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  • Explain Solution R.D.Sharma Class 12 Chapter 27 Straight Line in Space Exercise 27.4 Question 11 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 27 Straight Line in Space Exercise 27.4 Question 11 Maths Textbook Solution.

Answers (1)

Answer - The answer of the question is Q\left ( 1,1,-1 \right )

(Hint – By comparing with line equation).

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

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

Thus to find Distance PQ we have to first find co-ordinates of Q.

\begin{aligned} &\text { * } \frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1}=\gamma(\text { let }) \\ &\text { * }\mathrm{x}=2 \gamma-1, y=-2 \gamma+3, z=-\gamma \\ &:: \text { Co-ordinates of } \mathrm{Q}(2 \gamma-1,-2 \gamma+3,-\gamma) \end{aligned}

Hence,

Direction ratio of PQ is

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

And by comparing with given line equation, direction ratios of the given line are

( Hint – denominator terms of line equation).

= (2,-2,-1)

:: 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-2)+(-2)(-2 \gamma+1)-1(-\gamma+3)=0 \\ &=4 \gamma-4+4 \gamma-2+\gamma-3=0 \\ &=9 \gamma-9=0 \\ &=\gamma=1 \\ \end{aligned}

:: Co-ordinates of Q ,

By putting the value of \gamma in Q

 

\begin{aligned} &=Q\{2(1)-1,-2(1)+3,-1\} \\\\ &=Q\{1,1,-1\} \end{aligned}

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