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  • Explain solution RD Sharma class 12 chapter Straight Line in Space exercise 27.3 question 4 maths

Explain solution RD Sharma class 12 chapter Straight Line in Space exercise 27.3 question 4 maths

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Answer: The answer of the given equation is (10,14,4).

 

Hint: By substituting the value of  λ and  μ.

 

Given: A(0,-1,-1), B(4,5,1) , C(3,9,4) \; and \; D(-4,4,4).

 

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

\begin{aligned} &\frac{x-0}{4-0}=\frac{y+1}{5+1}=\frac{z+1}{1+1}=\lambda \\\\ &\Rightarrow x=4 \lambda \\\\ &\Rightarrow y=6 \lambda-1 \\\\ &\Rightarrow z=2 \lambda-1 \end{aligned}

The co-ordinates of general point on AB are

(4 \lambda, 6 \lambda-1,2 \lambda-1)

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

\begin{aligned} &\frac{x-3}{-4-3}=\frac{y-9}{4-9}=\frac{z-4}{4-4}=\mu \\\\ &\Rightarrow x=-7 \mu+3 \\\\ &\Rightarrow y=-5 \mu+9 \\\\ &\Rightarrow z=4 \end{aligned}

The co-ordinates of general point on CD are

(-7 \mu+3,-5 \mu+9,4)

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

We must have,

\begin{aligned} &4 \lambda=-7 \mu+3, \quad 6 \lambda-1=-5 \mu+9,2 \lambda-1=4\\\\ &4 \lambda+7 \mu=3 \quad \ldots(i)\\\\ &6 \lambda+5 \mu=10 \quad \ldots(ii)\\\\ &\lambda=\frac{5}{2} \quad \cdots(iii) \end{aligned}

Solving  (ii) and (iii), we get

\begin{aligned} &\lambda=\frac{5}{2} \\\\ &\mu=-1 \end{aligned}

Substituting\lambda=\frac{5}{2} \text { and } \mu=-1  in (i),we get

\begin{aligned} &\text { L.H.S } \\\\ &=4 \lambda-7 \mu \\\\ &=4\left(\frac{5}{2}\right)-7(1) \\\\ &=3 \\ &\text { R.H.S } \end{aligned}

Since  \lambda=\frac{5}{2} \text { and } \mu=1  satisfying eqn(iii),

Hence, the given line intersect.

Substituting the value of λ and μ in the Co- ordinates of a general point on the line AB and CD, we get

\begin{aligned} &x=10 \\ &y=14 \\ &z=4 \end{aligned}

 

Hence, AB and CD intersect at point (10, 14, 4).

 

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