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  • Please Solve R.D.Sharma class 12 Chapter 27 Straight Line in Space Exercise 27.4 Question 1 Maths Textbook Solution.

Please Solve R.D.Sharma class 12 Chapter 27 Straight Line in Space Exercise 27.4 Question 1 Maths Textbook Solution.

Answers (1)

Answer – The length of perpendicular is \sqrt{\frac{4901}{841}} unit.

(Hints – Denominator terms of line equation).

Given –\left ( 3,-1,11 \right );\frac{x}{z}=\frac{y-z}{3}=\frac{z-3}{4}

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

Thus to find distance PQ we have to first  find co-ordinator of Q .

\frac{x}{z}=\frac{y-2}{3}=\frac{z-3}{4}=\gamma (let)

 

Therefore ,Co-ordinates of Q \left ( 2\gamma ,-3\gamma +2,4\gamma +3 \right ).

Hence

Direction of PQ is

=\left ( 2y-3 \right ),\left ( -3y+2+1 \right ),\left ( 4y+3-11 \right )

=\left ( 2y-3 \right ),\left ( -3y+3 \right ),\left ( 4y-8 \right ).

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

 (Hint – denominator terms of line equation)

=2,-3,4

PQ is perpendicular to given line,

:: By “Condition 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} &\rightarrow 2(2 \gamma-3)+(-3)(-3 \gamma+3)+4(4 \gamma-8) \\ &\rightarrow 4 \gamma-6+9 \gamma-9+16 \gamma-32=0 \\ &\rightarrow 29 \gamma-47=0 \\ &\rightarrow \gamma=\frac{47}{29} \end{aligned}

::Co-ordinate of Q i.e foot of perpendicular , by putting the value of  γ 

In Q co-ordinate equation ,we got

                                                  =Q\left \{ 2\left ( \frac{47}{29} \right ),-3\left ( \frac{47}{29} \right )+2,4\left ( \frac{47}{29} \right )+3 \right \}

=Q\left ( \frac{94}{29},\frac{-83}{29},\frac{275}{29} \right )

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{94}{29}-3\right)^{2}+\left(\frac{-83}{29}+1\right)^{2}+\left(\frac{275}{29}-11\right)^{2}} \\ \end{aligned}

\begin{aligned} &=\sqrt{\left(\frac{94-87}{29}\right)^{2}+\left(\frac{-83+29}{29}\right)^{2}+\left(\frac{275-319}{29}\right)^{2}} \\ \end{aligned}

\begin{aligned} &=\sqrt{\frac{49}{841}+\frac{2016}{841}+\frac{1936}{841}} \\ &=\sqrt{\frac{4901}{841}} \text { unit. } \end{aligned}

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