Mathematics · 3D Geometry

JEE Main 2024 — 31 January, Shift 2 — Question 26

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 $\qquad$ .

Answer: 22

Numerical answer — enter this value.

Step-by-step solution

\begin{aligned} & \frac{x-4}{12}=\frac{x+6}{4}=\frac{z+2}{6} &\end{aligned}

$\frac{x-4}{\frac{6}{7}}=\frac{y+6}{\frac{2}{7}}=\frac{z+2}{\frac{3}{7}}=21$ $\left(21 \times \frac{6}{7}+4, \frac{2}{7} \times 21-6, \frac{3}{7} \times 21-2\right)$ $=(22,0,7)=(a, b, c)$ $\therefore \sqrt{324+144+16}=22$

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Exam: JEE Main 2024
Paper: 31 January, Shift 2
Subject: Mathematics
Chapter: 3D Geometry
Topic: Vector & Cartesian forms of lines and planes

Answer: 22

Step-by-step solution

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