CUET UG 2025 — Mathematics Calculus

The lines need to be rewritten in the standard $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ax−x1​​=by−y1​​=cz−z1​​ form to extract direction ratios.

Line 1: $\frac{x-1}{-3} = \frac{2y-2}{2p} = \frac{z-3}{2}$−3x−1​=2p2y−2​=2z−3​

$$= \frac{x-1}{-3} = \frac{2(y-1)}{2p} = \frac{z-3}{2}$$=−3x−1​=2p2(y−1)​=2z−3​

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

Direction ratios of Line 1: $(-3,\; p,\; 2)$(−3,p,2)

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Line 2: $\frac{x-1}{-3p} = \frac{y-1}{4} = \frac{6-z}{5}$−3px−1​=4y−1​=56−z​

Since $6 - z = -(z - 6)$6−z=−(z−6):

$$= \frac{x-1}{-3p} = \frac{y-1}{4} = \frac{z-6}{-5}$$=−3px−1​=4y−1​=−5z−6​

Direction ratios of Line 2: $(-3p,\; 4,\; -5)$(−3p,4,−5)

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Two lines are perpendicular when the dot product of their direction ratios equals zero:

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

$$(-3)(-3p) + (p)(4) + (2)(-5) = 0$$(−3)(−3p)+(p)(4)+(2)(−5)=0

$$9p + 4p - 10 = 0$$9p+4p−10=0

$$13p = 10$$13p=10

$$p = \dfrac{10}{13}$$p=1310​