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 $A B$ 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 (integer)
22
Solution
<p>$$\begin{aligned}
& \frac{x-4}{12}=\frac{x+6}{4}=\frac{z+2}{6} \\
& \frac{x-4}{\frac{6}{7}}=\frac{y+6}{\frac{2}{7}}=\frac{z+2}{\frac{3}{7}}=21 \\
& \left(21 \times \frac{6}{7}+4, \frac{2}{7} \times 21-6, \frac{3}{7} \times 21-2\right) \\
& =(22,0,7)=(a, b, c) \\
& \therefore \sqrt{324+144+16}=22
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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