Let A be the point of intersection of the lines L_1: (x-7)/(1)=(y-5)/(0)=(z-3)/(-1) and L_2: (x-1)/(3)=(y+3)/(4)=(z+7)/(5) . Let B and C be the point on the lines L_1 and L_2 respectively such that AB=AC= 15 . Then the

Correct answer: A. 54 Angle between both lines aligned & θ=| 3+0-5 2 50 | & θ=(2)/(10)=(1)/(5) & θ= 24 5 & area =(1)/(2) a b θ & (1)/(2) 15 15 24 5 aligned square of area (15.15 .24)/(4.25) option (1)

JEE Main 20254 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 4 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Three Dimensional Geometry

JEE Main 2025 · 4 April · Evening Shift · Mathematics · Three Dimensional Geometry

Let A be the point of intersection of the lines L_1:...

Let

A

be the point of intersection of the lines

L_1: \frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1}

and

L_2: \frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}

. Let

B

and

C

be the point on the lines

L_1

and

L_2

respectively such that

\mathrm{AB}=\mathrm{AC}=\sqrt{15}

. Then the square of the area of the triangle ABC is :

Options

A $54$
B $63$
C $57$
D $60$

Correct answer

A. $54$

Step-by-step solution

Angle between both lines

\begin{aligned}& \cos \theta=\left|\frac{3+0-5}{\sqrt{2} \sqrt{50}}\right| \\ & \sin \theta=\frac{2}{10}=\frac{1}{5} \\ & \sin \theta=\frac{\sqrt{24}}{5} \\ & \text{ area }=\frac{1}{2} a b \sin \theta \\ & \frac{1}{2} \sqrt{15} \sqrt{15} \frac{\sqrt{24}}{5}\end{aligned}

square of area

\frac{15.15 .24}{4.25}

option (1)

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