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  • Explain solution RD Sharma class 12 chapter Straight Line in Space exercise 27.5 question 2 sub question (iv) maths

Explain solution RD Sharma class 12 chapter Straight Line in Space exercise 27.5 question 2 sub question (iv) maths

Answers (1)

Answer: d=2 \sqrt{29} \text { units }

Hint: \text { Put } \lambda=0 \text { and } k=0

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

Solution:

                

Now, let’s take a point on first line as A(\lambda+3,2 \lambda+5, \lambda+7) and let B(7 k-1,-6 k-1, k-1) be point on the second line.

The direction ratio of the line;

(7 k-\lambda-4) \times 1+(-6 k+2 \lambda-6) \times(-2)+(k-\lambda-8) \times 1=0 \text { Line (1) and }

(7 k-\lambda-4) \times 7+(-6 k+2 \lambda-6) \times(-6)+(k-\lambda-8) \times 1=0 \text { Line(2) }

Solving equation (1) and (2) we get

\begin{aligned} &\lambda=0 \text { and } k=0 \\\\ &A=(3,5,7) \text { and } B=(-1,-1,-1) \\\\ &A B=\sqrt{(3+1)^{2}+(5+1)^{2}+(7+1)^{2}} \Rightarrow \sqrt{16+32+64} \Rightarrow \sqrt{116} \Rightarrow 2 \sqrt{29} \text { units } \end{aligned}

 

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