हिंदी

The lines andx1=y2=z3andx-1-2=y-2-4=z-3-6 are

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प्रश्न

The lines `x/1 = y/2 = z/3 and (x - 1)/-2 = (y - 2)/-4 = (z - 3)/-6` are

विकल्प

  • coincident 

  • skew

  • intersecting

  • parallel

MCQ
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उत्तर

(a) coincident
The equation of the given lines are 

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

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

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

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,

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

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

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

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 27: Straight Line in Space - MCQ [पृष्ठ ४२]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 27 Straight Line in Space
MCQ | Q 2 | पृष्ठ ४२

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