CUET UG 2025 — Mathematics Geometry

The line is given as $2x = 3y = z$2x=3y=z.

This means all three expressions are equal to each other.

---

Let $2x = 3y = z = k$2x=3y=z=k where $k$k is a parameter.

From $2x = k$2x=k:

$$x = \frac{k}{2}$$x=2k​

From $3y = k$3y=k:

$$y = \frac{k}{3}$$y=3k​

From $z = k$z=k:

$$z = k$$z=k

The direction ratios are $(\frac{1}{2}, \frac{1}{3}, 1)$(21​,31​,1).

Multiplying by 6 to simplify:

Direction ratios = $(3, 2, 6)$(3,2,6)

---

The $x$x-axis has direction ratios $(1, 0, 0)$(1,0,0).

The angle $\theta$θ between two lines is given by:

$$\cos \theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \times \sqrt{a_2^2 + b_2^2 + c_2^2}}$$cosθ=a12​+b12​+c12​​×a22​+b22​+c22​​∣a1​a2​+b1​b2​+c1​c2​∣​

where Line 1 has direction ratios $(3, 2, 6)$(3,2,6) and Line 2 has direction ratios $(1, 0, 0)$(1,0,0).

---

The dot product:

$$|3 \times 1 + 2 \times 0 + 6 \times 0| = |3| = 3$$∣3×1+2×0+6×0∣=∣3∣=3

---

For the line:

$$\sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$$32+22+62​=9+4+36​=49​=7

For the $x$x-axis:

$$\sqrt{1^2 + 0^2 + 0^2} = \sqrt{1} = 1$$12+02+02​=1​=1

---

$$\cos \theta = \frac{3}{7 \times 1}$$cosθ=7×13​

$$\cos \theta = \frac{3}{7}$$cosθ=73​

Therefore:

$$\theta = \cos^{-1}(\frac{3}{7})$$θ=cos−1(73​)