JEE Main 2025 23 January Evening Shift
Exam JEE Main
Year 2025
Date 23 January
Shift Evening Shift
Subject Mathematics
Chapter Three Dimensional Geometry
JEE Main 2025 · 23 January · Evening Shift · Mathematics · Three Dimensional Geometry
The distance of the line (x-2)/(2)=(y-6)/(3)=(z-3)/(4) from the...
The distance of the line
x − 2 2 = y − 6 3 = z − 3 4 \frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4} 2 x − 2 = 3 y − 6 = 4 z − 3 from the point
( 1 , 4 , 0 ) (1,4,0) ( 1 , 4 , 0 ) along the line
x 1 = y − 2 2 = z + 3 3 \frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3} 1 x = 2 y − 2 = 3 z + 3 is :
Options A 17 \sqrt{17} 17 B 15 \sqrt{15} 15 C 14 \sqrt{14} 14 D 13 \sqrt{13} 13
Correct answer C. 14 \sqrt{14} 14
Step-by-step solution
Line passing through
( 1 , 4 , 0 ) (1,4,0) ( 1 , 4 , 0 ) and parallel to
x 1 = y − 2 2 = z + 3 3 \frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3} 1 x = 2 y − 2 = 3 z + 3 is
L : x − 1 1 = y − 4 2 = z 3 L: \frac{x-1}{1}=\frac{y-4}{2}=\frac{z}{3} L : 1 x − 1 = 2 y − 4 = 3 z
Any point on
L : ( λ + 1 , 2 λ + 4 , 3 λ ) L:(\lambda+1,2 \lambda+4,3 \lambda) L : ( λ + 1 , 2 λ + 4 , 3 λ )
Any point on
x − 2 2 = y − 6 3 = z − 3 4 \frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4} 2 x − 2 = 3 y − 6 = 4 z − 3 is
(2 \mu+2,3 \mu+6,4 \mu+3) \left. λ + 1 = 2 μ + 2 2 λ + 4 = 3 μ + 6 3 λ = 4 μ + 3 \begin{array}{l}
\lambda+1=2 \mu+2 \\
2 \lambda+4=3 \mu+6 \\
3 \lambda=4 \mu+3
\end{array} λ + 1 = 2 μ + 2 2 λ + 4 = 3 μ + 6 3 λ = 4 μ + 3 \right\} \lambda=1 \mu=0
Point:
( 2 , 6 , 3 ) (2,6,3) ( 2 , 6 , 3 ) Distance = ( 2 − 1 ) 2 + ( 6 − 4 ) 2 + ( 3 − 0 ) 2 = 1 + 4 + 9 = 14 \begin{aligned}
& \text{ Distance }=\sqrt{(2-1)^2+(6-4)^2+(3-0)^2} \\
& =\sqrt{1+4+9}=\sqrt{14}
\end{aligned} Distance = ( 2 − 1 ) 2 + ( 6 − 4 ) 2 + ( 3 − 0 ) 2 = 1 + 4 + 9 = 14 $$
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