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  • Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.15 question 3

Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.15 question 3

Answers (1)

Answer:   (1,6,0), 2 \sqrt{6}

Hint:

Suppose \frac{x+1}{2}=\frac{y-3}{3}=\frac{3-1}{-1}=\lambda

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

Given:

Find the coordinates of the foot of the perpendicular drawn from the point (5,4,2) to the line

\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}  hence or otherwise deduce the length of the perpendicular.

\frac{x+1}{2}=\frac{y-3}{3}=\frac{3-1}{-1}

P(5,4,2)

                                    

Solution:

\begin{aligned} &x=2 \lambda+1 \\\\ &y=3 \lambda+3 \\\\ &z=-\lambda+1 \end{aligned}

Direction ratio of P as =(2 \lambda-1-5,3 \lambda+z-4,-\lambda+1-2)

\begin{aligned} &=(2 \lambda-6,3 \lambda-1,-\lambda-1) \\\ &a_{1} \quad b_{1} \quad c_{1} \\\\ &a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0 \end{aligned}

\begin{aligned} &=2(2 \lambda-6)+3(3 \lambda-1)-1(-\lambda-1)=0 \\\\ &=4 \lambda-12+9 \lambda-3+\lambda+1=0 \\\\ &=14 \lambda-14=0 \\\\ &\lambda=1 \end{aligned}

Putting the value of x in eq

\begin{aligned} &x=2 \lambda+1, \\\\ &y=3 \lambda+3, \\\\ &z=-\lambda+1 \end{aligned}

\begin{aligned} &x=3 \\ &y=6 \\ &z=0 \end{aligned}

co-ordinates (1,6,0)

\begin{aligned} &P Q=\sqrt{(5-1)^{2}+(4-8)^{2}+(2-0)^{2}} \\\\ &=\sqrt{16+14+4} \\\\ &=\sqrt{24} \\\\ &=2 \sqrt{6} \; \mathrm{Ans} \end{aligned}

 

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