If the lines (x-5)/7 = (y+2)/(-5) = z/() and x/1 = y/(2) = z/3 are perpendicular to each other, then is equal to Option 1: 1 Option 2: 2 Option 3: 3 Option 4: 4

Answer: 1. Solution: The lines (x-5)/7 = (y+2)/(-5) = z/() and x/1 = y/(2) = z/3 are perpendicular to each other. Two lines in 3D are perpendicular when the dot product of their direction vectors equals zero. Lines in 3D written as (x-a)/p = (y-b)/q = (z-c)/r have direction vector (p, q, r). For the first line (x-5)/7 = (y+2)/(-5) = z/(): Direction vector: d_1 = (7, -5, ) For the second line x/1 = y/(2) = z/3: Direction vector: d_2 = (1, 2, 3) The perpendicular condition requires: d_1 * d_2 = 0 (7)(1) + (-5)(2) + ()(3) = 0 7 - 10 + 3 = 0 7 - 7 = 0 7 = 7 = 1 Therefore, = 1

Geometry

3D Geometry

Medium

core

If the lines $\frac{x-5}{7} = \frac{y+2}{-5} = \frac{z}{\lambda}$ and $\frac{x}{1} = \frac{y}{2\lambda} = \frac{z}{3}$ are perpendicular to each other, then $\lambda$ is equal to

✅ Correct Option: 1

The lines $\frac{x-5}{7} = \frac{y+2}{-5} = \frac{z}{\lambda}$ and $\frac{x}{1} = \frac{y}{2\lambda} = \frac{z}{3}$ are perpendicular to each other.

Two lines in 3D are perpendicular when the dot product of their direction vectors equals zero.

Lines in 3D written as $\frac{x-a}{p} = \frac{y-b}{q} = \frac{z-c}{r}$ have direction vector $(p, q, r)$ .

For the first line $\frac{x-5}{7} = \frac{y+2}{-5} = \frac{z}{\lambda}$ :

Direction vector: $\vec{d_1} = (7, -5, \lambda)$

For the second line $\frac{x}{1} = \frac{y}{2\lambda} = \frac{z}{3}$ :

Direction vector: $\vec{d_2} = (1, 2\lambda, 3)$

The perpendicular condition requires:

$\vec{d_1} \cdot \vec{d_2} = 0$

$(7)(1) + (-5)(2\lambda) + (\lambda)(3) = 0$

$7 - 10\lambda + 3\lambda = 0$

$7 - 7\lambda = 0$

$7 = 7\lambda$

$\lambda = 1$

Therefore, $\lambda = 1$

2023: 15 June Shift 2

The equation of plane which cuts equal intercepts of unit length on the coordinate axes is :

2022: 4 Aug Shift 1

The line $\frac{x+2}{3} = \frac{y+3}{5} = \frac{z-6}{4}$ passes through (a, 2, c). The value of a and c are:

2025: 30 May Shift 1

Consider the line $\vec{r} = -2\hat{i} + 3\hat{j} + \hat{k} + \lambda(5\hat{i} - 3\hat{j} - \hat{k})$ . Match List-I with List-II

List-I	List-II
(A) A point on the given line	(I) $\left(\frac{5}{\sqrt{35}}, \frac{-3}{\sqrt{35}}, \frac{-1}{\sqrt{35}}\right)$
(B) Direction ratios of the given line	(II) (2, 3, 1)
(C) Direction cosines of the given line	(III) (5, -3, -1)
(D) Direction ratios of a line perpendicular to given line	(IV) (-2, 3, 1)

Choose the correct answer from the options given below:

2026: 19 May Shift 2

The angle between two lines whose direction ratio are proportional to $(-\theta, -2\theta, \theta)$ and $(-2\theta, \theta, \theta)$ , where $\theta \in \mathbb{R} \sim \{0\}$ , is

2023: 23 May Shift 3

The value of $\lambda$ , so that the lines $\frac{1-x}{3} = \frac{7y-14}{2\lambda} = \frac{z-3}{2}$ and $\frac{7-7x}{3\lambda} = \frac{y-5}{1} = \frac{6-z}{5}$ are perpendicular is:

2022: 17 Aug Shift 2

The distance of plane $\vec{r}\cdot(6\hat{i} - 3\hat{j} - 2\hat{k}) + 1 = 0$ from origin is :

If the lines x−57=y+2−5=zλ\frac{x-5}{7} = \frac{y+2}{-5} = \frac{z}{\lambda}7x−5​=−5y+2​=λz​ and x1=y2λ=z3\frac{x}{1} = \frac{y}{2\lambda} = \frac{z}{3}1x​=2λy​=3z​ are perpendicular to each other, then λ\lambdaλ is equal to

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

If the lines $\frac{x-5}{7} = \frac{y+2}{-5} = \frac{z}{\lambda}$ and $\frac{x}{1} = \frac{y}{2\lambda} = \frac{z}{3}$ are perpendicular to each other, then $\lambda$ is equal to