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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 27 Straight Line in Space Exercise 27.4 Question 10 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 27 Straight Line in Space  Exercise 27.4 Question 10  Maths Textbook Solution.

Answers (1)

Answer – The answer of the question is Q\left ( \frac{-3}{2},\frac{-1}{2},4 \right )

(Hint – By putting the value of \gamma ).

Given – Point (0,2,7) and equation of the line \frac{x+2}{-1}=\frac{y-1}{3}=\frac{z-3}{-2}.

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

Hence,

Direction of PQ is

\begin{aligned} &=(-\gamma-2-0),(3 \gamma+1-2),(-2 \gamma+3-7) \\\\ &=(-\gamma-2),(3 \gamma-1),(-2 \gamma-4) \end{aligned}

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

( Hint – denominator terms of line equation).

= (-1,3,-2)

:: 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} &=-1(-\gamma-2)+(3)(3 \gamma-1)-2(-2 \gamma-4)=0 \\ &=\gamma+2+9 \gamma-3+4 \gamma+8=0 \\ &=\gamma=\frac{-1}{2} . \end{aligned}

:: Co-ordinates of Q ,

By putting the value of \gamma  in Q

\begin{aligned} &=Q\left\{-\left(\frac{-1}{2}\right)-3\left(\frac{-1}{2}\right)+1,-2\left(\frac{-1}{2}\right)+3\right\} \\ &=Q\left\{\frac{-3}{2}, \frac{-1}{2}, 4\right\} \end{aligned}

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