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The Lines X 1 = Y 2 = Z 3 a N D X − 1 − 2 = Y − 2 − 4 = Z − 3 − 6 (A) Parallel (B) Intersecting (C) Skew (D) Coincident

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Question

The lines  \[\frac{x}{1} = \frac{y}{2} = \frac{z}{3} \text { and } \frac{x - 1}{- 2} = \frac{y - 2}{- 4} = \frac{z - 3}{- 6}\] 

 

Options

  • parallel

  • intersecting

  •  skew 

  •  coincident

     
MCQ
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Solution

 coincident

The equations of the given lines are

\[\frac{x}{1} = \frac{y}{2} = \frac{z}{3} . . . (1)\]

\[\frac{x - 1}{- 2} = \frac{y - 2}{- 4} = \frac{z - 3}{- 6}\]

\[ \Rightarrow \frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3} . . . \left( 2 \right)\] 

Thus, the two lines are parallel to the vector 

\[\overrightarrow{b} = \hat{i} + 2 \hat{j}  + 3 \hat{k} \] and pass through the points (0, 0, 0) and (1, 2, 3).

Now,

\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) \times \overrightarrow{b} = \left( \hat{i} + 2 \hat{j} + 3 \hat{k}  \right) \times \left( \hat{i}  + 2 \hat{j}  + 3 \hat{k}  \right)\]

\[ = \overrightarrow{0} \left[ \because \overrightarrow{a} \times \overrightarrow{a} = \overrightarrow{0} \right]\] 


Since, the distance between the two parallel lines is 0, the given two lines are coincident lines. 

 

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Chapter 27: Straight Line in Space - MCQ [Page 43]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 27 Straight Line in Space
MCQ | Q 12 | Page 43

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