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  • Need solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 27.3 question 3

Need solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 27.3 question 3

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Answer: The answer is \left(\frac{1}{2}, \frac{-1}{2}, \frac{-3}{2}\right)

 

Hint: \frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}=\lambda \text { and } \frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}=\mu(\text { say })

 

Given: \frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7} \text { and } \frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}

 

Solution: we have,

\begin{gathered} \frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}=\lambda(s a y) \\ x=3 \lambda-1, y=5 \lambda-3, z=7 \lambda-5 \end{gathered}

So, the co-ordinates of a general point on this line are (3 \lambda-1,5 \lambda-3,7 \lambda-5)

The equation of second line is given

\begin{aligned} &\frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}=\mu(\text { say }) \\ &\mathrm{x}=\mu+2, \quad y=3 \mu+4, \quad z=5 \mu+6 \end{aligned}

So, the co-ordinates of general point on this line are (\mu+2,3 \mu+4,5 \mu+6)

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

\begin{aligned} &3 \lambda-1=\mu+2, \quad 5 \lambda-3=3 \mu+4, \quad 7 \lambda-5=5 \mu+6\\ &3 \lambda-\mu=3\ldots(i)\\ &5 \lambda-3 \mu=7\ldots(ii)\\ &7 \lambda-5 \mu=11 \quad \ldots(i i i) \end{aligned}

Solving the first two equations (i) and (ii), we get

\begin{aligned} &\lambda=\frac{1}{2} \\\\ &\mu=\frac{-3}{2} . \end{aligned}

Since,\lambda=\frac{1}{2} \text { and } \mu=\frac{-3}{2}   satisfying the eqn(iii)  7 \lambda-5 \mu=11

 

Hence,the given lines intersect each other.

When \lambda=\frac{1}{2} \text { in }(3 \lambda-1,5 \lambda-3,7 \lambda-5)

the co-ordinates of the required point of intersection are \left(\frac{1}{2}, \frac{-1}{2}, \frac{-3}{2}\right)

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