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  • Please Solve R.D.Sharma class 12 Chapter 27 Straight Line in Space Exercise 27.4 Question 2 Maths Textbook Solution.

Please Solve R.D.Sharma class 12 Chapter 27 Straight Line in Space Exercise 27.4 Question 2 Maths Textbook Solution.

Answers (1)

Answer – The answer of the given question is 2\sqrt{6}.

(Hint – denominator terms of line equation).

Given – Point ( 1, 0 ,0 ) and equation of line \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z-10}{8}

Solution -  Let PQ be the perpendicular drawn from the P to given line whose endpoint/foot is Q point.

                                                                                                       \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z-10}{8}=\gamma (let)

 

:: Co-ordinates of Q \left ( 2\gamma +1,-3\gamma -1,8\gamma -10 \right )

Hence,

Direction ratio of PQ is

\begin{aligned} &=(2 \gamma+1-1),(-3 \gamma-1-0),(8 \gamma-10-0) \\ &=(2 \gamma),(-3 \gamma-1),(8 \gamma-10) \end{aligned}

and by comparing the given line equation, direction ratios of the given line are 

( Hint – denominator terms of line equation).

= (2 , -3 , 8)

Since PQ is perpendicular at the given line.

 Therefore, By “Conditions of Perpendicularity”.

                                                            a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0

; whereas a terms and b terms are direction ratio of lines which are perpendicular to each other.

\begin{aligned} &=2(2 \gamma)+(-3)(-3 \gamma-1)+(8)(8 \gamma-10)=0 \\ &=4 \gamma-9 \gamma+3+64 \gamma+80=0 \\ &=77 \gamma+77=0 \\ &=\gamma=1 \end{aligned}

::Co-ordinate of Q i.e foot of perpendicular , by putting the value of  γ  in Q co-ordinate equation ,we get

\begin{aligned} &\mathrm{Q}\{2(1)+1,-3(1)-1,8(1)-10\} \\ &\mathrm{Q}\{3,-4,-2\} \\ \end{aligned}

NOW

Distance between PQ

\begin{aligned} &=\sqrt{\left(x_{2}-x_{1}\right)+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}} \\ &=\sqrt{(1-3)^{2}+(0+4)^{2}+(-2-0)^{2}} \\ &=\sqrt{(-2)^{2}+(4)^{2}+(-2)^{2}} \\ &=\sqrt{4+16+4} \\ &=\sqrt{24} \\ &=2 \sqrt{6} \text { unit. } \end{aligned}

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