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  • Please Solve R.D.Sharma class 12 Chapter 27 Straight Line in Space Exercise 27.4 Question 4 Maths Textbook Solution.

Please Solve R.D.Sharma class 12 Chapter 27 Straight Line in Space Exercise 27.4 Question 4 Maths Textbook Solution.

Answers (1)

Answer- The answer of the given question is D\left ( \frac{22}{9},-\frac{11}{9},\frac{5}{9} \right )

(Hint – By using the formula of a line joining the two points.)

Given- Perpendicular from A (1,0,4)drawn at line joining points B(0,-11,3) and C(2,

3,1).

Solution – D be the foot of the perpendicular drawn from  A(1,0,4) to line joining

points B(0,-11,3) and C(2,-3,1).

Now, lets find the equation of the line which is formed by joining points B(0,-11,3) and C(2,-3,1).

\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-0}{2-0}=\frac{y+11}{-3+11}=\frac{z-3}{1-3} \\\\ &=\frac{x}{2}=\frac{y+11}{8}=\frac{z-3}{-2}=\gamma(\text { let }) \\\\&=(x=2 \gamma, y=8 \gamma-11, z=-2 \gamma+3 \end{aligned}

::Co-ordinates of D\left ( 2\gamma ,8\gamma -11,-2\gamma +3 \right ).

Hence,

Direction ratios of AD

\begin{aligned} &=(2 \gamma-1),(8 \gamma-11-0),(-2 \gamma+3-4) \\\ &=(2 \gamma-1),(8 \gamma-11),(-2 \gamma-1) \end{aligned}

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

( Hint – denominator terms of line equation).

= (2,8,-2)

AD 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} =2 &(2 \gamma-1)+(8)(8 \gamma-11)-2(-2 \gamma-1)=0 \\ \end{aligned}

\begin{aligned} =72 \gamma-88=0 \\ \end{aligned}

\begin{aligned} =& \gamma=\frac{88}{72} \\ \end{aligned}

\begin{aligned} =& \gamma=\frac{11}{9} \\ \end{aligned}

:: Co-ordinates of D i.e.foot of the perpendicular.

By putting the value of \gamma  in D co-ordinate equation, we get

                                                                    

\begin{aligned} & \bullet \quad D(2 \gamma, 8 \gamma-11,-2 \gamma+3\\ \end{aligned}

\begin{aligned} & \bullet & \quad D\left(2 \frac{11}{9}, 8\left(\frac{11}{9}\right)-11,-2\left(\frac{11}{9}\right)+3\right) \\ \end{aligned}

\begin{aligned} & \bullet & \quad D\left(\frac{22}{9}, \frac{-11}{9}, \frac{5}{9}\right) \end{aligned}
 

 

Posted by

infoexpert21

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