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Please solve RD Sharma class 12 chapter 27 Straight Line in Space exercise Multiple choice question 5 maths textbook solution

Answers (1)

Answer: Correct answer is A.

Hint: Solve Simultaneously.

 

Given: x-y+z-5=0 and  x-3y-6=0

 

Solution:x-y+z-5=0 and  x-3y-6=0

\begin{gathered} \Rightarrow x-y+z-5=0 \\\\ x-3 y-6=0 \end{gathered}

\begin{aligned} &\Rightarrow x-y+z-5=0 \quad\dots(1)\\ &x=3 y+6\quad\dots(2) \end{aligned}

From (1) and (2) we get,

 

\begin{aligned} &3 y+6-y+z-5=0 \\\\ &2 y+z+1=0 \\\\ &y=\frac{-z-1}{2} \end{aligned}

Also,

\begin{aligned} &y=\frac{x-6}{3} \rightarrow \text { from }(2) \\\\ &\therefore \frac{x-6}{3}=y=\frac{-z-1}{2} \end{aligned}
 

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.

So, the correct option is (a)  3, 1, -2.

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