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  • Provide solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 27.3 question 2

Provide solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 27.3 question 2

Answers (1)

Answer: The given line don’t intersect each other.

 

Hint: By substituting \lambda=-17 \text { and } \mu=-12

 

Given: \frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5} \text { and } \frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}

 

Solution: The co- ordinates of any point on the first line are given by

\begin{aligned} &\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}=\lambda(\text { say }) \\ \end{aligned}

\begin{aligned} &\bullet\text {} -x=3 \lambda+1 \\\\ &\bullet y=2 \lambda-1 \\\\ &\bullet z=5 \lambda+1 \end{aligned}

The co-ordinates of first line is (3 \lambda+1,2 \lambda-1,5 \lambda+1)

The co- ordinates of any point on the second line are given by

\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}=\mu(s a y)

\begin{aligned} &\text { } \bullet \mathrm{x}=4 \mu-2 \\\\ &\text { } \bullet \mathrm{y}=3 \mu+1 \\\\ &\text { } \bullet \mathrm{z}=-2 \mu-1 \end{aligned}

The co-ordinates of second line is (4 \mu-2,3 \mu+1,-2 \mu-1)

If the line intersect, then they have a common  point. So, for some value of λ and μ.

We must have,

3 \lambda+1=4 \mu-2, \quad 2 \lambda-1=3 \mu+1, \quad 5 \lambda+1=-2 \mu-1

\begin{aligned} &\Rightarrow 3 \lambda-4 \mu=-3\quad \ldots(i)\\ &\Rightarrow 2 \lambda-3 \mu=2 \quad \ldots(ii)\\ &\Rightarrow 5 \lambda+2 \mu=-2\quad \ldots(iii) \end{aligned}

Solving  (i)and (ii), we get

\begin{aligned} &\lambda=-17 \\ &\mu=-12 \end{aligned}

Substituting  \lambda=-17 \text { and } \mu=-12 in (iii)

L.H.S

\begin{aligned} &=5 \lambda+2 \mu \\ &=5(-17)+2(-12) \end{aligned}

\begin{aligned} &=-85-24 \\ &=-109 \\ &\neq-2 \\ &\text { L.H.S } \neq R . H . S \end{aligned}

 

 This shows that the given line will not intersect each other.

 

Posted by

infoexpert26

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