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  • Explain solution RD Sharma class 12 chapter Straight Line in Space exercise 27.3 question 6 maths sub question (iii)

Explain solution RD Sharma class 12 chapter Straight Line in Space exercise 27.3 question 6 maths sub question (iii)

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Answer: The line intersect at the point of intersection is (4,0,-1).

 

Hint: Equate the coefficient of vector equation of line.

 

Given:   \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0} \text { and } \frac{x-4}{2}=\frac{y-0}{0}=\frac{z+1}{3}

 

Solution: co-ordinate of first equation of line

\begin{aligned} &\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}=\lambda(\text { let })\quad \dots (i)\\\\ &\mathrm{x}=3 \lambda+1, y=-\lambda+1, z=-1 \end{aligned}

Co-ordinate of second equation of line

\begin{aligned} &\frac{x-4}{2}=\frac{y-0}{0}=\frac{z+1}{3}=\mu(\text { let })\quad\dots(ii)\\\\ &x=2 \mu+4, y=0, z=3 \mu-1 \end{aligned}

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

\begin{aligned} &\text { So, } 3 \lambda+1=2 \mu+4\quad\dots(iii)\\\\ &-\lambda+1=0 \Rightarrow \lambda=1\quad\dots(iv)\\\\ &3 \mu-1=-1 \Rightarrow \mu=0\quad\dots(v)\\\\ &\lambda=1 \text { and } \mu=0 \end{aligned}

Satisfy equation (iii). So, the given line intersect and the point of intersection is (4,0,-1).

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