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  • Please solve RD Sharma class 12 chapter Straight Line in Space exercise 27.5 question 1 sub question (i) maths textbook solution

Please solve RD Sharma class 12 chapter Straight Line in Space exercise 27.5 question 1 sub question (i) maths textbook solution

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Answer: d=3 \sqrt[2]{30}

Hint: Using formula a \frac{\left|\begin{array}{ccc} x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}\right|}{\sqrt{\left(a_{1} b-a_{2} b_{1}\right)^{2}+\left(b_{1} c_{2}-b_{2} c_{1}\right)^{2}+\left(a_{1} c_{2}-a_{2} c_{1}\right)^{2}}}

 

Given: \text { } \overrightarrow{\mathrm{\gamma}}=(3 \hat{\imath}+8 \hat{\jmath}+3 \hat{k})+\lambda(3 \hat{\imath}-\hat{\jmath}+\hat{k}) \text { and }(-3 \hat{\imath}-\hat{\jmath}+\hat{k})+\mu(7 \hat{\imath}-6 \hat{\jmath}+\hat{k})

 

Solution: From the given pair of equation

\begin{aligned} &\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1} \Rightarrow 1 \\\\ &\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{-4} \Rightarrow 2 \end{aligned}

d=\frac{\left|\begin{array}{ccc} 6 & 15 & -3 \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{array}\right|}{\sqrt{(6-3)^{2}+(-4-2)^{2}+(12+3)^{2}}}

    \begin{aligned} &=\left|\frac{6(-4-2)-15(12+3)-3(6-3)}{\sqrt{9+36+225}}\right| \\\\ &=\frac{-36-225-9}{\sqrt{270}} \end{aligned}

    \begin{aligned} &=\frac{270}{\sqrt{270}} \\\\ &=\sqrt{270} \\\\ &=3 \sqrt{30} \Rightarrow \mathrm{ANS} \end{aligned}

 

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