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The Direction Ratios of the Line X − Y + Z − 5 = 0 = X − 3y − 6 Are Proportional to (A) 3, 1, −2 (B) 2, −4, 1 (C) 3 √ 14 , 1 √ 14 , − 2 √ 14 (D) 2 √ 41 , − 4 √ 41 , 1 √ 41

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Question

The direction ratios of the line x − y + z − 5 = 0 = x − 3y − 6 are proportional to

 

 

Options

  • 3, 1, −2

  • 2, −4, 1

  •  \[\frac{3}{\sqrt{14}}, \frac{1}{\sqrt{14}}, \frac{- 2}{\sqrt{14}}\]

  •  \[\frac{2}{\sqrt{41}}, \frac{- 4}{\sqrt{41}}, \frac{1}{\sqrt{41}}\]

MCQ
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Solution

 3, 1, −2

We have ,

x − y + z − 5 = 0 = x − 3y − 6 

\[\Rightarrow x - 3y - 6 = 0 \]

\[ x - y + z - 5 = 0\]

\[ \Rightarrow x = 3y + 6 . . . \left( 1 \right) \]

\[ x - y + z - 5 = 0 . . . \left( 2 \right)\]

From (1) and (2),

we get ,

\[3y + 6 - y + z - 5 = 0\]

\[ \Rightarrow 2y + z + 1 = 0\]

\[ \Rightarrow y = \frac{- z - 1}{2} \]

\[y = \frac{x - 6}{3} \left[\text { From } \left( 1 \right) \right]\]

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

So, the given equation can be re-written as 

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

Hence, the direction ratios of the given line are proportional to 3, 1, -2 . 

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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 5 | Page 43

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