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  • Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27.1 question 14 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27.1 question 14 maths textbook solution

Answers (1)

Answer :-

\left ( x,y,z \right )=\left ( -2,-1,3 \right ) or \left ( 4,3,7 \right )

Hint :-

Distance Formula : \sqrt{\left(x_{1-} x_{2}\right)^{2}+\left(y_{1-} y_{2}\right)^{2}+\left(z_{1-} z_{2}\right)^{2}} 

Given :-

Point are P (1 , 3 , 3)

And line are  \frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}

Solution :-

\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}  -------- (1)

Let the point be x_{1},y_{1},z_{1}

\therefore \frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}=\lambda \left ( say \right )

x_{1}=3\lambda -2       ,      y_{1}=2\lambda -1      ,     z_{1}=2\lambda +3

The distance of \left ( x_{1},y_{1},z_{1} \right )  and P (1 , 3 , 3) = 5

                        \sqrt{(3 \lambda-2-1)^{2}+(2 \lambda-1-3)^{2}+(2 \lambda+3-3)^{2}}=5

Squaring both sides

     \begin{aligned} &(3 \lambda-3)^{2}+(2 \lambda-4)^{2}+(2 \lambda)^{2}=25 \\ &\Rightarrow 9 \lambda^{2}-18 \lambda+9+4 \lambda^{2}-16 \lambda+16+4 \lambda^{2}=25 \\ &\Rightarrow 17 \lambda^{2}-34 \lambda=0 \\ &\Rightarrow 17 \lambda(\lambda-2)=0 \\ &\lambda=0, \lambda=2 \end{aligned}

∴ when \lambda =0;x=-2        ,      y=-1     ,      z=3

& when \lambda =2;x=4        ,      y=3     ,      z=7

\left ( x,y,z \right )=\left ( -2,-1,3 \right ) or \left ( 4,3,7 \right )

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