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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 27 Straight Line in Space Exercise 27.4 Question 5 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 27 Straight Line in Space Exercise 27.4 Question 5 Maths Textbook Solution.

Answers (1)

Answer - The answer of this question is \frac{3}{7}\sqrt{101}Units ,Q\left ( \frac{170}{49},\frac{78}{49},\frac{10}{49} \right )

(Hint – By using the distance between two point formula).

Given – Point (2,3,4) and equation of the line \frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}

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

Now,

\begin{aligned} &* \frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}=\gamma(\text { let }) \\ &* \mathrm{x}=4-2 \gamma, y=6 \gamma, z=1-3 \gamma \\ \end{aligned}

:: Co-ordinates of  \begin{aligned} &\mathrm{Q}(-2 \gamma, 6 \gamma,-3 \gamma+1) \\ \end{aligned}

Hence

Direction of PQ is

\begin{aligned} &=(-2 \gamma+4-2),(6 \gamma-3),(-3 \gamma+1-4) \\ &=(-2 \gamma+2),(6 \gamma-3),(-3 \gamma-3) \end{aligned}

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

( Hint – denominator terms of line equation).

= (-2,6,-3)

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+2)+(6)(6 \gamma-3)-3(-3 \gamma-3)=0 \\ &=4 \gamma-4+36 \gamma-18+9 \gamma+9=0 \\ &=49 \gamma-13=0 \\ &=\gamma=\frac{13}{49} \end{aligned}

:: Co-ordinates of Q.,

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

\begin{aligned} &=\mathrm{Q}\left\{-2 \frac{13}{49}+4,6 \frac{13}{49},-3 \frac{13}{49}+1\right\} \\ &=\mathrm{Q} \gamma\left\{\frac{170}{49}, \frac{78}{49}, \frac{10}{49}\right\} \end{aligned}

Now,

Distance between PQ

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

\begin{aligned} &=\sqrt{\left(\frac{170}{49}-2\right)^{2}+\left(\frac{78}{49}-3\right)^{2}+\left(\frac{10}{49}-4\right)^{2}} \\ &=\sqrt{\left(\frac{72}{49}\right)^{2}+\left(\frac{69}{49}\right)^{2}+\left(\frac{168}{49}\right)^{2}} \\ \end{aligned}

\begin{aligned} &=\sqrt{\frac{5184}{2401}+\frac{4761}{2401}+\frac{34596}{2401}} \\ \end{aligned}

\begin{aligned} &=\sqrt{\frac{44541}{2401}} \\ \end{aligned}

\begin{aligned} &=\sqrt{\frac{909}{49}} \\ \end{aligned}

\begin{aligned} &=\frac{3}{7} \sqrt{101} \text { units. } \end{aligned}

 

Posted by

infoexpert21

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