CUET UG 2025 — Mathematics Geometry

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

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

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

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

Direction vector: $\vec{d_1} = (7, -5, \lambda)$d1​​=(7,−5,λ)

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

Direction vector: $\vec{d_2} = (1, 2\lambda, 3)$d2​​=(1,2λ,3)

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The perpendicular condition requires:

$$\vec{d_1} \cdot \vec{d_2} = 0$$d1​​⋅d2​​=0

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

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

$$7 - 7\lambda = 0$$7−7λ=0

$$7 = 7\lambda$$7=7λ

$$\lambda = 1$$λ=1

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Therefore, $\lambda = 1$λ=1