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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 27 Straight Line in Space Exercise 27.4 Question 7 Maths Textbook Solution.

Provide Solution For  R.D.Sharma Maths Class 12 Chapter 27 Straight Line in Space Exercise 27.4 Question 7 Maths Textbook Solution.

Answers (1)

Answer - The answer of the given question is \sqrt{\frac{2109}{110}}Units ,Q\left ( \frac{188}{110},\frac{351}{110},\frac{195}{110} \right )

Hint – By using two point formula).

Given – Point (5,4,-1) and equation of the line \hat{r}=\hat{i}+\gamma \left ( 2\hat{i}+9\hat{j}+5\hat{x} \right )

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

As we know positive vector is given by

                                                         \hat{r}=x\hat{i}+y\hat{j}+z\hat{x}

:: position vector of point P is

                                             5\hat{i}+4\hat{j}-1\hat{x}

 Ad from a given line we get

                                        \Rightarrow \hat{r}=\hat{i}+\gamma ( z\hat{i}+9\hat{j}+5\hat{x}

                                       \begin{gathered} \Rightarrow x \hat{\imath}+y \hat{\jmath}+z \hat{x} \\ \Rightarrow \hat{i}+\gamma(z \hat{i}+9 \hat{j}+5 \hat{x} \\ \Rightarrow x \hat{\imath}+y \hat{\jmath}+z \hat{x} \\ \Rightarrow(1+2 \gamma) \hat{i}+(9 \gamma) \hat{j}+(5 \gamma) \hat{x} \end{gathered}

On comparing both side we get,

\begin{aligned} &=\mathrm{x}=(1+2 \gamma), y=(9 \gamma), z=(5 \gamma)\\ &=\frac{x-1}{2}=\frac{y}{9}=\frac{z}{5}=\gamma ; \text { equation of line }\\ \end{aligned}

Thus, Co-ordinates of Q i.e. General point of the given line.

\begin{aligned} &\mathrm{Q}\{(1+2 \gamma),(9 \gamma),(5 \gamma)\} \end{aligned}

Hence

Direction ratio of PQ is

\begin{aligned} &=(2 \gamma+1-5),(9 \gamma-4),(5 \gamma+1) \\ &=(2 \gamma-4),(9 \gamma-4),(5 \gamma+1) \end{aligned}

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

( Hint – denominator terms of line equation).

= (2,9,5).

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-4)+(9)(9 \gamma-4)+5(5 \gamma+1)=0 \\\\ &=4 \gamma-8+81 \gamma-36+25 \gamma+5=0 \\\\ &=110 \gamma-39=0 \\\\ &=\gamma=\frac{39}{110} \\ \end{aligned}

:: Co-ordinates of Q.,

By putting the value of γ  in Q co-ordinate equation, we get

\begin{aligned} &=Q\{(1)-5,4(1)-3,-9(1)+6\} \\\\ &=Q \gamma\{-4,1,-3\} \\ \end{aligned}

:: Co-ordinates of Q i.e. foot of perpendicular

By putting the value of γ in Q

\begin{aligned} &=Q\left\{2\left(\frac{39}{110}\right)+1,9\left(\frac{39}{110}\right), 5\left(\frac{39}{110}\right)\right\} \\\\ &=Q\left\{\left(\frac{188}{110}, \frac{351}{110}, \frac{195}{110}\right)\right\} \\ \end{aligned}

Now,

Distance between PQ

\begin{aligned} &=\sqrt{\left(x_{2}-x_{1}\right)+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}} \end{aligned}

\begin{aligned} &=\sqrt{\left(\frac{188}{110}-5\right)^{2}+\left(\frac{351}{110}-4\right)^{2}+\left(\frac{195}{110}+1\right)^{2}} \\ &=\sqrt{\left(\frac{-362}{110}\right)^{2}+\left(\frac{-89}{110}\right)^{2}+\left(\frac{305}{110}\right)^{2}} \\ \end{aligned}

\begin{aligned} &=\sqrt{\frac{131044}{12100}+\frac{7921}{12100}+\frac{93025}{12100}} \\ &=\sqrt{\frac{231990}{12100}} \\ &=\sqrt{\frac{2109}{110} \text { units. }} \end{aligned}

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