Three Dimensional Geometry Question 24
Question 24 - 2024 (31 Jan Shift 2)
A line passes through $A(4,-6,-2)$ and $B(16,-2,4)$. The point $\mathrm{P}(\mathrm{a}, \mathrm{b}, \mathrm{c})$ where $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are non-negative integers, on the line $A B$ lies at a distance of 21 units, from the point $\mathrm{A}$. The distance between the points $P(a, b, c)$ and $Q(4,-12,3)$ is equal to
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Answer (22)
Solution
$\frac{x-4}{12}=\frac{x+6}{4}=\frac{z+2}{6}$
$\frac{x-4}{\frac{6}{7}\frac{2}{7}}=\frac{y+6}{\frac{3}{7}}=21$
$\left(21 \times \frac{6}{7}+4, \frac{2}{7} \times 21-6, \frac{3}{7} \times 21-2\right)$
$=(22,0,7)=(a, b, c)$
$\therefore \sqrt{324+144+16}=22$
