CUET · MATHS · PYQ PAPER 2025
If the lines \(\frac{1-x}{3}=\frac{3 y-6}{k}=\frac{3-z}{-2}\) and \(\frac{1-x}{2 k}=\frac{y-5}{3}=\frac{6-z}{5}\) are perpendicular, then \(k\) is equal to :
- A \(\frac{10}{7}\)
- B \(\frac{10}{9}\)
- C \(-\frac{10}{9}\)
- D \(-\frac{10}{7}\)
Answer & Solution
Correct Answer
(A) \(\frac{10}{7}\)
Step-by-step Solution
Detailed explanation
\(\vec{d_1} = \langle -3, k/3, 2 \rangle\) \(\vec{d_2} = \langle -2k, 3, -5 \rangle\) \(\vec{d_1} \cdot \vec{d_2} = 0\) \((-3)(-2k) + (k/3)(3) + (2)(-5) = 0\) \(6k + k - 10 = 0\) \(7k = 10\) \(k = \frac{10}{7}\)
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