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  • Please Solve R.D.Sharma class 12 Chapter 27 Straight Line in Space Exercise 27.4 Question 3 Maths Textbook Solution.

Please Solve R.D.Sharma class 12 Chapter 27 Straight Line in Space Exercise 27.4 Question 3 Maths Textbook Solution.

Answers (1)

Answer -  The answer of this question is D\left ( \frac{5}{3},\frac{7}{3},\frac{17}{3} \right )

(Hint – By using \frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}})

Given – Perpendicular from A ( 1,0,3)drawn at line joining points B (4,7,1) and (3,5,3).

Solution – Let D be the foot of the perpendicular drawn from A (1,0,3) to line joining points B (4,7,1) and C(3,5,3).

Now,

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

                                                            \begin{aligned} &=\frac{x-4}{3-4}=\frac{y-7}{5-7}=\frac{z-1}{3-1} \\ &=\frac{x-4}{-1}=\frac{y-7}{-2}=\frac{z-1}{2} \end{aligned}

Now,

\begin{aligned} &=\frac{x-4}{-1}=\frac{y-7}{-2}=\frac{z-1}{2} =\gamma (let) \end{aligned}

=X=-\gamma +4,y=-2\gamma +7,z=2\gamma +1

:: Co-ordinates of D (-\gamma +4,-2y+7,2y+1)

Hence

Direction ratios pf AD

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

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

( Hint – denominator terms of line equation).

= (-1,-2,2)

Since PQ 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} =&-1(-\gamma+3)+(-2)(-2 \gamma+7)+2(2 \gamma-2)=0 \\ =& \gamma-3+4 \gamma-14+4 \gamma-4=0 \\ =& 9 \gamma-21=0 \\ \end{aligned}

\begin{aligned} =& \gamma=\frac{21}{9} \\ =& \gamma=\frac{7}{3} . \\ \end{aligned}

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

By putting value of γ  in D co-ordinate equation. we get,

\begin{aligned} \quad & \bullet \operatorname{D}(-\gamma+4,-2 \gamma+7,2 \gamma+1) \\ \quad & \quad D\left(\frac{-7}{3}+4,-2 \frac{7}{3}+7,2 \frac{7}{3}+1\right) \\ \quad & \text { - }\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right) \end{aligned}

                                                                                

Posted by

infoexpert21

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