If the equation of the line passing through the point (0,-1/2, 0 ) and perpendicular to the lines vec r =λ(hat i +a hat j +b hat k ) and vec r =(hat i -hat j -6 hat k )+μ (-b hat i +a hat j +5 hat k ) is x-1/-2=y+4/d=z-c/-4 , then a+b +c+d is equal to:

Mathematics · 3D Geometry

JEE Main 2025 — 7 April, Evening Shift — Question 24

If the equation of the line passing through the point $\left(0,-\frac{1}{2}, 0\right)$ and perpendicular to the lines $\vec{r}=\lambda(\hat{i}+a \hat{j}+b \hat{k}) \quad$ and $\quad \vec{r}=(\hat{i}-\hat{j}-6 \hat{k})+\mu$ $(-b \hat{i}+a \hat{j}+5 \hat{k})$ is $\frac{x-1}{-2}=\frac{y+4}{d}=\frac{z-c}{-4}$ , then $a+b$ $+c+d$ is equal to:

Option A:
$14$
Correct
Option B:
$10$
Option C:
$12$
Option D:
$13$

Answer: A

Step-by-step solution

Direction ratio of the given line are $-2, d$ and -4

$\Rightarrow-2+a d-4 b=0 …(i)$

and $2 b+a d-20=0 …(ii)$

subtracting equation (i) and (ii)

$6 b-18=0$

$b=3$

also, line passes through the point $\left(0, \frac{-1}{2}, 0\right)$

$\Rightarrow \frac{0-1}{-2}=\frac{-\frac{1}{2}+4}{d}=\frac{0-c}{-4}$

$\Rightarrow d=7, c=2$

From equation (i)

$-2+7 a-12=0$

$\Rightarrow a=2$

$\boxed{a+b+c+d=14}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 7 April, Evening Shift
Subject: Mathematics
Chapter: 3D Geometry
Topic: Straight Lines in 3D Geometry

Consider the lines L 1: x-1=y-2=z and L 2: x- 2=y=z-1 . Let the feet of the perpendiculars from the point P(5,1,-3) on the lines L 1 and L…

Answer: A

Step-by-step solution

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