Get Answers to all your Questions

header-bg qa

Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.15 question 14

Answers (1)

Answer: \left(0, \frac{5}{2}, 0\right) ; \sqrt{6}

Hint:

Direction ratio on (2,-2,4)

Equation of line =\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}

                            

                                2 x-2 y+4 z-6=0

Given:

Find the length and foot of the perpendicular from the point (1,3/2,2) to the plane

2x-2y+4z-6=0

Solution:

2x-2y+4z+5=0

\begin{aligned} &D R^{\prime} \Delta=(2,-2,4) \\\\ &A B D R^{\prime} S=(2,-2,4) \end{aligned}

\begin{aligned} &\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{3-3}{c} \\\\ &\frac{x-1}{2}=\frac{y-3 / 2}{-2}=\frac{2-2}{4}=1 \end{aligned}

\begin{aligned} &x=2 \lambda+1 \\\\ &y=-2 \lambda+\frac{3}{2} \\\\ &z=4 \lambda+2 \end{aligned}

\begin{aligned} &\Rightarrow 2(2 \lambda+1)-2\left(-2 \lambda+\frac{3}{2}\right)+4(4 \lambda+2)+5=0 \\\\ &\Rightarrow 4 \lambda+2+4 \lambda-3+16 \lambda+8+5=0 \end{aligned}

\begin{aligned} &24 \lambda=-12 \\\\ &\lambda=-\frac{12}{24} \\\\ &=-\frac{1}{2} \end{aligned}

Foot of the perpendicular,

\begin{aligned} &x=2 \lambda+1 \\\\ &=2\left(\frac{-1}{2}\right)+1=0 \\\\ &y=-2\left(\frac{-1}{2}\right)+\frac{3}{2}=\frac{5}{2} \\\\ &z=4\left(\frac{-1}{2}\right)+2=0 \end{aligned}

Foot of the perpendicular \left(0, \frac{5}{2}, 0\right)

 

A B=\sqrt{1+1+4}=\sqrt{6}

Posted by

infoexpert26

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads