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MHT CET · Maths · Three Dimensional Geometry

If the shortest distance between the lines \(\frac{x-\mathrm{k}}{2}=\frac{\mathrm{y}-4}{3}=\frac{\mathrm{z}-3}{4}\) and \(\frac{x-2}{4}=\frac{\mathrm{y}-4}{6}=\frac{\mathrm{z}-7}{8}\) is \(\frac{13}{\sqrt{29}}\), then \(\mathrm{k}=\)

  1. A \(1\)
  2. B \(-1\)
  3. C \(-\left(\cos x-\frac{2}{3} \cos ^2 x+\frac{\cos ^5 x}{5}+c\right)\), where \(c\) is the constant of integration
  4. D \(\left(\cos x-\frac{2}{3} \cos ^2 x+\frac{\cos ^5 x}{5}\right)+\mathrm{c}\), where c is the constant of integration
Verified Solution

Answer & Solution

Correct Answer

(A) \(1\)

Step-by-step Solution

Detailed explanation

Lines are parallel. Let \(\vec{a_1} = \left\langle \mathrm{k}, 4, 3 \right\rangle\), \(\vec{a_2} = \left\langle 2, 4, 7 \right\rangle\), \(\vec{b} = \left\langle 2, 3, 4 \right\rangle\). \(\vec{a_2} - \vec{a_1} = \left\langle 2-\mathrm{k}, 0, 4 \right\rangle\).