CUET UG 2025 — Mathematics Geometry

The angle between two lines in 3D requires finding the direction vectors of each line, then using the dot product formula.

Given: $l_1: \frac{x + 1}{1} = \frac{2 - y}{2} = \frac{z - 1}{1}$l1​:1x+1​=22−y​=1z−1​

The middle term has "2 - y" instead of "y - 2". Rewriting:

$$\frac{x + 1}{1} = \frac{-(y - 2)}{2} = \frac{z - 1}{1}$$1x+1​=2−(y−2)​=1z−1​

$$\frac{x + 1}{1} = \frac{y - 2}{-2} = \frac{z - 1}{1}$$1x+1​=−2y−2​=1z−1​

Direction vector of $l_1$l1​: $\vec{b_1} = (1, -2, 1)$b1​​=(1,−2,1)

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Given: $l_2: \frac{x - 1}{4} = \frac{2y - 4}{6} = \frac{z - 1}{2}$l2​:4x−1​=62y−4​=2z−1​

The middle term simplifies: $2y - 4 = 2(y - 2)$2y−4=2(y−2)

$$\frac{x - 1}{4} = \frac{2(y - 2)}{6} = \frac{z - 1}{2}$$4x−1​=62(y−2)​=2z−1​

$$\frac{x - 1}{4} = \frac{y - 2}{3} = \frac{z - 1}{2}$$4x−1​=3y−2​=2z−1​

Direction vector of $l_2$l2​: $\vec{b_2} = (4, 3, 2)$b2​​=(4,3,2)

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$$\vec{b_1} \cdot \vec{b_2} = (1)(4) + (-2)(3) + (1)(2)$$b1​​⋅b2​​=(1)(4)+(−2)(3)+(1)(2)

$$= 4 - 6 + 2$$=4−6+2

$$= 0$$=0

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When the dot product of two direction vectors is zero, the lines are perpendicular to each other.

The angle between perpendicular lines is $\frac{\pi}{2}$2π​.

Therefore, the angle between the lines is $\frac{\pi}{2}$2π​.