CUET UG 2025 — Mathematics Geometry

The lines need to be rewritten in standard form $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$ax−x0​​=by−y0​​=cz−z0​​, where $(a, b, c)$(a,b,c) are the direction ratios.

For Line 1: $\frac{2x - 3}{4} = \frac{3 - y}{k} = \frac{z - 2}{-2}$42x−3​=k3−y​=−2z−2​

$$\frac{2(x - \frac{3}{2})}{4} = \frac{-(y - 3)}{k} = \frac{z - 2}{-2}$$42(x−23​)​=k−(y−3)​=−2z−2​

$$\frac{x - \frac{3}{2}}{2} = \frac{y - 3}{-k} = \frac{z - 2}{-2}$$2x−23​​=−ky−3​=−2z−2​

Direction ratios of Line 1: $(a_1, b_1, c_1) = (2,\ -k,\ -2)$(a1​,b1​,c1​)=(2, −k, −2)

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For Line 2: $\frac{x - 2}{1} = \frac{y}{4} = \frac{5 - z}{3}$1x−2​=4y​=35−z​

$$\frac{x - 2}{1} = \frac{y - 0}{4} = \frac{-(z - 5)}{3} = \frac{z - 5}{-3}$$1x−2​=4y−0​=3−(z−5)​=−3z−5​

Direction ratios of Line 2: $(a_2, b_2, c_2) = (1,\ 4,\ -3)$(a2​,b2​,c2​)=(1, 4, −3)

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For two lines to be perpendicular, the dot product of their direction ratios must equal zero:

$$a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$$a1​a2​+b1​b2​+c1​c2​=0

$$(2)(1) + (-k)(4) + (-2)(-3) = 0$$(2)(1)+(−k)(4)+(−2)(−3)=0

$$2 - 4k + 6 = 0$$2−4k+6=0

$$8 - 4k = 0$$8−4k=0

$$4k = 8$$4k=8

$$k = 2$$k=2