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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 27 Straight Line in Space Exercise 27.4 Question 9 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 27 Straight Line in Space  Exercise 27.4 Question 9 Maths Textbook Solution.

Answers (1)

Answer- The answer of the given question is \left(\frac{-4}{7}, \frac{12}{7}, \frac{15}{7}\right) \cdot \hat{r}=(-\hat{\imath}+3 \hat{\jmath}+2 \hat{k})+\gamma\left\{\left(\frac{-4}{7} \hat{\imath}+\frac{12}{7} \hat{\jmath}+\frac{15}{7} \hat{k}\right)-(\hat{\imath}+3 \hat{\jmath}+2 \hat{k})\right\}

(Hint- By comparing both the equation).

Given – Point P(-1,3,2) and equation of line \hat{r}=\left ( 2\hat{j}+3\hat{k} \right )+\gamma \left ( 2\hat{i}+\hat{j}+3\hat{k} \right ).

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

As we know position vector is given by

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

:: Position vector is given by

=-\hat{i}=3\hat{j}+2\hat{k}

 And, from a given line, we get

\begin{aligned} &=\hat{r}=(2 \hat{\jmath}+3 \hat{k})+\gamma(2 \hat{\imath}+\hat{\jmath}+3 \hat{k}) \\ &=x \hat{\imath}+y \hat{\jmath}+z \hat{k}=(2 \hat{\jmath}+3 \hat{k})+\gamma(2 \hat{\imath}+\hat{\jmath}+3 \hat{k}) \\ &=x \hat{\imath}+y \hat{\jmath}+z \hat{k}=(2 \gamma) \hat{\imath}+(\gamma+2) \hat{\jmath}+(3 \gamma+3) \hat{k} \end{aligned}

On comparing both sides we get,

=\frac{x}{2}=\frac{y-2}{1}=\frac{z-3}{3}= \gamma ;equation of line

=x=2 \gamma, y=\gamma+2, z=3 \gamma+3

Thus, co-ordinate of Q i.e. general point on the given line.

\begin{aligned} &\mathrm{Q}(\gamma, \gamma+2,3 \gamma+3) \\ \end{aligned} 

Hence

Direction of PQ is

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

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

( Hint – denominator terms of line equation).

= (2,1,3).

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

                                                         \begin{aligned} &\qquad 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)+(\gamma-1)+3(3 \gamma+1)=0 \\ &=14 \gamma+4=0 \\ &=\gamma=\frac{-4}{14} \\ &=\gamma=\frac{-2}{7} \\ \end{aligned}

:: Co-ordinates of Q,

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

\begin{aligned} &=2 \gamma, \gamma+2,3 \gamma+3 \\ &=Q\left\{2\left(\frac{-2}{7}\right),\left(\frac{-2}{7}\right)+2,3\left(\frac{-2}{7}\right)+3\right\} \\ &=Q\left(\frac{-4}{7}, \frac{12}{7}, \frac{15}{7}\right) \\ \end{aligned}

Position vector of Q

\begin{aligned} &=\frac{-4}{7} \hat{\imath}+\frac{12}{7} \hat{\jmath}+\frac{15}{7} \hat{k} \end{aligned}

Now, Equation of line passing through two points with position vector \hat{a}\; and\; \hat{b} \: is \: given \: by

                                                                              \begin{gathered} \quad \hat{r}=\hat{a}+\gamma(\hat{b}-\hat{a}) \\ \end{gathered}

= Here,

\begin{gathered} =\hat{a}=-\hat{\imath}+3 \hat{\jmath}+2 \hat{k} \\ \end{gathered}

\begin{gathered} =\text { and } \hat{b}=\frac{-4}{7} \hat{\imath}+\frac{12}{7} \hat{\jmath}+\frac{15}{7} \hat{k} \\ \end{gathered}

\begin{gathered} =\hat{r}=(-\hat{\imath}+3 \hat{\jmath}+2 \hat{k})+\gamma\left[\left(\frac{-4}{7} \hat{\imath}+\frac{12}{7} \hat{\jmath}+\frac{15}{7} \hat{k}\right)-(-\hat{\imath}+3 \hat{\jmath}+2 \hat{k})\right] . \end{gathered}

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