CUET · MATHS · PYQ PAPER 2025
If the line \(\frac{-x+1}{3}=\frac{-y-2}{-2 k}=\frac{z+3}{2}\) and \(\frac{-1+x}{3 k}=\frac{-1+y}{1}=\frac{-z+6}{5}\) are perpendicular, then the value of \(k\) is :
- A \(\frac{10}{7}\)
- B \(-\frac{10}{7}\)
- C \(\frac{1}{2}\)
- D \(\frac{4}{3}\)
Answer & Solution
Correct Answer
(B) \(-\frac{10}{7}\)
Step-by-step Solution
Detailed explanation
\(\vec{d_1} = \langle -3, 2k, 2 \rangle\), \(\vec{d_2} = \langle 3k, 1, -5 \rangle\) \(\vec{d_1} \cdot \vec{d_2} = 0 \implies (-3)(3k) + (2k)(1) + (2)(-5) = 0\) \(-9k + 2k - 10 = 0\) \(-7k = 10\) \(k = -\frac{10}{7}\)
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