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  • Please solve RD Sharma class 12 chapter Straight Line in Space exercise 27.3 question 6 sub question (iv) maths textbook solution

Please solve RD Sharma class 12 chapter Straight Line in Space exercise 27.3 question 6 sub question (iv) maths textbook solution

Answers (1)

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Answer: The line intersect each other and the point of intersection is (1,3,2).

 

Hint: Equate the coefficient of vector equation of line.

 

Given:   Two equation of line.

 

Solution: The equation of the first line,

\begin{aligned} &\frac{x-5}{4}=\frac{y-7}{4}=\frac{z+3}{-5}=\lambda(l e t) \\ \\&x=4 \lambda+5, \quad y=4 \lambda+7, \quad z=-5 \lambda-3 \end{aligned}

So, the co-ordinates of a general point on this line are (4 \lambda+5,4 \lambda+7,-5 \lambda-3)

The equation of the second line,

\begin{aligned} &\frac{x-8}{7}=\frac{y-4}{1}=\frac{z-5}{3}=\mu(\text { let }) \\\\ &\mathrm{x}=7 \mu+8, \quad y=\mu+4, \quad z=3 \mu+5 \end{aligned}

So, the co-ordinates of a general point on this line are ((7 \mu+8, \mu+4,3 \mu+5)

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

We have,

\begin{aligned} &4 \lambda+5=7 \mu+8,4 \lambda+7=\mu+4,-5 \lambda-3=3 \mu+5\\ &4 \lambda-7 \mu=3\quad\dots(i)\\\\ &\mu=4 \lambda+3\quad\dots(ii)\\\\\ &-5 \lambda-3 \mu=8\quad\dots(iii)\\\\ \end{aligned}

Putting the value of μ from equation (ii) to (i), we get

\begin{aligned} &4 \lambda-7 \mu=3 \\\\ &4 \lambda-7(4 \lambda+3)=3 \\\\ &4 \lambda-28 \lambda-21=3 \end{aligned}

\begin{aligned} &-24 \lambda=24 \\\\ &\lambda=-1 \end{aligned}

Now putting the value of λ in equation (ii), we get

\begin{aligned} &\mu=4 \lambda+3 \\\\ &\mu=4(-1)+3 \\\\ &\mu=-1 \end{aligned}

As we can see by putting the value of λ and μ in equation (iii), that it satisfy the equation

Check, \begin{aligned} &\text { }-5 \lambda-3 \mu=8 \\ \end{aligned}

-5(-1)-3(-1)=8 \\\\

5+3=8 \

8=8

L.H.S.=R.H.S.

Hence intersection point exists or line of intersects.

Thus, point of intersection (4 \lambda+5,4 \lambda+7,-5 \lambda-3)

\begin{aligned} &=(4(-1)+5,4(-1)+7,-5(-1)-3) \\\\ &=(1,3,2) \end{aligned}

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