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

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 $Q(4, -12, 3)$:

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