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 \(\_\_\_\_\_\).
Correct Answer: 22
Approach Solution - 1
To find the point \( P(a, b, c) \) on the line \( AB \) that is 21 units away from \( A(4, -6, -2) \), we first need the parametric equations of the line \( AB \).
Given points: \( A(4, -6, -2) \), \( B(16, -2, 4) \).
Direction vector \( \overrightarrow{AB} = (16-4, -2+6, 4+2) = (12, 4, 6) \).
Parametric equations of line \( AB \):
\[ x = 4 + 12t \]
\[ y = -6 + 4t \]
\[ z = -2 + 6t \]
Distance between \( A \) and \( P(a, b, c) \) is 21:
\[\sqrt{(12t)^2+(4t)^2+(6t)^2}=21\]
Simplify:
\[\sqrt{144t^2+16t^2+36t^2}=21\]
\[\sqrt{196t^2}=21\]
\[14|t|=21\Rightarrow |t|=\frac{3}{2}\]
Since \( a, b, \) and \( c \) are non-negative, take \( t = \frac{3}{2} \):
\[a = 4 + 12\left(\frac{3}{2}\right) = 22\]
\[b = -6 + 4\left(\frac{3}{2}\right) = 0\]
\[c = -2 + 6\left(\frac{3}{2}\right) = 7\]
So, \( P(22, 0, 7) \).
Next, find the distance between \( P(22, 0, 7) \) and \( Q(4, -12, 3) \):
\[\sqrt{(22-4)^2+(0+12)^2+(7-3)^2}\]
\[\sqrt{18^2+12^2+4^2}\]
\[\sqrt{324+144+16}\]
\[\sqrt{484}=22\]
Therefore, the distance is 22, which falls within the specified range of 22,22.
Approach Solution -2
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 \( Q(4, -12, 3) \):
\[ \text{Distance} = \sqrt{(22 - 4)^2 + (0 - (-12))^2 + (7 - 3)^2} \] \[ = \sqrt{18^2 + 12^2 + 4^2} \] \[ = \sqrt{324 + 144 + 16} \] \[ = \sqrt{484} = 22 \]
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