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

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

Answers (1)

Answer - The answer of the given question is  \frac{x-7}{4}=\frac{y-4}{1}=\frac{z+1}{2}

(Hint – By using two point formula).

Given –  Line passing through the point A(0,6,-9) and B(-3,-6,3) and C (7,4,-1).

Solution – \frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}

Hence equation of line AB

\begin{aligned} &=\frac{x-0}{-3-0}=\frac{y-6}{-6-6}=\frac{z+9}{3+9} \\\\ &=\frac{x}{-3}=\frac{y-6}{-12}=\frac{z+9}{12}=\gamma(l e t) \\\\ &=x=-3 \gamma, y=-12 \gamma+6, z=12 \gamma-9 \\\\ &:: \text { Co-ordinates of point } D(-3 \gamma,(-12 \gamma+6),(12 \gamma-9) \end{aligned}

Hence,

Direction ratio of CD is

\begin{aligned} &=(-3 \gamma-7),(-12 \gamma+6-4),(12 \gamma-9+1) \\ &=(-3 \gamma-7),(-12 \gamma+2),(12 \gamma-8) \\ \end{aligned}

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

( Hint – denominator terms of line equation).

= (-3,-12,12)

CD  is perpendicular at the given line. Therefore, By “ Conditions of Perpendicularity”.

                                                                \begin{aligned} &a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0 \\ \end{aligned}

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

\begin{aligned} &=-3(-3 \gamma-7)+(-12)(-12 \gamma+2)+12(12 \gamma-8)=0 \\ &=9 \gamma+21+144 \gamma-24+144 \gamma-96=0 \\ &=297 \gamma-99=0 \\ &=\gamma=\frac{1}{3} \\ \end{aligned}

:: Co-ordinates of D,

By putting the value of \gamma  in D , we get

\begin{aligned} &(-3 \gamma),(-12 \gamma+6),(12 \gamma-9) \\ &=D\left[-3\left(\frac{1}{3}\right),\left(-12\left(\frac{1}{3}\right)+6\right),\left(12\left(\frac{1}{3}\right)-9\right)\right. \\ &=D\{-1,2,-5\} \end{aligned}

Hence, Equation of line

\begin{aligned} &=\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}} \\ &=\frac{x-7}{-1-7}=\frac{y-4}{2-4}=\frac{z+1}{-5+1} \\ &=\frac{x-7}{-8}=\frac{y-4}{-2}=\frac{z+1}{-4} \\ &=\frac{x-7}{4}=\frac{y-4}{1}=\frac{z+1}{2} \end{aligned}

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