Key Concepts and Formulas

This problem involves finding a point on a line at a specific distance from a given point and then calculating the distance between two points in 3D space. The key concepts and formulas we will use are:

1. Direction Ratios and Direction Cosines of a Line: If a line passes through two points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$, its direction ratios are $(x_2-x_1, y_2-y_1, z_2-z_1)$. The direction cosines $(l, m, n)$ are obtained by dividing the direction ratios by the magnitude of the direction vector, i.e., $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.
2. Equation of a Line in Parametric Form (Distance Form): The coordinates of any point $P(x, y, z)$ on a line passing through a point $A(x_1, y_1, z_1)$ with direction cosines $(l, m, n)$ can be expressed as: $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n} = r$ Here, $r$ represents the signed distance of the point $P(x, y, z)$ from the point $A(x_1, y_1, z_1)$. If $r>0$, $P$ is in the direction of $(l,m,n)$ from $A$; if $r<0$, $P$ is in the opposite direction.
3. Distance Formula in 3D: The distance between two points $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$ is given by: $D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$

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Step-by-Step Solution

1. Determine the Direction Ratios and Magnitude of Line AB

• Why this step? To define the orientation of the line in space, we first need its direction. Direction ratios provide this information, and their magnitude is needed to find direction cosines.
• Given points are $A(4,-6,-2)$ and $B(16,-2,4)$.
• The direction ratios of the line $AB$, denoted as $(DR_x, DR_y, DR_z)$, are calculated by subtracting the coordinates of $A$ from $B$: $DR_x = 16 - 4 = 12$ $DR_y = -2 - (-6) = -2 + 6 = 4$ $DR_z = 4 - (-2) = 4 + 2 = 6$
• So, the direction ratios of line $AB$ are $(12, 4, 6)$.
• Now, we find the magnitude of this direction vector: $d = \sqrt{12^2 + 4^2 + 6^2} = \sqrt{144 + 16 + 36} = \sqrt{196} = 14$

2. Calculate the Direction Cosines of Line AB

• Why this step? The distance form of the line equation requires direction cosines, which are normalized direction ratios. This normalization makes the parameter $r$ directly represent distance.
• We divide each direction ratio by the magnitude $d=14$ to get the direction cosines $(l, m, n)$: $l = \frac{12}{14} = \frac{6}{7}$ $m = \frac{4}{14} = \frac{2}{7}$ $n = \frac{6}{14} = \frac{3}{7}$
• The direction cosines of line $AB$ are $\left(\frac{6}{7}, \frac{2}{7}, \frac{3}{7}\right)$.

3. Write the Equation of Line AB in Parametric (Distance) Form

• Why this step? We are given that point $P$ is at a specific distance from $A$. The parametric form of the line equation directly uses this distance as a parameter, simplifying the calculation of $P$'s coordinates.
• Using point $A(4,-6,-2)$ as $(x_1, y_1, z_1)$ and the calculated direction cosines: $\frac{x-4}{6/7} = \frac{y-(-6)}{2/7} = \frac{z-(-2)}{3/7} = r$ $\frac{x-4}{6/7} = \frac{y+6}{2/7} = \frac{z+2}{3/7} = r$ Here, $r$ represents the signed distance of a point $(x,y,z)$ on the line from point $A$.

4. Determine the Coordinates of Point P(a, b, c)

• Why this step? The problem states that point $P$ lies on line $AB$ at a distance of 21 units from point $A$. We use this information to find $P$'s coordinates by setting $r = \pm 21$. The condition that $a,b,c$ are non-negative integers will help us choose the correct value for $r$.

• We are given that the distance of $P$ from $A$ is 21 units. So, $|r|=21$. We consider two cases for $r$:

  • Case 1: $r=21$ $\frac{a-4}{6/7} = 21 \implies a-4 = 21 \times \frac{6}{7} = 3 \times 6 = 18 \implies a = 4 + 18 = 22$ $\frac{b+6}{2/7} = 21 \implies b+6 = 21 \times \frac{2}{7} = 3 \times 2 = 6 \implies b = -6 + 6 = 0$ $\frac{c+2}{3/7} = 21 \implies c+2 = 21 \times \frac{3}{7} = 3 \times 3 = 9 \implies c = -2 + 9 = 7$ Thus, $P(a, b, c) = (22, 0, 7)$. These coordinates ($a=22, b=0, c=7$) are all non-negative integers, satisfying the condition.
  • Case 2: $r=-21$ $\frac{a-4}{6/7} = -21 \implies a-4 = -21 \times \frac{6}{7} = -3 \times 6 = -18 \implies a = 4 - 18 = -14$ $\frac{b+6}{2/7} = -21 \implies b+6 = -21 \times \frac{2}{7} = -3 \times 2 = -6 \implies b = -6 - 6 = -12$ $\frac{c+2}{3/7} = -21 \implies c+2 = -21 \times \frac{3}{7} = -3 \times 3 = -9 \implies c = -2 - 9 = -11$ Thus, $P(a, b, c) = (-14, -12, -11)$. These coordinates are negative, so this point does not satisfy the condition that $a, b, c$ are non-negative integers.

• Therefore, the only valid coordinates for point $P(a, b, c)$ are $(22, 0, 7)$.

5. Calculate the Distance Between Points P and Q

• Why this step? This is the final objective of the problem: to find the distance between the determined point $P$ and the given point $Q$.
• We have $P(22,0,7)$ and $Q(4,-12,3)$.
• Using the 3D distance formula: $PQ = \sqrt{(x_P-x_Q)^2 + (y_P-y_Q)^2 + (z_P-z_Q)^2}$ $PQ = \sqrt{(22-4)^2 + (0-(-12))^2 + (7-3)^2}$ $PQ = \sqrt{(18)^2 + (12)^2 + (4)^2}$ $PQ = \sqrt{324 + 144 + 16}$ $PQ = \sqrt{484}$ $PQ = 22$

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Common Mistakes & Tips

• Confusing Distance with Signed Parameter: Remember that 'distance' is usually non-negative, but the parameter 'r' in the parametric form of a line equation can be negative, indicating a direction opposite to the chosen direction vector. Always check both positive and negative values for 'r' when finding a point at a specific distance.
• Forgetting Non-Negative Conditions: Carefully apply all given conditions, such as "non-negative integers" for coordinates. This often helps in uniquely identifying a point from multiple possibilities.
• Calculation Errors: 3D geometry problems often involve multiple steps and numerical calculations. Double-check each step, especially squares and sums, to avoid arithmetic mistakes.
• Understanding "on the line AB": This implies the infinite line passing through A and B, not just the line segment AB, unless specified.

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Summary

We first determined the direction ratios and direction cosines of the line AB. Then, using the parametric form of the line equation starting from point A, we found the coordinates of point P by considering a distance of 21 units. The condition that P's coordinates must be non-negative integers allowed us to uniquely identify P as $(22,0,7)$. Finally, we calculated the Euclidean distance between this point P and the given point Q using the 3D distance formula. The calculated distance between P and Q is 22 units.

The final answer is $\boxed{4}$.