) = 7 \]

Thus, \( P(a, b, c) = (22, 0, 7) \).

Next, we find the distance between points \( P(22, 0, 7) \) and \( Q(4, -12, 3) \):

\[ \text{Distance} = \sqrt{(22 - 4)^2 + (0 - (-12))^2 + (7 - 3)^2} \] \[ = \sqrt{18^2 + 12^2 + 4^2} \] \[ = \sqrt{324 + 144 + 16} \] \[ = \sqrt{484} = 22 \]